Question

我有以下的LaTeX代码，并希望删除\ NEW {“跨越多行的文本”}的所有出现。需要保留“跨越多行的文本”，只需删除“\ NEW {”和文件“}”中的某个位置，新的括号内应保持不变。需要保留选项卡，空格和换行符。我已经尝试编写python应用程序，但无法生成合适的输出。最困难的部分是你删除括号的位置（可以在下一行）。

输入：

\chapter{A \NEW{very} small \NEW{chapter}}

\begin{itemize}
    \item \NEW{Bla}
    \item Dusse
    \item Mekker
\end{itemize}

\NEW{This is new
    multiline \texttt{text} with some things \TBD{TBD} in between
  } The end

输出（预期）：

\chapter{A very small chapter}

\begin{itemize}
    \item Bla
    \item Dusse
    \item Mekker
\end{itemize}

This is new
    multiline \texttt{text} with some things \TBD{TBD} in between
   The end

python中的自有解决方案有效：

读行
- 用标记替换\ NEW {出现{char 0xff）
- 在线阅读c
  - 检查c是否为标记，设置标记= True，为嵌套括号设置marked_cnt，读取下一个字符
  - 否则检查：c =='{'，增加marked_cnt
  - 另外检查：c =='}'并标记为== True，减去marked_cnt
  - 如果marked_cnt == -1，则重置标记= False，marked_cnt = 0，读取下一个字符
  - 打印“有效”字符

    #!/usr/bin/env python2.7
    import sys

    marker=chr(255)
    marked=False
    marked_cnt=0

    fin = open("file.tex", "r")
    fout = open("file.tex.out", "w")

    for line in fin:
        l = line.replace("\NEW{", marker)
        for c in l:
            if c == marker:
                marked = True
                marked_cnt = 0
                continue
            elif c == '{':
                marked_cnt += 1
            elif ((c == '}') and
                (marked == True)):
                marked_cnt -= 1

            if marked_cnt == -1:
                marked = False
                marked_cnt = 0
                continue

            fout.write(c)

    fin.close()
    fout.close()

Answer 1

尝试使用正则表达式：

import re
myRe = re.compile(r'\\NEW{\w+}')
for match in myRe.findall(myString):
    newstring = match.replace('\NEW{','')
    newstring = newstring.replace('}','')
    myString.replace(match,newstring)

然而，这并没有摆脱多线问题。要解决这个问题，请直接查看字符串，然后检查括号的打开和关闭：

while s.find('\\NEW{')>-1:
    position = s.find('\\NEW{')
    print(position, s[position:position+4])
    s = s[0:position]+s[position+5:]
    trailexist = True
    openbrackets = 0

    while trailexist and position<len(s):
        position +=1
        print(len(s), position,s[position])
        if s[position] == '}' and openbrackets == 0:
            s = s[:position]+s[position+1:]
            trailexist = False
            print("Removed!", position)
        elif s[position] == '{':
            openbrackets += 1
            print('Openbrackets:',openbrackets)
        elif s[position] == '}' and openbrackets>0:
            openbrackets -= 1

删除跨越多行的LaTeX宏

1 个答案: