Python regex会立即将组替换为组名

Question

以下是re：

import re
s = "the blue dog and blue cat wore 7 blue hats 9 days ago"
p = re.compile(r'blue (?P<animal>dog|cat)')
p.sub(r'\1',s)

结果是，

'the dog and cat wore 7 blue hats 9 days ago'

有没有可能编写这样一个re.sub：

import re
s = "the blue dog and blue cat wore 7 blue hats 9 days ago"
p = re.compile(r'blue (?P<animal>dog|cat)|(?P<numberBelowSeven>[0-7])|(?P<numberNotSeven>[8-9])')

结果是，

'the animal and animal wore numberBelowSeven blue hats numberNotSeven days ago"

奇怪的是，有很多关于replace strings和getting group names的文档，但没有一种很好的方法来做这两件事。

Answer 1

您可以使用返回matchobj.lastgroup的re.sub with a callback

import re

s = "the blue dog and blue cat wore 7 blue hats 9 days ago"
p = re.compile(r'blue (?P<animal>dog|cat)|(?P<numberBelowSeven>[0-7])|(?P<numberNotSeven>[8-9])')

def callback(matchobj):
    return matchobj.lastgroup

result = p.sub(callback, s)
print(result)

收益率

the animal and animal wore numberBelowSeven blue hats numberNotSeven days ago

请注意，如果您使用的是Pandas，则可以使用Series.str.replace

import pandas as pd

def callback(matchobj):
    return matchobj.lastgroup

df = pd.DataFrame({'foo':["the blue dog", "and blue cat wore 7 blue", "hats 9", 
                          "days ago"]})
pat = r'blue (?P<animal>dog|cat)|(?P<numberBelowSeven>[0-7])|(?P<numberNotSeven>[8-9])'
df['result'] = df['foo'].str.replace(pat, callback)
print(df)

收益率

                        foo                                 result
0              the blue dog                             the animal
1  and blue cat wore 7 blue  and animal wore numberBelowSeven blue
2                    hats 9                    hats numberNotSeven
3                  days ago                               days ago

如果您有嵌套的命名组，则可能需要一个更复杂的回调，该回调遍历matchobj.groupdict().items()以收集所有相关的组名：

import pandas as pd

def callback(matchobj):
    names = [groupname for groupname, matchstr in matchobj.groupdict().items()
             if matchstr is not None]
    names = sorted(names, key=lambda name: matchobj.span(name))
    result = ' '.join(names)
    return result

df = pd.DataFrame({'foo':["the blue dog", "and blue cat wore 7 blue", "hats 9", 
                          "days ago"]})

pat=r'blue (?P<animal>dog|cat)|(?P<numberItem>(?P<numberBelowSeven>[0-7])|(?P<numberNotSeven>[8-9]))'

# pat=r'(?P<someItem>blue (?P<animal>dog|cat)|(?P<numberBelowSeven>[0-7])|(?P<numberNotSeven>[8-9]))'

df['result'] = df['foo'].str.replace(pat, callback)
print(df)

收益率

                        foo                                            result
0              the blue dog                                        the animal
1  and blue cat wore 7 blue  and animal wore numberItem numberBelowSeven blue
2                    hats 9                    hats numberItem numberNotSeven
3                  days ago                                          days ago

Answer 2

为什么不多次调用re.sub()：

>>> s = re.sub(r"blue (dog|cat)", "animal", s)
>>> s = re.sub(r"\b[0-7]\b", "numberBelowSeven", s)
>>> s = re.sub(r"\b[8-9]\b", "numberNotSeven", s)
>>> s
'the animal and animal wore numberBelowSeven blue hats numberNotSeven days ago'

然后，您可以将其放入“更改列表”中，并逐一应用：

>>> changes = [
...     (re.compile(r"blue (dog|cat)"), "animal"),
...     (re.compile(r"\b[0-7]\b"), "numberBelowSeven"),
...     (re.compile(r"\b[8-9]\b"), "numberNotSeven")
... ]
>>> s = "the blue dog and blue cat wore 7 blue hats 9 days ago"
>>> for pattern, replacement in changes:
...     s = pattern.sub(replacement, s)
... 
>>> s
'the animal and animal wore numberBelowSeven blue hats numberNotSeven days ago'

请注意，我另外添加了边界检查(\b)一词。