python - 如何在 Python 中删除或替换由起点和终点确定的子字符串？

Question

有时我会删除或替换一个长字符串的子字符串。因此，我将确定一个开始模式和一个结束模式，这将确定子字符串的起点和终点：

long_string = "lorem ipsum..white chevy..blah,blah...lot of text..beer bottle....and so to the end"
removed_substr_start = "white chevy"
removed_substr_end = "beer bott"

# this is pseudo method down
STRresult = long_string.replace( [from]removed_substr_start [to]removed_substr_end, "")

Answer 1

I guess you want something like that, without regex:

def replace_between(text, begin, end, alternative=''):
    middle = text.split(begin, 1)[1].split(end, 1)[0]
    return text.replace(middle, alternative)

Not tested and you should protected the first line from exception (if begin or end is not found), but the idea is here :)

Answer 2

You can use regex:

>>> import re
>>> strs = "lorem ipsum..white chevy..blah,blah...lot of text..beer bottle....and so to the end"
>>> sub_start = "white chevy"
>>> sub_end = "beer bott"
>>> re.sub(r'{}.*?{}'.format(re.escape(sub_start),re.escape(sub_end)),'',strs)
'lorem ipsum..le....and so to the end'

If you only want to remove the sub-string between "white chevy" and "beer bott" but not these words:

>>> re.sub(r'({})(.*?)({})'.format(re.escape(sub_start),
                                               re.escape(sub_end)),r'\1\3',strs)
'lorem ipsum..white chevybeer bottle....and so to the end'

Answer 3

使用获取起始索引string.find()和使用最后一个索引string.rfind()，然后使用以下方法删除内部部分：

lindex = string.find(long_string, removed_substr_start)
rindex = string.find(long_string, removed_substr_end, lindex)
result = long_string[0:lindex] + longstring[rindex:]

请参阅：http ://docs.python.org/2/library/string.html#string.find

Answer 4

import re
regexp = "white chevy.*?beer bott"
long_string = "lorem ipsum..white chevy..blah,blah...lot of text..beer bottle....and so to the end"
re.sub(regexp, "", long_string)

gives:

'lorem ipsum..le....and so to the end'

Answer 5

在使用了许多方法后，我发现这个解决方案是没有正则表达式的最佳解决方案：

def getString( str, _from, _to ):
    end_from = str.find( _from ) +len( _from)
    return str[ end_from : str.find( _to, end_from ) ]