使用regex中的findall和sub函数来搜索和替换精确的字符串

Question

基于此论坛Replacing a line in a file based on a keyword search, by line from another file我在我的真实档案中遇到了一点困难。如下图所示，我想搜索关键字＆＃34; PBUSH后跟数字（不断增加）＆＃34;并且基于该关键字，如果存在或不存在，则在另一个文件中搜索。如果它存在，则替换行中的数据＆＃34; PBUSH数字K一些小数＆＃34;到另一个文件中找到的行，保持搜索关键字相同。它一直持续到文件末尾，看起来像

和我修改的代码（请注意findall和sub格式）如下所示：

import re
path1 = "C:\Users\sony\Desktop\PBUSH1.BDF"
path2 = "C:\Users\sony\Desktop\PBUSH2.BDF"

with open(path1) as f1, open(path2) as f2:
    dat1 = f1.read()
    dat2 = f2.read()

    matches = re.findall('^PBUSH \s [0-9] \s K [0-9 ]+', dat1, flags=re.MULTILINE)
    for match in matches:
        dat2 = re.sub('^{} \s [0-9] \s K \s'.format(match.split(' ')[0]), match, dat2, flags=re.MULTILINE)

with open(path2, 'w') as f:
    f.write(dat2)

这里我的搜索关键字是PBUSH空格数，然后其余部分如PBUSH行所示。我无法使它工作。可能是什么原因！

Answer 1

最好在这种情况下使用组，并将整个字符串分成两个，一个用于匹配短语，另一个用于数据。

import re
# must use raw strings for paths, otherwise we need to
# escape \ characters
input1 = r"C:\Users\sony\Desktop\PBUSH1.BDF" 
input2 = r"C:\Users\sony\Desktop\PBUSH2.BDF"

with open(input1) as f1, open(input2) as f2:
    dat1 = f1.read()
    dat2 = f2.read()

# use finditer instead of findall so that we will get 
# a match object for each match.
# For each matching line we also have one subgroup, containing the
# "PBUSH   NNN     " part, whereas the whole regex matches until
# the next end of line
matches = re.finditer('^(PBUSH\s+[0-9]+\s+).*$', dat1, flags=re.MULTILINE)

for match in matches:
    # for each match we construct a regex that looks like
    # "^PBUSH   123      .*$", then replace all matches thereof
    # with the contents of the whole line
    dat2 = re.sub('^{}.*$'.format(match.group(1)), match.group(0), dat2, flags=re.MULTILINE)

with open(input2, 'w') as outf:
    outf.write(dat2)