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我有一个项目需要使用 python 将十几个正则表达式应用于大约 100 个文件。在网上搜索了 4 个多小时的各种组合，包括“(merge|concatenate|stack|join|compile) multiple regex in python”，但我还没有找到任何关于我需要的帖子。

这对我来说是一个中等规模的项目。我需要几个较小的正则表达式项目，它们只需要 5-6 个正则表达式模式，仅应用于十几个文件。虽然这些将对我的工作有很大帮助，但祖父项目是一个应用 100 多个搜索的文件，将字符串替换为我得到的任何新文件。（某些语言的拼写约定不是标准化的，能够快速处理文件将提高工作效率。）

理想情况下，正则表达式字符串需要可由非程序员更新，但这可能超出了本文的范围。

这是我到目前为止所拥有的：

import os, re, sys # Is "sys" necessary?

path = "/Users/mypath/testData"
myfiles = os.listdir(path)

for f in myfiles:

    # split the filename and file extension for use in renaming the output file
    file_name, file_extension = os.path.splitext(f)
    generated_output_file = file_name + "_regex" + file_extension

    # Only process certain types of files.
    if re.search("txt|doc|odt|htm|html")

    # Declare input and output files, open them, and start working on each line.
        input_file = os.path.join(path, f)
        output_file = os.path.join(path, generated_output_file)

        with open(input_file, "r") as fi, open(output_file, "w") as fo:
            for line in fi:

    # I realize that the examples are not regex, but they are in my real data.
    # The important thing, is that each of these is a substitution.
                line = re.sub(r"dog","cat" , line)
                line = re.sub(r"123", "789" , line)
                # Etc.

    # Obviously this doesn't work, because it is only writing the last instance of line.
                fo.write(line)
                fo.close()
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1 回答 1

3

这是你要找的吗？

不幸的是，您没有指定如何知道应该应用哪些正则表达式，所以我将它们放入元组列表中（第一个元素是正则表达式，第二个是替换文本）。

import os, os.path, re

path = "/Users/mypath/testData"
myfiles = os.listdir(path)
# its much faster if you compile your regexes before you
# actually use them in a loop
REGEXES = [(re.compile(r'dog'), 'cat'),
           (re.compile(r'123'), '789')]
for f in myfiles:
    # split the filename and file extension for use in
    # renaming the output file
    file_name, file_extension = os.path.splitext(f)
    generated_output_file = file_name + "_regex" + file_extension

    # As l4mpi said ... if odt is zipped, you'd need to unzip it first
    # re.search is slower than a simple if statement
    if file_extension in ('.txt', '.doc', '.odt', '.htm', '.html'):

        # Declare input and output files, open them,
        # and start working on each line.
        input_file = os.path.join(path, f)
        output_file = os.path.join(path, generated_output_file)

        with open(input_file, "r") as fi, open(output_file, "w") as fo:
            for line in fi:
                for search, replace in REGEXES:
                    line = search.sub(replace, line)
                fo.write(line)
        # both the input and output files are closed automatically
        # after the with statement closes
于 2012-09-23T11:15:01.453 回答