私は LaTeX の初心者で、このような単純な Python コードを LaTeX でプロフェッショナルな方法で表現する方法がわかりません。あなたの解決策と例があれば、学習と進歩に大いに役立ちます。
if transactions: Transaction.create_transactions() # if transactions = "true"
node.generate_emptyState() # empty state for all nodes
S.initial_events() # initiate initial events to start with
while not queue.isEmpty() and clock <= targetTime:
next_e = queue.get_next_event()
clock = next_e.time # move clock to the time of the event
Event.execute_event(next_e)
Queue.remove_event(next_e)
print results
答え1
これは私が好んで行う方法です。もちろん、色、フォント サイズ、境界線などを変更できます。
必要なのはピグメント使用するminted
。
ムウェ
\documentclass[11pt,a4paper]{report}
\usepackage{tcolorbox}
\tcbuselibrary{minted,breakable,xparse,skins}
\definecolor{bg}{gray}{0.95}
\DeclareTCBListing{mintedbox}{O{}m!O{}}{%
breakable=true,
listing engine=minted,
listing only,
minted language=#2,
minted style=default,
minted options={%
linenos,
gobble=0,
breaklines=true,
breakafter=,,
fontsize=\small,
numbersep=8pt,
#1},
boxsep=0pt,
left skip=0pt,
right skip=0pt,
left=25pt,
right=0pt,
top=3pt,
bottom=3pt,
arc=5pt,
leftrule=0pt,
rightrule=0pt,
bottomrule=2pt,
toprule=2pt,
colback=bg,
colframe=orange!70,
enhanced,
overlay={%
\begin{tcbclipinterior}
\fill[orange!20!white] (frame.south west) rectangle ([xshift=20pt]frame.north west);
\end{tcbclipinterior}},
#3}
\begin{document}
\begin{mintedbox}{python}
if transactions: Transaction.create_transactions() # if transactions = "true"
node.generate_emptyState() # empty state for all nodes
S.initial_events() # initiate initial events to start with
while not queue.isEmpty() and clock <= targetTime:
next_e = queue.get_next_event()
clock = next_e.time # move clock to the time of the event
Event.execute_event(next_e)
Queue.remove_event(next_e)
print results
\end{mintedbox}
\end{document}
答え2
私は を好みますminted
が、 も使用できますpythonhighlight
。これは、 -package の優れたフロントエンドですlistings
。
\documentclass{article}
\usepackage{pythonhighlight}
\begin{document}
\begin{python}
if transactions: Transaction.create_transactions() # if transactions = "true"
node.generate_emptyState() # empty state for all nodes
S.initial_events() # initiate initial events to start with
while not queue.isEmpty() and clock <= targetTime:
next_e = queue.get_next_event()
clock = next_e.time # move clock to the time of the event
Event.execute_event(next_e)
Queue.remove_event(next_e)
print results
\end{python}
\end{document}
listings
あまりいじらずに良い結果を得るもう一つの方法は、solarzied
-パッケージ:
\documentclass{article}
\usepackage{solarized-light}
\begin{document}
\begin{lstlisting}[language=python]
if transactions: Transaction.create_transactions() # if transactions = "true"
node.generate_emptyState() # empty state for all nodes
S.initial_events() # initiate initial events to start with
while not queue.isEmpty() and clock <= targetTime:
next_e = queue.get_next_event()
clock = next_e.time # move clock to the time of the event
Event.execute_event(next_e)
Queue.remove_event(next_e)
print results
\end{lstlisting}
\end{document}
答え3
{Piton}
拡張機能 を使用できますpiton
。これは LuaLaTeX の Lua ライブラリ LPEG を使用します (LuaLaTeX が必要です)。外部プログラムは必要ありません。
\documentclass{article}
\usepackage{xcolor}
\usepackage{piton}
\begin{document}
\begin{Piton}
from math import pi
def arctan(x,n=10):
"""Compute the mathematical value of arctan(x)
n is the number of terms in the sum
"""
if x < 0:
return -arctan(-x) # recursive call
elif x > 1:
return pi/2 - arctan(1/x)
#> (we have used that $\arctan(x)+\arctan(1/x)=\frac{\pi}{2}$ for $x>0$)
else:
s = 0
for k in range(n):
s += (-1)**k/(2*k+1)*x**(2*k+1)
return s
\end{Piton}
\end{document}
と組み合わせて使用することも可能ですtcolorbox
。
\documentclass{article}
\usepackage{xcolor}
\usepackage{piton}
\usepackage{tcolorbox}
\NewPitonEnvironment{Python}{}
{\begin{tcolorbox}}
{\end{tcolorbox}}
\begin{document}
\begin{Python}
from math import pi
def arctan(x,n=10):
"""Compute the mathematical value of arctan(x)
n is the number of terms in the sum
"""
if x < 0:
return -arctan(-x) # recursive call
elif x > 1:
return pi/2 - arctan(1/x)
#> (we have used that $\arctan(x)+\arctan(1/x)=\frac{\pi}{2}$ for $x>0$)
else:
s = 0
for k in range(n):
s += (-1)**k/(2*k+1)*x**(2*k+1)
return s
\end{Python}
\end{document}