Question

我想剪掉匹配的字符串。

我考虑在re.finditer中使用＆＃34; [m.start（）for m（＆＃39;＆＃39;）]＆＃34;获得索引。

但我认为存在比这更好的方式。

例如，我想在＆＃34;标题＆＃34;之间剪切字符串。和＆＃34;页脚＆＃34;。

str = "header1svdijfooter1ccsdheader2cdijhfooter2"
headers = ["one": "header1", "two": "header2"]
footers = ["one": "footer1", "two": "footer2"]

#I want to get ["header1svdijfooter1", "header2cdijhfooter2"]

请建议我。

Answer 1

import re

def returnmatches(text,headers,footers):
    """headers is a list of headers
footers is a list of footers
text is the text to search"""
    for header,footer in zip(headers,footers):
        pattern = r"{}\w+?{}".format(header,footer)
        try:
            yield re.search(pattern,input_text).group()
        except AttributeError:
            # handle no match
            pass

或者：

text = "header1svdijfooter1ccsdheader2cdijhfooter2"
headers = ["header1", "header2"]
footers = ["footer1", "footer2"]

import re

matches = [re.search(r"{}\w+?{}".format(header,footer),text).group() for header,footer in zip(headers,footers) if re.search(r"{}\w+?{}".format(header,footer),text)]

Answer 2

import re

# as a general rule you shouldn't call variables str in python as it's a builtin function name.
str = "header1svdijfooter1ccsdheader2cdijhfooter2" 

# this is how you declare dicts.. but if you're only going to have "one"
# and "two" for the keys why not use a list?  (you need the {} for dicts).
#headers = {"one": "header1", "two": "header2"}  
#footers = {"one": "footer1", "two": "footer2"}  
delimiters = [("header1", "footer1"), ("header2", "footer2")]

results = []
for header, footer in delimiters:

    regex = re.compile("({header}.*?{footer})".format(header = header, footer = footer))

    matches = regex.search(str)
    if matches is not None:
        for group in matches.groups():
            results.append(group)

print results

Answer 3

可以使用列表推导在一行中完成计算：

s = "header1svdijfooter1ccsdheader2cdijhfooter2"
headers = {"one": "header1", "two": "header2"}
footers = {"one": "footer1", "two": "footer2"}
out = [re.search('({}.*?{})'.format(headers[k], footers[k]), s).group(0) for k in sorted(headers.keys())]

以上假设，根据示例，只有一个匹配组。

或者，如果有人喜欢循环：

s = "header1svdijfooter1ccsdheader2cdijhfooter2"
headers = {"one": "header1", "two": "header2"}
footers = {"one": "footer1", "two": "footer2"}
out=[]
for k in sorted(headers.keys()):
    out.extend(re.search('({}.*?{})'.format(headers[k], footers[k]), s).groups())
print out

以上产生输出：

['header1svdijfooter1', 'header2cdijhfooter2']

Answer 4

没有重新：

str = "header1svdijfooter1ccsdheader2cdijhfooter2"
result = []
capture=False
currentCapture = ""
for i in range(len(str)):
    if str[i:].startswith("header1") or str[i:].startswith("header2"):
        currentCapture = ""
        capture=True
    elif str[:i].endswith("footer1") or str[:i].endswith("footer2"):
        capture=False
        result.append(currentCapture)
        currentCapture = ""
    if capture:
        currentCapture = currentCapture+str[i]
if currentCapture:
    result.append(currentCapture)

print result

输出：

['header1svdijfooter1', 'header2cdijhfooter2']

如何剪出匹配的字符串

4 个答案: