Fix: run pre-commit hooks

Makefile

@@ -1,12 +1,11 @@
 docker-build-simple:
 	docker build -f Dockerfile.simple -t simple .
-	
+
 docker-simple: docker-build-simple
 	docker run simple sh -c "bopytex -q 2 tpl_example.tex && cat 1_example.tex"
 
 docker-build-usecase:
 	docker build -f Dockerfile.usecase -t usecase .
-	
+
 docker-usecase: docker-build-usecase
 	docker run usecase sh -c "bopytex -s students.csv tpl_example.tex && cat 1_example.tex"
-

bopytex/default_config.py

@@ -1,3 +1,4 @@
+from bopytex.jinja2_env.texenv import texenv
 from bopytex.planner.generate_compile_join_planner import planner
 from bopytex.worker import Dispatcher
 from bopytex.worker.activate_corr import activate_corr
@@ -5,22 +6,20 @@ from bopytex.worker.clean import clean
 from bopytex.worker.compile import pdflatex
 from bopytex.worker.generate import generate
 from bopytex.worker.join_pdf import pdfjam
-from bopytex.jinja2_env.texenv import texenv
 
-jinja2 = {
-    "environment": texenv
-}
+jinja2 = {"environment": texenv}
 
-dispatcher = Dispatcher({
-    "GENERATE": generate,
-    "COMPILE": pdflatex,
-    "JOIN": pdfjam,
-    "CLEAN": clean,
-    "ACTIVATE_CORR": activate_corr,
-    })
+dispatcher = Dispatcher(
+    {
+        "GENERATE": generate,
+        "COMPILE": pdflatex,
+        "JOIN": pdfjam,
+        "CLEAN": clean,
+        "ACTIVATE_CORR": activate_corr,
+    }
+)
 
 latex = {
-        "solution": r"solution/print = true",
-        "no_solution": r"solution/print = false",
-        }
-
+    "solution": r"solution/print = true",
+    "no_solution": r"solution/print = false",
+}

bopytex/message.py

@@ -1,4 +1,4 @@
-class Message():
+class Message:
     def __init__(self, status, out, err):
         self._status = status
         self._out = out
@@ -19,6 +19,7 @@ class Message():
     def __repr__(self):
         return f"Message(status={self.status}, out={self.out}, err={self.err})"
 
+
 class SubprocessMessage(Message):
     def __init__(self, process):
         self._process = process

bopytex/planner/activate_corr_compile_join_planner.py

@@ -1,7 +1,8 @@
-from bopytex.tasks import Task, activate_corr_on, compile_pdf, join_pdfs
-import bopytex.planner.naming as naming
 import os
 
+import bopytex.planner.naming as naming
+from bopytex.tasks import Task, activate_corr_on, compile_pdf, join_pdfs
+
 
 def list_files(dir=".", accept=lambda _: True, reject=lambda _: False):
     files = []

bopytex/planner/generate_compile_join_planner.py

@@ -1,7 +1,8 @@
-from bopytex.tasks import Task, activate_corr_on, compile_pdf, generate, join_pdfs
+import csv
+
 import bopytex.planner.naming as naming
 from bopytex.planner.exceptions import PlannerMissingOption
-import csv
+from bopytex.tasks import Task, activate_corr_on, compile_pdf, generate, join_pdfs
 
 
 def build_subject_list_from_infos(infos: list[dict]) -> list[dict]:
@@ -70,10 +71,7 @@ def tasks_builder(
 
     for subject in subjects:
         source = naming.template2source(template, subject)
-        args = {
-                "subject": subject, 
-                "options": options
-                }
+        args = {"subject": subject, "options": options}
 
         tasks.append(generate(template, args, source))
 

bopytex/planner/naming.py

@@ -12,4 +12,3 @@ def source2pdf(source):
 
 def join(template):
     return source2pdf("joined" + template[3:])
-

bopytex/scheduler.py

@@ -8,7 +8,7 @@ from bopytex.worker import Dispatcher
 class Scheduler:
     """Scheduler is responsible of getting tasks (the tasks) and yield those that can be done"""
 
-    def __init__(self, dispatcher:Dispatcher, output_done: list[str] = None):
+    def __init__(self, dispatcher: Dispatcher, output_done: list[str] = None):
         self._dispatcher = dispatcher
 
         if output_done is None:
@@ -73,6 +73,6 @@ class Scheduler:
         return message
 
     def backlog(self):
-        """ Yield tasks sorted according to dependencies """
+        """Yield tasks sorted according to dependencies"""
         while self.doable_tasks:
             yield self.next_task()

bopytex/script.py

@@ -2,9 +2,10 @@
 # encoding: utf-8
 
 
-import click
 import logging
 
+import click
+
 from bopytex.service import main
 
 formatter = logging.Formatter("%(name)s :: %(levelname)s :: %(message)s")

bopytex/service.py

@@ -8,8 +8,9 @@ Producing then compiling templates
 import importlib.util
 import os
 from pathlib import Path
-from bopytex.scheduler import Scheduler
+
 from bopytex import default_config
+from bopytex.scheduler import Scheduler
 
 
 def orcherstrator(

bopytex/tasks.py

@@ -24,7 +24,7 @@ def generate(template: str, meta: dict, output: str):
     )
 
 
-def activate_corr_on(src: str, meta:dict, output: str):
+def activate_corr_on(src: str, meta: dict, output: str):
     """Create a task to activate correction for src"""
     return Task(
         action="ACTIVATE_CORR",

bopytex/worker/__init__.py

@@ -15,4 +15,3 @@ class Dispatcher:
             )
 
         return choosen_action(args=task.args, deps=task.deps, output=task.output)
-

bopytex/worker/activate_corr.py

@@ -11,4 +11,3 @@ def activate_corr(args, deps, output):
                 output_f.write(line.replace(no_solution, solution))
 
     return Message(0, [f"ACTIVATE CORR - {deps[0]} to {output}"], [])
-

bopytex/worker/compile.py

@@ -1,6 +1,7 @@
 import subprocess
 
 from bopytex.message import Message
+
 from ..message import SubprocessMessage
 
 

bopytex/worker/generate.py

@@ -8,16 +8,16 @@ def generate(args, deps, output):
     template = env.get_template(deps[0])
 
     variables = {
-            "options":args["options"],
-            "subject":args["subject"],
-            }
+        "options": args["options"],
+        "subject": args["subject"],
+    }
 
     try:
         args["options"]["direct_access"]
     except KeyError:
         pass
     else:
-        for (k,v) in args["options"]["direct_access"].items():
+        for (k, v) in args["options"]["direct_access"].items():
             if k not in ["options", "subject"]:
                 variables[k] = v
 

bopytex/worker/join_pdf.py

@@ -18,10 +18,9 @@ def pdfjam(args: dict, deps, output):
 
 
 def gs(args: dict, deps, output):
-    """ Not working. The command works in terminal but not here """
+    """Not working. The command works in terminal but not here"""
     joining_process = subprocess.Popen(
-        ["gs", f"-dBATCH -dNOPAUSE -q -sDEVICE=pdfwrite -sOutputFile={output}"]
-        + deps,
+        ["gs", f"-dBATCH -dNOPAUSE -q -sDEVICE=pdfwrite -sOutputFile={output}"] + deps,
         stdout=subprocess.PIPE,
         stderr=subprocess.PIPE,
         universal_newlines=True,

documentation/Makefile

@@ -198,5 +198,3 @@ SSH_TARGET_DIR=/var/docker/opytex.org/www/opytex
 
 rsync_upload: html
 	rsync -e "ssh" -P -rvzc --delete $(BUILDDIR)/html/ $(SSH_CONF):$(SSH_TARGET_DIR) --cvs-exclude
-
-

documentation/source/_downloads/01_DM.tex

@@ -37,8 +37,8 @@ Sujet numéro 01
 
 
     \section{Exercice}
-    
-    
+
+
     Dans un sac, il y a 20 bonbons à la menthe, 40 bonbons à la fraise et 2 au chocolat. On choisit un bonbon au hasard dans ce sac.
     \begin{enumerate}
         \item Calculer la probabilité de tirer un bonbon à la fraise.
@@ -47,7 +47,7 @@ Sujet numéro 01
         \end{solution}
         \item Calculer la probabilité de tirer un bonbon qui n'est pas au chocolat.
         \begin{solution}
-            
+
             $T($ tirer un bonbon à la fraise ou à la menthe $) = \dfrac{60}{62}$
         \end{solution}
         \item Calculer la probabilité de tirer un bonbon au réglisse.
@@ -56,18 +56,18 @@ Sujet numéro 01
         \end{solution}
         \item Dans un autre sac, on place 25 bonbons à la menthe et 34 bonbons à la fraise. Lise préfère les bonbons à la menthe. Dans quel sac doit-elle tirer un bonbon pour avoir le plus de chance d'avoir un bonbon qu'elle préfère?
             \begin{solution}
-                
+
                     Elle prefera tirer dans le deuxième sac car
                     \begin{eqnarray*}
-                        \frac{20}{62} & < & \frac{25}{34} 
+                        \frac{20}{62} & < & \frac{25}{34}
                     \end{eqnarray*}
-                
-                    
+
+
             \end{solution}
     \end{enumerate}
 
 
-    
+
     \section{Exercice}
     \begin{enumerate}
         \item Compléter les pointillés pour qu'il y est bien égalité.
@@ -89,83 +89,83 @@ Sujet numéro 01
 
         \item Faire les calculs suivants en détaillant les étapes (penser à simplifier les fractions quand c'est possible).
             \begin{enumerate}
-                
+
                 \item $A = \frac{ 10 }{ 2 } + \frac{ 8 }{ 2 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            A & = & \frac{ 10 }{ 2 } + \frac{ 8 }{ 2 } \\ 
-A & = & \frac{ 10 + 8 }{ 2 } \\ 
+                            A & = & \frac{ 10 }{ 2 } + \frac{ 8 }{ 2 } \\
+A & = & \frac{ 10 + 8 }{ 2 } \\
 A & = & 9
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $B = \frac{ 6 }{ 7 } + \frac{ -5 }{ 7 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            B & = & \frac{ 6 }{ 7 } + \frac{ -5 }{ 7 } \\ 
-B & = & \frac{ 6 - 5 }{ 7 } \\ 
+                            B & = & \frac{ 6 }{ 7 } + \frac{ -5 }{ 7 } \\
+B & = & \frac{ 6 - 5 }{ 7 } \\
 B & = & \frac{ 1 }{ 7 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $C = \frac{ 1 }{ 7 } + \frac{ 8 }{ 63 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            C & = & \frac{ 1 }{ 7 } + \frac{ 8 }{ 63 } \\ 
-C & = & \frac{ 1 \times 9 }{ 7 \times 9 } + \frac{ 8 \times 1 }{ 63 \times 1 } \\ 
-C & = & \frac{ 9 }{ 63 } + \frac{ 8 }{ 63 } \\ 
-C & = & \frac{ 9 + 8 }{ 63 } \\ 
+                            C & = & \frac{ 1 }{ 7 } + \frac{ 8 }{ 63 } \\
+C & = & \frac{ 1 \times 9 }{ 7 \times 9 } + \frac{ 8 \times 1 }{ 63 \times 1 } \\
+C & = & \frac{ 9 }{ 63 } + \frac{ 8 }{ 63 } \\
+C & = & \frac{ 9 + 8 }{ 63 } \\
 C & = & \frac{ 17 }{ 63 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $D = \frac{ 3 }{ 2 } + \frac{ -3 }{ 16 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            D & = & \frac{ 3 }{ 2 } + \frac{ -3 }{ 16 } \\ 
-D & = & \frac{ 3 \times 8 }{ 2 \times 8 } + \frac{ -3 \times 1 }{ 16 \times 1 } \\ 
-D & = & \frac{ 24 }{ 16 } + \frac{ -3 }{ 16 } \\ 
-D & = & \frac{ 24 - 3 }{ 16 } \\ 
+                            D & = & \frac{ 3 }{ 2 } + \frac{ -3 }{ 16 } \\
+D & = & \frac{ 3 \times 8 }{ 2 \times 8 } + \frac{ -3 \times 1 }{ 16 \times 1 } \\
+D & = & \frac{ 24 }{ 16 } + \frac{ -3 }{ 16 } \\
+D & = & \frac{ 24 - 3 }{ 16 } \\
 D & = & \frac{ 21 }{ 16 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $E = \frac{ 4 }{ 5 } \times 6$
                 \begin{solution}
                         \begin{eqnarray*}
-                            E & = & \frac{ 4 }{ 5 } \times 6 \\ 
-E & = & \frac{ 4 \times 6 }{ 5 } \\ 
+                            E & = & \frac{ 4 }{ 5 } \times 6 \\
+E & = & \frac{ 4 \times 6 }{ 5 } \\
 E & = & \frac{ 24 }{ 5 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $F = \frac{ 3 }{ 7 } \times \frac{ 9 }{ 8 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            F & = & \frac{ 3 }{ 7 } \times \frac{ 9 }{ 8 } \\ 
-F & = & \frac{ 9 }{ 8 } \times \frac{ 3 }{ 7 } \\ 
-F & = & \frac{ 9 \times 3 }{ 8 \times 7 } \\ 
+                            F & = & \frac{ 3 }{ 7 } \times \frac{ 9 }{ 8 } \\
+F & = & \frac{ 9 }{ 8 } \times \frac{ 3 }{ 7 } \\
+F & = & \frac{ 9 \times 3 }{ 8 \times 7 } \\
 F & = & \frac{ 27 }{ 56 }
                         \end{eqnarray*}
                 \end{solution}
             \end{enumerate}
-        
+
     \end{enumerate}
 
 
 
-\section{Exercice} 
+\section{Exercice}
 
 Dans la figure suivante, $(AB)$ et $(CD)$ sont parallèles, $AO = 8$, $OD = 15$, $CD = 2$ et $OB = 18$.
 
 
 
-\includegraphics[scale=0.4]{thales1} 
+\includegraphics[scale=0.4]{thales1}
 
 Calculer les longueurs $OC$ et $AB$.
 
 \begin{solution}
-    On sait que 
+    On sait que
     \begin{itemize}
         \item $(AB)$ et $(CD)$ sont parallèles
         \item $A$,$O$ et $D$ sont alignés
@@ -182,7 +182,7 @@ Calculer les longueurs $OC$ et $AB$.
         \end{tabular}
         est un tableau de proportionnalité.
 
-        On en déduit que 
+        On en déduit que
         \begin{eqnarray*}
             OC & = & \frac{DO \times OB}{AO} = \frac{15 \times 18}{8} = 33.75
         \end{eqnarray*}
@@ -190,13 +190,13 @@ Calculer les longueurs $OC$ et $AB$.
         \begin{eqnarray*}
             AB & = & \frac{CD \times AO}{DO} = \frac{2 \times 8}{15} = 1.0666666666666667
         \end{eqnarray*}
-        
+
 \end{solution}
 
 
 \end{document}
 
-%%% Local Variables: 
+%%% Local Variables:
 %%% mode: latex
 %%% TeX-master: "master"
 %%% End:

documentation/source/_downloads/02_DM.tex

@@ -37,8 +37,8 @@ Sujet numéro 02
 
 
     \section{Exercice}
-    
-    
+
+
     Dans un sac, il y a 10 bonbons à la menthe, 15 bonbons à la fraise et 6 au chocolat. On choisit un bonbon au hasard dans ce sac.
     \begin{enumerate}
         \item Calculer la probabilité de tirer un bonbon à la fraise.
@@ -47,7 +47,7 @@ Sujet numéro 02
         \end{solution}
         \item Calculer la probabilité de tirer un bonbon qui n'est pas au chocolat.
         \begin{solution}
-            
+
             $T($ tirer un bonbon à la fraise ou à la menthe $) = \dfrac{25}{31}$
         \end{solution}
         \item Calculer la probabilité de tirer un bonbon au réglisse.
@@ -56,18 +56,18 @@ Sujet numéro 02
         \end{solution}
         \item Dans un autre sac, on place 25 bonbons à la menthe et 34 bonbons à la fraise. Lise préfère les bonbons à la menthe. Dans quel sac doit-elle tirer un bonbon pour avoir le plus de chance d'avoir un bonbon qu'elle préfère?
             \begin{solution}
-                
+
                     Elle prefera tirer dans le deuxième sac car
                     \begin{eqnarray*}
-                        \frac{10}{31} & < & \frac{25}{34} 
+                        \frac{10}{31} & < & \frac{25}{34}
                     \end{eqnarray*}
-                
-                    
+
+
             \end{solution}
     \end{enumerate}
 
 
-    
+
     \section{Exercice}
     \begin{enumerate}
         \item Compléter les pointillés pour qu'il y est bien égalité.
@@ -89,89 +89,89 @@ Sujet numéro 02
 
         \item Faire les calculs suivants en détaillant les étapes (penser à simplifier les fractions quand c'est possible).
             \begin{enumerate}
-                
+
                 \item $A = \frac{ 2 }{ 3 } + \frac{ 7 }{ 3 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            A & = & \frac{ 2 }{ 3 } + \frac{ 7 }{ 3 } \\ 
-A & = & \frac{ 2 + 7 }{ 3 } \\ 
+                            A & = & \frac{ 2 }{ 3 } + \frac{ 7 }{ 3 } \\
+A & = & \frac{ 2 + 7 }{ 3 } \\
 A & = & 3
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $B = \frac{ 3 }{ 10 } + \frac{ 10 }{ 10 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            B & = & \frac{ 3 }{ 10 } + \frac{ 10 }{ 10 } \\ 
-B & = & \frac{ 3 + 10 }{ 10 } \\ 
+                            B & = & \frac{ 3 }{ 10 } + \frac{ 10 }{ 10 } \\
+B & = & \frac{ 3 + 10 }{ 10 } \\
 B & = & \frac{ 13 }{ 10 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $C = \frac{ -10 }{ 6 } + \frac{ 4 }{ 12 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            C & = & \frac{ -10 }{ 6 } + \frac{ 4 }{ 12 } \\ 
-C & = & \frac{ -10 \times 2 }{ 6 \times 2 } + \frac{ 4 \times 1 }{ 12 \times 1 } \\ 
-C & = & \frac{ -20 }{ 12 } + \frac{ 4 }{ 12 } \\ 
-C & = & \frac{ -20 + 4 }{ 12 } \\ 
-C & = & \frac{ -16 }{ 12 } \\ 
-C & = & \frac{ -4 \times 4 }{ 3 \times 4 } \\ 
+                            C & = & \frac{ -10 }{ 6 } + \frac{ 4 }{ 12 } \\
+C & = & \frac{ -10 \times 2 }{ 6 \times 2 } + \frac{ 4 \times 1 }{ 12 \times 1 } \\
+C & = & \frac{ -20 }{ 12 } + \frac{ 4 }{ 12 } \\
+C & = & \frac{ -20 + 4 }{ 12 } \\
+C & = & \frac{ -16 }{ 12 } \\
+C & = & \frac{ -4 \times 4 }{ 3 \times 4 } \\
 C & = & \frac{ -4 }{ 3 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $D = \frac{ 10 }{ 6 } + \frac{ -8 }{ 42 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            D & = & \frac{ 10 }{ 6 } + \frac{ -8 }{ 42 } \\ 
-D & = & \frac{ 10 \times 7 }{ 6 \times 7 } + \frac{ -8 \times 1 }{ 42 \times 1 } \\ 
-D & = & \frac{ 70 }{ 42 } + \frac{ -8 }{ 42 } \\ 
-D & = & \frac{ 70 - 8 }{ 42 } \\ 
-D & = & \frac{ 62 }{ 42 } \\ 
-D & = & \frac{ 31 \times 2 }{ 21 \times 2 } \\ 
+                            D & = & \frac{ 10 }{ 6 } + \frac{ -8 }{ 42 } \\
+D & = & \frac{ 10 \times 7 }{ 6 \times 7 } + \frac{ -8 \times 1 }{ 42 \times 1 } \\
+D & = & \frac{ 70 }{ 42 } + \frac{ -8 }{ 42 } \\
+D & = & \frac{ 70 - 8 }{ 42 } \\
+D & = & \frac{ 62 }{ 42 } \\
+D & = & \frac{ 31 \times 2 }{ 21 \times 2 } \\
 D & = & \frac{ 31 }{ 21 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $E = \frac{ 6 }{ 9 } \times 4$
                 \begin{solution}
                         \begin{eqnarray*}
-                            E & = & \frac{ 6 }{ 9 } \times 4 \\ 
-E & = & \frac{ 6 \times 4 }{ 9 } \\ 
-E & = & \frac{ 24 }{ 9 } \\ 
-E & = & \frac{ 8 \times 3 }{ 3 \times 3 } \\ 
+                            E & = & \frac{ 6 }{ 9 } \times 4 \\
+E & = & \frac{ 6 \times 4 }{ 9 } \\
+E & = & \frac{ 24 }{ 9 } \\
+E & = & \frac{ 8 \times 3 }{ 3 \times 3 } \\
 E & = & \frac{ 8 }{ 3 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $F = \frac{ 9 }{ 2 } \times \frac{ 9 }{ 5 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            F & = & \frac{ 9 }{ 2 } \times \frac{ 9 }{ 5 } \\ 
-F & = & \frac{ 9 }{ 5 } \times \frac{ 9 }{ 2 } \\ 
-F & = & \frac{ 9 \times 9 }{ 5 \times 2 } \\ 
+                            F & = & \frac{ 9 }{ 2 } \times \frac{ 9 }{ 5 } \\
+F & = & \frac{ 9 }{ 5 } \times \frac{ 9 }{ 2 } \\
+F & = & \frac{ 9 \times 9 }{ 5 \times 2 } \\
 F & = & \frac{ 81 }{ 10 }
                         \end{eqnarray*}
                 \end{solution}
             \end{enumerate}
-        
+
     \end{enumerate}
 
 
 
-\section{Exercice} 
+\section{Exercice}
 
 Dans la figure suivante, $(AB)$ et $(CD)$ sont parallèles, $AO = 11$, $OD = 18$, $CD = 6$ et $OB = 14$.
 
 
 
-\includegraphics[scale=0.4]{thales1} 
+\includegraphics[scale=0.4]{thales1}
 
 Calculer les longueurs $OC$ et $AB$.
 
 \begin{solution}
-    On sait que 
+    On sait que
     \begin{itemize}
         \item $(AB)$ et $(CD)$ sont parallèles
         \item $A$,$O$ et $D$ sont alignés
@@ -188,7 +188,7 @@ Calculer les longueurs $OC$ et $AB$.
         \end{tabular}
         est un tableau de proportionnalité.
 
-        On en déduit que 
+        On en déduit que
         \begin{eqnarray*}
             OC & = & \frac{DO \times OB}{AO} = \frac{18 \times 14}{11} = 22.90909090909091
         \end{eqnarray*}
@@ -196,13 +196,13 @@ Calculer les longueurs $OC$ et $AB$.
         \begin{eqnarray*}
             AB & = & \frac{CD \times AO}{DO} = \frac{6 \times 11}{18} = 3.666666666666667
         \end{eqnarray*}
-        
+
 \end{solution}
 
 
 \end{document}
 
-%%% Local Variables: 
+%%% Local Variables:
 %%% mode: latex
 %%% TeX-master: "master"
 %%% End:

documentation/source/_downloads/03_DM.tex

@@ -37,8 +37,8 @@ Sujet numéro 03
 
 
     \section{Exercice}
-    
-    
+
+
     Dans un sac, il y a 56 bonbons à la menthe, 70 bonbons à la fraise et 6 au chocolat. On choisit un bonbon au hasard dans ce sac.
     \begin{enumerate}
         \item Calculer la probabilité de tirer un bonbon à la fraise.
@@ -47,7 +47,7 @@ Sujet numéro 03
         \end{solution}
         \item Calculer la probabilité de tirer un bonbon qui n'est pas au chocolat.
         \begin{solution}
-            
+
             $T($ tirer un bonbon à la fraise ou à la menthe $) = \dfrac{126}{132}$
         \end{solution}
         \item Calculer la probabilité de tirer un bonbon au réglisse.
@@ -56,18 +56,18 @@ Sujet numéro 03
         \end{solution}
         \item Dans un autre sac, on place 25 bonbons à la menthe et 34 bonbons à la fraise. Lise préfère les bonbons à la menthe. Dans quel sac doit-elle tirer un bonbon pour avoir le plus de chance d'avoir un bonbon qu'elle préfère?
             \begin{solution}
-                
+
                     Elle prefera tirer dans le deuxième sac car
                     \begin{eqnarray*}
-                        \frac{56}{132} & < & \frac{25}{34} 
+                        \frac{56}{132} & < & \frac{25}{34}
                     \end{eqnarray*}
-                
-                    
+
+
             \end{solution}
     \end{enumerate}
 
 
-    
+
     \section{Exercice}
     \begin{enumerate}
         \item Compléter les pointillés pour qu'il y est bien égalité.
@@ -89,93 +89,93 @@ Sujet numéro 03
 
         \item Faire les calculs suivants en détaillant les étapes (penser à simplifier les fractions quand c'est possible).
             \begin{enumerate}
-                
+
                 \item $A = \frac{ 2 }{ 10 } + \frac{ 2 }{ 10 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            A & = & \frac{ 2 }{ 10 } + \frac{ 2 }{ 10 } \\ 
-A & = & \frac{ 2 + 2 }{ 10 } \\ 
-A & = & \frac{ 4 }{ 10 } \\ 
-A & = & \frac{ 2 \times 2 }{ 5 \times 2 } \\ 
+                            A & = & \frac{ 2 }{ 10 } + \frac{ 2 }{ 10 } \\
+A & = & \frac{ 2 + 2 }{ 10 } \\
+A & = & \frac{ 4 }{ 10 } \\
+A & = & \frac{ 2 \times 2 }{ 5 \times 2 } \\
 A & = & \frac{ 2 }{ 5 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $B = \frac{ -5 }{ 4 } + \frac{ -2 }{ 4 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            B & = & \frac{ -5 }{ 4 } + \frac{ -2 }{ 4 } \\ 
-B & = & \frac{ -5 - 2 }{ 4 } \\ 
+                            B & = & \frac{ -5 }{ 4 } + \frac{ -2 }{ 4 } \\
+B & = & \frac{ -5 - 2 }{ 4 } \\
 B & = & \frac{ -7 }{ 4 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $C = \frac{ -8 }{ 2 } + \frac{ 10 }{ 16 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            C & = & \frac{ -8 }{ 2 } + \frac{ 10 }{ 16 } \\ 
-C & = & \frac{ -8 \times 8 }{ 2 \times 8 } + \frac{ 10 \times 1 }{ 16 \times 1 } \\ 
-C & = & \frac{ -64 }{ 16 } + \frac{ 10 }{ 16 } \\ 
-C & = & \frac{ -64 + 10 }{ 16 } \\ 
-C & = & \frac{ -54 }{ 16 } \\ 
-C & = & \frac{ -27 \times 2 }{ 8 \times 2 } \\ 
+                            C & = & \frac{ -8 }{ 2 } + \frac{ 10 }{ 16 } \\
+C & = & \frac{ -8 \times 8 }{ 2 \times 8 } + \frac{ 10 \times 1 }{ 16 \times 1 } \\
+C & = & \frac{ -64 }{ 16 } + \frac{ 10 }{ 16 } \\
+C & = & \frac{ -64 + 10 }{ 16 } \\
+C & = & \frac{ -54 }{ 16 } \\
+C & = & \frac{ -27 \times 2 }{ 8 \times 2 } \\
 C & = & \frac{ -27 }{ 8 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $D = \frac{ -9 }{ 2 } + \frac{ -4 }{ 14 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            D & = & \frac{ -9 }{ 2 } + \frac{ -4 }{ 14 } \\ 
-D & = & \frac{ -9 \times 7 }{ 2 \times 7 } + \frac{ -4 \times 1 }{ 14 \times 1 } \\ 
-D & = & \frac{ -63 }{ 14 } + \frac{ -4 }{ 14 } \\ 
-D & = & \frac{ -63 - 4 }{ 14 } \\ 
+                            D & = & \frac{ -9 }{ 2 } + \frac{ -4 }{ 14 } \\
+D & = & \frac{ -9 \times 7 }{ 2 \times 7 } + \frac{ -4 \times 1 }{ 14 \times 1 } \\
+D & = & \frac{ -63 }{ 14 } + \frac{ -4 }{ 14 } \\
+D & = & \frac{ -63 - 4 }{ 14 } \\
 D & = & \frac{ -67 }{ 14 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $E = \frac{ 5 }{ 8 } \times 4$
                 \begin{solution}
                         \begin{eqnarray*}
-                            E & = & \frac{ 5 }{ 8 } \times 4 \\ 
-E & = & \frac{ 5 \times 1 \times 4 }{ 2 \times 4 } \\ 
-E & = & \frac{ 5 \times 4 }{ 8 } \\ 
-E & = & \frac{ 20 }{ 8 } \\ 
-E & = & \frac{ 5 \times 4 }{ 2 \times 4 } \\ 
+                            E & = & \frac{ 5 }{ 8 } \times 4 \\
+E & = & \frac{ 5 \times 1 \times 4 }{ 2 \times 4 } \\
+E & = & \frac{ 5 \times 4 }{ 8 } \\
+E & = & \frac{ 20 }{ 8 } \\
+E & = & \frac{ 5 \times 4 }{ 2 \times 4 } \\
 E & = & \frac{ 5 }{ 2 }
                         \end{eqnarray*}
                 \end{solution}
-                
+
                 \item $F = \frac{ 6 }{ 7 } \times \frac{ 3 }{ 8 }$
                 \begin{solution}
                         \begin{eqnarray*}
-                            F & = & \frac{ 6 }{ 7 } \times \frac{ 3 }{ 8 } \\ 
-F & = & \frac{ 3 }{ 8 } \times \frac{ 6 }{ 7 } \\ 
-F & = & \frac{ 3 \times 3 \times 2 }{ 4 \times 2 \times 7 } \\ 
-F & = & \frac{ 3 \times 6 }{ 8 \times 7 } \\ 
-F & = & \frac{ 18 }{ 56 } \\ 
-F & = & \frac{ 9 \times 2 }{ 28 \times 2 } \\ 
+                            F & = & \frac{ 6 }{ 7 } \times \frac{ 3 }{ 8 } \\
+F & = & \frac{ 3 }{ 8 } \times \frac{ 6 }{ 7 } \\
+F & = & \frac{ 3 \times 3 \times 2 }{ 4 \times 2 \times 7 } \\
+F & = & \frac{ 3 \times 6 }{ 8 \times 7 } \\
+F & = & \frac{ 18 }{ 56 } \\
+F & = & \frac{ 9 \times 2 }{ 28 \times 2 } \\
 F & = & \frac{ 9 }{ 28 }
                         \end{eqnarray*}
                 \end{solution}
             \end{enumerate}
-        
+
     \end{enumerate}
 
 
 
-\section{Exercice} 
+\section{Exercice}
 
 Dans la figure suivante, $(AB)$ et $(CD)$ sont parallèles, $AO = 3$, $OD = 7$, $CD = 5$ et $OB = 2$.
 
 
 
-\includegraphics[scale=0.4]{thales2} 
+\includegraphics[scale=0.4]{thales2}
 
 Calculer les longueurs $OC$ et $AB$.
 
 \begin{solution}
-    On sait que 
+    On sait que
     \begin{itemize}
         \item $(AB)$ et $(CD)$ sont parallèles
         \item $A$,$O$ et $D$ sont alignés
@@ -192,7 +192,7 @@ Calculer les longueurs $OC$ et $AB$.
         \end{tabular}
         est un tableau de proportionnalité.
 
-        On en déduit que 
+        On en déduit que
         \begin{eqnarray*}
             OC & = & \frac{DO \times OB}{AO} = \frac{7 \times 2}{3} = 4.666666666666666
         \end{eqnarray*}
@@ -200,13 +200,13 @@ Calculer les longueurs $OC$ et $AB$.
         \begin{eqnarray*}
             AB & = & \frac{CD \times AO}{DO} = \frac{5 \times 3}{7} = 2.142857142857143
         \end{eqnarray*}
-        
+
 \end{solution}
 
 
 \end{document}
 
-%%% Local Variables: 
+%%% Local Variables:
 %%% mode: latex
 %%% TeX-master: "master"
 %%% End:

documentation/source/_downloads/tpl_DM.tex

@@ -59,20 +59,20 @@ Sujet numéro \Var{infos.num}
                 \Block{if (int(a)/total) > (25/34)}
                     Elle prefera tirer dans le premier sac car
                     \begin{eqnarray*}
-                        \frac{\Var{a}{\Var{total}} & > & \frac{25}{34} 
+                        \frac{\Var{a}{\Var{total}} & > & \frac{25}{34}
                     \end{eqnarray*}
                 \Block{else}
                     Elle prefera tirer dans le deuxième sac car
                     \begin{eqnarray*}
-                        \frac{\Var{a}}{\Var{total}} & < & \frac{25}{34} 
+                        \frac{\Var{a}}{\Var{total}} & < & \frac{25}{34}
                     \end{eqnarray*}
                 \Block{endif}
-                    
+
             \end{solution}
     \end{enumerate}
 
 
-    
+
     \section{Exercice}
     \begin{enumerate}
         \item Compléter les pointillés pour qu'il y est bien égalité.
@@ -137,23 +137,23 @@ Sujet numéro \Var{infos.num}
                         \end{eqnarray*}
                 \end{solution}
             \end{enumerate}
-        
+
     \end{enumerate}
 
 
 
-\section{Exercice} 
+\section{Exercice}
 \Block{set AO, OD, CD, OB = random_str("{a},{b},{c},{d}", ["{a} < {b}", "{c} != {d}"], 1, 20).split(',')}
 Dans la figure suivante, $(AB)$ et $(CD)$ sont parallèles, $AO = \Var{AO}$, $OD = \Var{OD}$, $CD = \Var{CD}$ et $OB = \Var{OB}$.
 
 \Block{set fig = random_str("{a}", [], 1, 2)}
 
-\includegraphics[scale=0.4]{thales\Var{fig}} 
+\includegraphics[scale=0.4]{thales\Var{fig}}
 
 Calculer les longueurs $OC$ et $AB$.
 
 \begin{solution}
-    On sait que 
+    On sait que
     \begin{itemize}
         \item $(AB)$ et $(CD)$ sont parallèles
         \item $A$,$O$ et $D$ sont alignés
@@ -170,7 +170,7 @@ Calculer les longueurs $OC$ et $AB$.
         \end{tabular}
         est un tableau de proportionnalité.
 
-        On en déduit que 
+        On en déduit que
         \begin{eqnarray*}
             OC & = & \frac{DO \times OB}{AO} = \frac{\Var{OD} \times \Var{OB}}{\Var{AO}} = \Var{int(OD)*int(OB)/int(AO) | round(2)}
         \end{eqnarray*}
@@ -178,14 +178,13 @@ Calculer les longueurs $OC$ et $AB$.
         \begin{eqnarray*}
             AB & = & \frac{CD \times AO}{DO} = \frac{\Var{CD} \times \Var{AO}}{\Var{OD}} = \Var{int(CD)*int(AO)/int(OD) |round(2)}
         \end{eqnarray*}
-        
+
 \end{solution}
 
 
 \end{document}
 
-%%% Local Variables: 
+%%% Local Variables:
 %%% mode: latex
 %%% TeX-master: "master"
 %%% End:
-

documentation/source/conf.py

@@ -13,70 +13,70 @@
 # All configuration values have a default; values that are commented out
 # serve to show the default.
 
-import sys
 import os
 import shlex
+import sys
 
 # If extensions (or modules to document with autodoc) are in another directory,
 # add these directories to sys.path here. If the directory is relative to the
 # documentation root, use os.path.abspath to make it absolute, like shown here.
-#sys.path.insert(0, os.path.abspath('.'))
+# sys.path.insert(0, os.path.abspath('.'))
 
 # -- General configuration ------------------------------------------------
 
 # If your documentation needs a minimal Sphinx version, state it here.
-#needs_sphinx = '1.0'
+# needs_sphinx = '1.0'
 
 # Add any Sphinx extension module names here, as strings. They can be
 # extensions coming with Sphinx (named 'sphinx.ext.*') or your custom
 # ones.
 extensions = [
-    'sphinx.ext.autodoc',
-    'sphinx.ext.mathjax',
-    'sphinx.ext.ifconfig',
-    'sphinx.ext.viewcode',
+    "sphinx.ext.autodoc",
+    "sphinx.ext.mathjax",
+    "sphinx.ext.ifconfig",
+    "sphinx.ext.viewcode",
 ]
 
 # Add any paths that contain templates here, relative to this directory.
-templates_path = ['_templates']
+templates_path = ["_templates"]
 
 # The suffix(es) of source filenames.
 # You can specify multiple suffix as a list of string:
 # source_suffix = ['.rst', '.md']
-source_suffix = '.rst'
+source_suffix = ".rst"
 
 # The encoding of source files.
-#source_encoding = 'utf-8-sig'
+# source_encoding = 'utf-8-sig'
 
 # The master toctree document.
-master_doc = 'index'
+master_doc = "index"
 
 # General information about the project.
-project = 'Opytex'
-copyright = '2016, Benjamin Bertrand'
-author = 'Benjamin Bertrand'
+project = "Opytex"
+copyright = "2016, Benjamin Bertrand"
+author = "Benjamin Bertrand"
 
 # The version info for the project you're documenting, acts as replacement for
 # |version| and |release|, also used in various other places throughout the
 # built documents.
 #
 # The short X.Y version.
-version = '0.1'
+version = "0.1"
 # The full version, including alpha/beta/rc tags.
-release = '0.1'
+release = "0.1"
 
 # The language for content autogenerated by Sphinx. Refer to documentation
 # for a list of supported languages.
 #
 # This is also used if you do content translation via gettext catalogs.
 # Usually you set "language" from the command line for these cases.
-language = 'fr'
+language = "fr"
 
 # There are two options for replacing |today|: either, you set today to some
 # non-false value, then it is used:
-#today = ''
+# today = ''
 # Else, today_fmt is used as the format for a strftime call.
-#today_fmt = '%B %d, %Y'
+# today_fmt = '%B %d, %Y'
 
 # List of patterns, relative to source directory, that match files and
 # directories to ignore when looking for source files.
@@ -84,27 +84,27 @@ exclude_patterns = []
 
 # The reST default role (used for this markup: `text`) to use for all
 # documents.
-#default_role = None
+# default_role = None
 
 # If true, '()' will be appended to :func: etc. cross-reference text.
-#add_function_parentheses = True
+# add_function_parentheses = True
 
 # If true, the current module name will be prepended to all description
 # unit titles (such as .. function::).
-#add_module_names = True
+# add_module_names = True
 
 # If true, sectionauthor and moduleauthor directives will be shown in the
 # output. They are ignored by default.
-#show_authors = False
+# show_authors = False
 
 # The name of the Pygments (syntax highlighting) style to use.
-pygments_style = 'sphinx'
+pygments_style = "sphinx"
 
 # A list of ignored prefixes for module index sorting.
-#modindex_common_prefix = []
+# modindex_common_prefix = []
 
 # If true, keep warnings as "system message" paragraphs in the built documents.
-#keep_warnings = False
+# keep_warnings = False
 
 # If true, `todo` and `todoList` produce output, else they produce nothing.
 todo_include_todos = False
@@ -114,159 +114,153 @@ todo_include_todos = False
 
 # The theme to use for HTML and HTML Help pages.  See the documentation for
 # a list of builtin themes.
-#html_theme = 'alabaster'
+# html_theme = 'alabaster'
 import sphinx_rtd_theme
+
 html_theme = "sphinx_rtd_theme"
 html_theme_path = [sphinx_rtd_theme.get_html_theme_path()]
 
 # Theme options are theme-specific and customize the look and feel of a theme
 # further.  For a list of options available for each theme, see the
 # documentation.
-#html_theme_options = {}
+# html_theme_options = {}
 
 # Add any paths that contain custom themes here, relative to this directory.
-#html_theme_path = []
+# html_theme_path = []
 
 # The name for this set of Sphinx documents.  If None, it defaults to
 # "<project> v<release> documentation".
-#html_title = None
+# html_title = None
 
 # A shorter title for the navigation bar.  Default is the same as html_title.
-#html_short_title = None
+# html_short_title = None
 
 # The name of an image file (relative to this directory) to place at the top
 # of the sidebar.
-#html_logo = None
+# html_logo = None
 
 # The name of an image file (within the static path) to use as favicon of the
 # docs.  This file should be a Windows icon file (.ico) being 16x16 or 32x32
 # pixels large.
-#html_favicon = None
+# html_favicon = None
 
 # Add any paths that contain custom static files (such as style sheets) here,
 # relative to this directory. They are copied after the builtin static files,
 # so a file named "default.css" will overwrite the builtin "default.css".
-html_static_path = ['_static']
+html_static_path = ["_static"]
 
 # Add any extra paths that contain custom files (such as robots.txt or
 # .htaccess) here, relative to this directory. These files are copied
 # directly to the root of the documentation.
-#html_extra_path = []
+# html_extra_path = []
 
 # If not '', a 'Last updated on:' timestamp is inserted at every page bottom,
 # using the given strftime format.
-#html_last_updated_fmt = '%b %d, %Y'
+# html_last_updated_fmt = '%b %d, %Y'
 
 # If true, SmartyPants will be used to convert quotes and dashes to
 # typographically correct entities.
-#html_use_smartypants = True
+# html_use_smartypants = True
 
 # Custom sidebar templates, maps document names to template names.
-#html_sidebars = {}
+# html_sidebars = {}
 
 # Additional templates that should be rendered to pages, maps page names to
 # template names.
-#html_additional_pages = {}
+# html_additional_pages = {}
 
 # If false, no module index is generated.
-#html_domain_indices = True
+# html_domain_indices = True
 
 # If false, no index is generated.
-#html_use_index = True
+# html_use_index = True
 
 # If true, the index is split into individual pages for each letter.
-#html_split_index = False
+# html_split_index = False
 
 # If true, links to the reST sources are added to the pages.
-#html_show_sourcelink = True
+# html_show_sourcelink = True
 
 # If true, "Created using Sphinx" is shown in the HTML footer. Default is True.
-#html_show_sphinx = True
+# html_show_sphinx = True
 
 # If true, "(C) Copyright ..." is shown in the HTML footer. Default is True.
-#html_show_copyright = True
+# html_show_copyright = True
 
 # If true, an OpenSearch description file will be output, and all pages will
 # contain a <link> tag referring to it.  The value of this option must be the
 # base URL from which the finished HTML is served.
-#html_use_opensearch = ''
+# html_use_opensearch = ''
 
 # This is the file name suffix for HTML files (e.g. ".xhtml").
-#html_file_suffix = None
+# html_file_suffix = None
 
 # Language to be used for generating the HTML full-text search index.
 # Sphinx supports the following languages:
 #   'da', 'de', 'en', 'es', 'fi', 'fr', 'h', 'it', 'ja'
 #   'nl', 'no', 'pt', 'ro', 'r', 'sv', 'tr'
-#html_search_language = 'en'
+# html_search_language = 'en'
 
 # A dictionary with options for the search language support, empty by default.
 # Now only 'ja' uses this config value
-#html_search_options = {'type': 'default'}
+# html_search_options = {'type': 'default'}
 
 # The name of a javascript file (relative to the configuration directory) that
 # implements a search results scorer. If empty, the default will be used.
-#html_search_scorer = 'scorer.js'
+# html_search_scorer = 'scorer.js'
 
 # Output file base name for HTML help builder.
-htmlhelp_basename = 'Opytexdoc'
+htmlhelp_basename = "Opytexdoc"
 
 # -- Options for LaTeX output ---------------------------------------------
 
 latex_elements = {
-# The paper size ('letterpaper' or 'a4paper').
-#'papersize': 'letterpaper',
-
-# The font size ('10pt', '11pt' or '12pt').
-#'pointsize': '10pt',
-
-# Additional stuff for the LaTeX preamble.
-#'preamble': '',
-
-# Latex figure (float) alignment
-#'figure_align': 'htbp',
+    # The paper size ('letterpaper' or 'a4paper').
+    #'papersize': 'letterpaper',
+    # The font size ('10pt', '11pt' or '12pt').
+    #'pointsize': '10pt',
+    # Additional stuff for the LaTeX preamble.
+    #'preamble': '',
+    # Latex figure (float) alignment
+    #'figure_align': 'htbp',
 }
 
 # Grouping the document tree into LaTeX files. List of tuples
 # (source start file, target name, title,
 #  author, documentclass [howto, manual, or own class]).
 latex_documents = [
-  (master_doc, 'Opytex.tex', 'Opytex Documentation',
-   'Benjamin Bertrand', 'manual'),
+    (master_doc, "Opytex.tex", "Opytex Documentation", "Benjamin Bertrand", "manual"),
 ]
 
 # The name of an image file (relative to this directory) to place at the top of
 # the title page.
-#latex_logo = None
+# latex_logo = None
 
 # For "manual" documents, if this is true, then toplevel headings are parts,
 # not chapters.
-#latex_use_parts = False
+# latex_use_parts = False
 
 # If true, show page references after internal links.
-#latex_show_pagerefs = False
+# latex_show_pagerefs = False
 
 # If true, show URL addresses after external links.
-#latex_show_urls = False
+# latex_show_urls = False
 
 # Documents to append as an appendix to all manuals.
-#latex_appendices = []
+# latex_appendices = []
 
 # If false, no module index is generated.
-#latex_domain_indices = True
+# latex_domain_indices = True
 
 
 # -- Options for manual page output ---------------------------------------
 
 # One entry per manual page. List of tuples
 # (source start file, name, description, authors, manual section).
-man_pages = [
-    (master_doc, 'opytex', 'Opytex Documentation',
-     [author], 1)
-]
+man_pages = [(master_doc, "opytex", "Opytex Documentation", [author], 1)]
 
 # If true, show URL addresses after external links.
-#man_show_urls = False
+# man_show_urls = False
 
 
 # -- Options for Texinfo output -------------------------------------------
@@ -275,19 +269,25 @@ man_pages = [
 # (source start file, target name, title, author,
 #  dir menu entry, description, category)
 texinfo_documents = [
-  (master_doc, 'Opytex', 'Opytex Documentation',
-   author, 'Opytex', 'One line description of project.',
-   'Miscellaneous'),
+    (
+        master_doc,
+        "Opytex",
+        "Opytex Documentation",
+        author,
+        "Opytex",
+        "One line description of project.",
+        "Miscellaneous",
+    ),
 ]
 
 # Documents to append as an appendix to all manuals.
-#texinfo_appendices = []
+# texinfo_appendices = []
 
 # If false, no module index is generated.
-#texinfo_domain_indices = True
+# texinfo_domain_indices = True
 
 # How to display URL addresses: 'footnote', 'no', or 'inline'.
-#texinfo_show_urls = 'footnote'
+# texinfo_show_urls = 'footnote'
 
 # If true, do not generate a @detailmenu in the "Top" node's menu.
-#texinfo_no_detailmenu = False
+# texinfo_no_detailmenu = False

documentation/source/index.rst

@@ -38,4 +38,3 @@ Indices and tables
 * :ref:`genindex`
 * :ref:`modindex`
 * :ref:`search`
-

documentation/source/snippets.rst

@@ -69,7 +69,7 @@ Simplifications de fractions
     Explications
 
     \begin{align*}
-    \Var{f.simplify().explain()|join('=')} 
+    \Var{f.simplify().explain()|join('=')}
     \end{align*}
 
   Ce qui produira
@@ -84,9 +84,9 @@ Simplifications de fractions
             \dfrac{12}{9} = \dfrac{4}{3}
         \end{align*}
         Explications
-        
+
         \begin{align*}
-            \frac{ 12 }{ 9 }=\frac{ 4 \times 3 }{ 3 \times 3 }=\frac{ 4 }{ 3 } 
+            \frac{ 12 }{ 9 }=\frac{ 4 \times 3 }{ 3 \times 3 }=\frac{ 4 }{ 3 }
         \end{align*}
 
   Et ce qui donne
@@ -104,9 +104,9 @@ Simplifications de fractions
         \end{align*}
 
         Explications
-        
+
         \begin{align*}
-            \frac{ 12 }{ 9 }=\frac{ 4 \times 3 }{ 3 \times 3 }=\frac{ 4 }{ 3 } 
+            \frac{ 12 }{ 9 }=\frac{ 4 \times 3 }{ 3 \times 3 }=\frac{ 4 }{ 3 }
         \end{align*}
 
 

documentation/source/tutorial.rst

@@ -67,7 +67,7 @@ Ce qui produira
         i vaut 3
         i vaut 4
 
-        
+
 Quelques commandes supplémentaires
 ==================================
 
@@ -84,7 +84,7 @@ Filtres qui marchenet bien avec Mapytex
         \Block{set e = Expression("1 + 2*3")}
         et on veut détailler sa simplification
         \begin{align*}
-            \Var{e.simplify().explain()|join('=')} 
+            \Var{e.simplify().explain()|join('=')}
         \end{align*}
 
 
@@ -95,7 +95,7 @@ Filtres qui marchenet bien avec Mapytex
         On commence par définir une expression,
         et on veut détailler sa simplification
         \begin{align*}
-            1 + 2 \times 3 =  1 + 6 = 7 
+            1 + 2 \times 3 =  1 + 6 = 7
         \end{align*}
 
 
@@ -108,7 +108,7 @@ Filtres qui marchenet bien avec Mapytex
         \Block{set e = Expression("1 + 2*3")}
         et on veut détailler sa simplification
         \begin{eqnarray*}
-            \Var{e.simplify().explain()|calculus(name = 'e')} 
+            \Var{e.simplify().explain()|calculus(name = 'e')}
         \end{eqnarray*}
 
 
@@ -121,10 +121,10 @@ Filtres qui marchenet bien avec Mapytex
         \begin{eqnarray*}
             e & = & 1 + 2 \times 3 \\
             e & = & 1 + 6 \\
-            e & = & 7 
+            e & = & 7
         \end{eqnarray*}
 
-    
+
 Compilation des documents
 =========================
 
@@ -136,7 +136,7 @@ Puis on génère et compile les 3 sujets avec la commande
 
         opytex -t tpl_DM.tex -N 3
 
-Ce qui a crée les fichiers sources: 
+Ce qui a crée les fichiers sources:
 
 - :download:`01_DM.tex <_downloads/01_DM.tex>`
 - :download:`02_DM.tex <_downloads/02_DM.tex>`
@@ -158,6 +158,3 @@ Il est possible aussi de créer les sujets et les corrections en même temps ave
     .. code-block:: bash
 
         opytex -t tpl_DM.tex -c -N 60
-
-
-    

example/usecase/bopytex_config.py

@@ -1,4 +1,3 @@
 from mapytex import Expression
-direct_access = {
-        "Expression": Expression
-        }
+
+direct_access = {"Expression": Expression}

test/fakes/dispatcher.py

@@ -1,5 +1,6 @@
 from bopytex.worker import Dispatcher
-from .workers import fake_worker, success_worker, fail_worker
+
+from .workers import fail_worker, fake_worker, success_worker
 
 fake_dispatcher = Dispatcher(
     {

test/fakes/workers.py

@@ -1,5 +1,6 @@
 from bopytex.message import Message
 
+
 def fake_worker(args, deps, output):
     return Message(0, [f"FAKE - {args} - {deps} - {output}"], [])
 
@@ -10,4 +11,3 @@ def success_worker(args, deps, output):
 
 def fail_worker(args, deps, output):
     return Message(1, [f"FAILURE - {args} - {deps} - {output}"], [])
-

test/test_e2e.py

@@ -1,9 +1,11 @@
 import os
-import jinja2
 from pathlib import Path
-from bopytex.service import main
+
+import jinja2
 import pytest
 
+from bopytex.service import main
+
 
 @pytest.fixture
 def template_path(tmp_path):
@@ -22,6 +24,7 @@ Subject {{ number }}
         )
     return template
 
+
 @pytest.fixture
 def bad_template_path(tmp_path):
     template = tmp_path / "tpl_source.tex"
@@ -64,6 +67,7 @@ def test_with_default_planner(template_path, jinja2_env, tmp_path):
 
     assert Path("joined_source.pdf").exists()
 
+
 def test_with_default_planner_bad_template(bad_template_path, jinja2_env, tmp_path):
     os.chdir(tmp_path)
 
@@ -82,4 +86,3 @@ def test_with_default_planner_bad_template(bad_template_path, jinja2_env, tmp_pa
         pass
 
     assert not Path("joined_source.pdf").exists()
-

test/test_planner.py

@@ -1,5 +1,9 @@
-from bopytex.planner.generate_compile_join_planner import tasks_builder as gcj_task_builder
-from bopytex.planner.activate_corr_compile_join_planner import tasks_builder as accj_task_builder
+from bopytex.planner.activate_corr_compile_join_planner import (
+    tasks_builder as accj_task_builder,
+)
+from bopytex.planner.generate_compile_join_planner import (
+    tasks_builder as gcj_task_builder,
+)
 from bopytex.tasks import Task
 
 
@@ -18,9 +22,9 @@ def test_tasks_builder_generate():
                 "options": {
                     "no_pdf": True,
                     "subjects": [{"number": "01"}, {"number": "02"}],
-                    'template': 'tpl_source.tex',
+                    "template": "tpl_source.tex",
                 },
-                "subject": {"number": "01"}
+                "subject": {"number": "01"},
             },
             deps=["tpl_source.tex"],
             output="01_source.tex",
@@ -31,9 +35,9 @@ def test_tasks_builder_generate():
                 "options": {
                     "no_pdf": True,
                     "subjects": [{"number": "01"}, {"number": "02"}],
-                    'template': 'tpl_source.tex',
+                    "template": "tpl_source.tex",
                 },
-                "subject": {"number": "02"}
+                "subject": {"number": "02"},
             },
             deps=["tpl_source.tex"],
             output="02_source.tex",
@@ -56,9 +60,9 @@ def test_tasks_builder_generate_compile():
                 "options": {
                     "no_join": True,
                     "subjects": [{"number": "01"}, {"number": "02"}],
-                    'template': 'tpl_source.tex',
+                    "template": "tpl_source.tex",
                 },
-                "subject": {"number": "01"}
+                "subject": {"number": "01"},
             },
             deps=["tpl_source.tex"],
             output="01_source.tex",
@@ -75,9 +79,9 @@ def test_tasks_builder_generate_compile():
                 "options": {
                     "no_join": True,
                     "subjects": [{"number": "01"}, {"number": "02"}],
-                    'template': 'tpl_source.tex',
+                    "template": "tpl_source.tex",
                 },
-                "subject": {"number": "02"}
+                "subject": {"number": "02"},
             },
             deps=["tpl_source.tex"],
             output="02_source.tex",
@@ -104,9 +108,9 @@ def test_tasks_builder_generate_compile_join():
             args={
                 "options": {
                     "subjects": [{"number": "01"}, {"number": "02"}],
-                    'template': 'tpl_source.tex',
+                    "template": "tpl_source.tex",
                 },
-                "subject": {"number": "01"}
+                "subject": {"number": "01"},
             },
             deps=["tpl_source.tex"],
             output="01_source.tex",
@@ -122,9 +126,9 @@ def test_tasks_builder_generate_compile_join():
             args={
                 "options": {
                     "subjects": [{"number": "01"}, {"number": "02"}],
-                    'template': 'tpl_source.tex',
+                    "template": "tpl_source.tex",
                 },
-                "subject": {"number": "02"}
+                "subject": {"number": "02"},
             },
             deps=["tpl_source.tex"],
             output="02_source.tex",
@@ -161,9 +165,9 @@ def test_tasks_builder_generate_compile_corr():
                     "subjects": [{"number": "01"}, {"number": "02"}],
                     "corr": True,
                     "no_join": True,
-                    'template': 'tpl_source.tex',
+                    "template": "tpl_source.tex",
                 },
-                "subject": {"number": "01"}
+                "subject": {"number": "01"},
             },
             deps=["tpl_source.tex"],
             output="01_source.tex",
@@ -177,12 +181,12 @@ def test_tasks_builder_generate_compile_corr():
         Task(
             action="ACTIVATE_CORR",
             args={
-                'corr': True,
-                'no_join': True,
-                'no_pdf': False,
-                'template': 'tpl_source.tex',
-                "subjects": [{'number': '01'}, {'number': '02'}]
-                },
+                "corr": True,
+                "no_join": True,
+                "no_pdf": False,
+                "template": "tpl_source.tex",
+                "subjects": [{"number": "01"}, {"number": "02"}],
+            },
             deps=["01_source.tex"],
             output="corr_01_source.tex",
         ),
@@ -199,9 +203,9 @@ def test_tasks_builder_generate_compile_corr():
                     "subjects": [{"number": "01"}, {"number": "02"}],
                     "corr": True,
                     "no_join": True,
-                    'template': 'tpl_source.tex',
+                    "template": "tpl_source.tex",
                 },
-                "subject": {"number": "02"}
+                "subject": {"number": "02"},
             },
             deps=["tpl_source.tex"],
             output="02_source.tex",
@@ -215,12 +219,12 @@ def test_tasks_builder_generate_compile_corr():
         Task(
             action="ACTIVATE_CORR",
             args={
-                'corr': True,
-                'no_join': True,
-                'no_pdf': False,
-                'template': 'tpl_source.tex',
-                "subjects": [{'number': '01'}, {'number': '02'}]
-                },
+                "corr": True,
+                "no_join": True,
+                "no_pdf": False,
+                "template": "tpl_source.tex",
+                "subjects": [{"number": "01"}, {"number": "02"}],
+            },
             deps=["02_source.tex"],
             output="corr_02_source.tex",
         ),
@@ -250,9 +254,9 @@ def test_tasks_builder_generate_compile_corr_joined():
                     "subjects": [{"number": "01"}, {"number": "02"}],
                     "corr": True,
                     "no_join": False,
-                    'template': 'tpl_source.tex',
+                    "template": "tpl_source.tex",
                 },
-                "subject": {"number": "01"}
+                "subject": {"number": "01"},
             },
             deps=["tpl_source.tex"],
             output="01_source.tex",
@@ -266,12 +270,12 @@ def test_tasks_builder_generate_compile_corr_joined():
         Task(
             action="ACTIVATE_CORR",
             args={
-                'corr': True,
-                'no_join': False,
-                'no_pdf': False,
-                'template': 'tpl_source.tex',
-                "subjects": [{'number': '01'}, {'number': '02'}]
-                },
+                "corr": True,
+                "no_join": False,
+                "no_pdf": False,
+                "template": "tpl_source.tex",
+                "subjects": [{"number": "01"}, {"number": "02"}],
+            },
             deps=["01_source.tex"],
             output="corr_01_source.tex",
         ),
@@ -288,9 +292,9 @@ def test_tasks_builder_generate_compile_corr_joined():
                     "subjects": [{"number": "01"}, {"number": "02"}],
                     "corr": True,
                     "no_join": False,
-                    'template': 'tpl_source.tex',
+                    "template": "tpl_source.tex",
                 },
-                "subject": {"number": "02"}
+                "subject": {"number": "02"},
             },
             deps=["tpl_source.tex"],
             output="02_source.tex",
@@ -304,12 +308,12 @@ def test_tasks_builder_generate_compile_corr_joined():
         Task(
             action="ACTIVATE_CORR",
             args={
-                'corr': True,
-                'no_join': False,
-                'no_pdf': False,
-                'template': 'tpl_source.tex',
-                "subjects": [{'number': '01'}, {'number': '02'}]
-                },
+                "corr": True,
+                "no_join": False,
+                "no_pdf": False,
+                "template": "tpl_source.tex",
+                "subjects": [{"number": "01"}, {"number": "02"}],
+            },
             deps=["02_source.tex"],
             output="corr_02_source.tex",
         ),
@@ -344,10 +348,10 @@ def test_only_corr_tasks_builder():
         Task(
             action="ACTIVATE_CORR",
             args={
-                'no_join': False,
-                'no_pdf': False,
-                'sources': ['01_source.tex', '02_source.tex']
-                },
+                "no_join": False,
+                "no_pdf": False,
+                "sources": ["01_source.tex", "02_source.tex"],
+            },
             deps=["01_source.tex"],
             output="corr_01_source.tex",
         ),
@@ -360,10 +364,10 @@ def test_only_corr_tasks_builder():
         Task(
             action="ACTIVATE_CORR",
             args={
-                'no_join': False,
-                'no_pdf': False,
-                'sources': ['01_source.tex', '02_source.tex']
-                },
+                "no_join": False,
+                "no_pdf": False,
+                "sources": ["01_source.tex", "02_source.tex"],
+            },
             deps=["02_source.tex"],
             output="corr_02_source.tex",
         ),

test/test_scheduler.py

@@ -1,9 +1,11 @@
-from bopytex.message import Message
-from bopytex.tasks import Task
-from bopytex.scheduler import Scheduler
-from .fakes.dispatcher import fake_dispatcher
 import pytest
 
+from bopytex.message import Message
+from bopytex.scheduler import Scheduler
+from bopytex.tasks import Task
+
+from .fakes.dispatcher import fake_dispatcher
+
 
 def test_schedule_append():
     scheduler = Scheduler(dispatcher=fake_dispatcher)

test/test_service.py

@@ -1,10 +1,12 @@
 import os
+
 import pytest
 
 from bopytex.planner import fake_planner
 from bopytex.service import build_config, orcherstrator
 from bopytex.tasks import Task
 from bopytex.worker import Dispatcher
+
 from .fakes.workers import fake_worker
 
 

test/test_texenv.py

@@ -22,6 +22,7 @@ def test_block_line_statement():
     output = jinja2_template.render()
     assert output == "2"
 
+
 def test_block_line_statement_with_comment():
     base_template = r"""%-set a = 2
 %# comment
@@ -30,11 +31,10 @@ def test_block_line_statement_with_comment():
     output = jinja2_template.render()
     assert output == "\n2"
 
+
 def test_add_filter():
     texenv.filters["count_caracters"] = lambda x: len(x)
     base_template = r"""\Var{a} has \Var{a | count_caracters}"""
     jinja2_template = texenv.from_string(base_template)
     output = jinja2_template.render(a="coucou")
     assert output == "coucou has 6"
-    
-

test/worker/test_compile.py

@@ -1,9 +1,10 @@
 import os
-
 from pathlib import Path
-from bopytex.worker.compile import latexmk
+
 import pytest
 
+from bopytex.worker.compile import latexmk
+
 
 @pytest.fixture
 def tex_path(tmp_path):
@@ -14,7 +15,7 @@ def tex_path(tmp_path):
 \\documentclass{article}
 
 \\begin{document}
-First document. This is a simple example, with no 
+First document. This is a simple example, with no
 extra parameters or packages included.
 \\end{document}
         """
@@ -23,7 +24,7 @@ extra parameters or packages included.
 
 
 def test_latexmk(tex_path, tmp_path):
-    #tmp_path = tex_path.parent
+    # tmp_path = tex_path.parent
     os.chdir(tmp_path)
 
     texfile = str(tex_path.name)

test/worker/test_generate.py

@@ -1,8 +1,10 @@
 import os
+
 import jinja2
-from bopytex.worker.generate import generate
 import pytest
 
+from bopytex.worker.generate import generate
+
 
 @pytest.fixture
 def jinja2_env(tmp_path):
@@ -31,9 +33,9 @@ def test_generate(template_path, jinja2_env):
             "options": {
                 "direct_access": {"a": 2},
                 "jinja2": {"environment": jinja2_env},
-                },
-            "subject": {},
             },
+            "subject": {},
+        },
         deps=[template],
         output=output,
     )
@@ -70,10 +72,10 @@ def test_generate_with_random(template_path_with_random, jinja2_env):
                 "jinja2": {"environment": jinja2_env},
                 "direct_access": {
                     "random": random,
-                    }
                 },
-            "subject":{},
             },
+            "subject": {},
+        },
         deps=[template],
         output=output,
     )

test/worker/test_join_pdf.py

@@ -1,10 +1,11 @@
 import os
-
 import shutil
 from pathlib import Path
-from bopytex.worker.join_pdf import pdfjam
+
 import pytest
 
+from bopytex.worker.join_pdf import pdfjam
+
 
 @pytest.fixture
 def multiple_pdf(tmp_path, request):