findall返回指定捕获组的命令?

Inspired by a now-deleted question; given a regex with named groups, is there a method like findall which returns a list of dict with the named capturing groups instead of a list of tuple?

受到一个现在被删除的问题的启发;给定一个带命名组的regex，是否有像findall这样的方法，它返回带有命名捕获组的dict列表，而不是tuple列表?

Given:

考虑到:

>>> import re
>>> text = "bob sue jon richard harry"
>>> pat = re.compile('(?P<name>[a-z]+)\s+(?P<name2>[a-z]+)')
>>> pat.findall(text)
[('bob', 'sue'), ('jon', 'richard')]

Should instead give:

应该给:

[{'name': 'bob', 'name2': 'sue'}, {'name': 'jon', 'name2': 'richard'}]

4 个解决方案

#1

>>> import re
>>> s = "bob sue jon richard harry"
>>> r = re.compile('(?P<name>[a-z]+)\s+(?P<name2>[a-z]+)')
>>> [m.groupdict() for m in r.finditer(s)]
[{'name2': 'sue', 'name': 'bob'}, {'name2': 'richard', 'name': 'jon'}]

#2

you could switch to finditer

你可以切换到finditer

>>> import re
>>> text = "bob sue jon richard harry"
>>> pat = re.compile('(?P<name>[a-z]+)\s+(?P<name2>[a-z]+)')
>>> for m in pat.finditer(text):
...     print m.groupdict()
... 
{'name2': 'sue', 'name': 'bob'}
{'name2': 'richard', 'name': 'jon'}

#3

If you are using match :

如果您正在使用match:

r = re.match('(?P<name>[a-z]+)\s+(?P<name2>[a-z]+)', text)
r.groupdict()

documentation here

文件在这里

#4

There's no built-in method for doing this, but the expected result can be achieved by using list comprehensions.

没有内置的方法来实现这一点，但是通过使用列表理解可以实现预期的结果。

[dict([[k, i if isinstance(i, str) else i[v-1]] for k,v in pat.groupindex.items()]) for i in pat.findall(text)]

With friendly formatting:

和友好的格式:

>>> [
...     dict([
...         [k, i if isinstance(i, str) else i[v-1]]
...         for k,v in pat.groupindex.items()
...     ])
...     for i in pat.findall(text)
... ]

We construct a list using a list comprehension, iterate over the result from findall which is either a list of strings or a list of tuples (0 or 1 capturing groups result in a list of str).

我们使用列表理解构造一个列表，遍历findall的结果，它要么是字符串列表，要么是元组列表(0或1捕获组导致str列表)。

For each item in the result we construct a dict from another list comprehension which is generated from the groupindex field of the compiled pattern, which looks like:

对于结果中的每一项，我们从另一个列表理解中构造一条命令，这个理解来自编译模式的groupindex字段，它看起来如下:

>>> pat.groupindex
{'name2': 2, 'name': 1}

A list is constructed for each item in the groupindex and if the item from findall was a tuple, the group number from groupindex is used to find the correct item, otherwise the item is assigned to the (only extant) named group.

为groupindex中的每个项构造一个列表，如果findall中的项是一个元组，则使用groupindex中的组号来查找正确的项，否则该项被分配给(仅存在的)命名组。

[k, i if isinstance(i, str) else i[v-1]]

Finally, a dict is constructed from the list of lists of strings.

最后，一个命令由字符串列表组成。

Note that groupindex contains only named groups, so non-named capturing groups will be omitted from the resulting dict.

注意groupindex只包含已命名的组，因此在生成的命令中将忽略未命名的捕获组。

And the result:

结果:

[dict([[k, i if isinstance(i, str) else i[v-1]] for k,v in pat.groupindex.items()])  for i in pat.findall(text)]
[{'name2': 'sue', 'name': 'bob'}, {'name2': 'richard', 'name': 'jon'}]

#1

>>> import re
>>> s = "bob sue jon richard harry"
>>> r = re.compile('(?P<name>[a-z]+)\s+(?P<name2>[a-z]+)')
>>> [m.groupdict() for m in r.finditer(s)]
[{'name2': 'sue', 'name': 'bob'}, {'name2': 'richard', 'name': 'jon'}]

#2