如何从字符串列表中删除未使用的字符

Question

For example, my list is:

list_1 =  [ '[1234,', '4567,', '19234,', '786222]' ]

and the expected output is:

list_1 = [1234, 4567, 19234, 786222]

Answer 1

Frankly, the easiest way to get a list of integers back out of that is to put it back together as a string representing a list:

>>> list_1 = ['[1234,', '4567,', '19234,', '786222]']

>>> list_repr = ' '.join(list_1)
>>> list_repr
'[1234, 4567, 19234, 786222]'

And then feed it through ast.literal_eval :

>>> from ast import literal_eval
>>> literal_eval(list_repr)
[1234, 4567, 19234, 786222]

In the likely event that you got list_1 by using the split method on a string that already represented a list in the first place, I'm sure you can figure out how to shorten this process...

Answer 2

You canstrip the bad characters and convert to int .

>>> [int(s.strip("[],")) for s in list_1]
[1234, 4567, 19234, 786222]

Answer 3

Using Python 2:

>>> map(int, ''.join(list_1).strip('[]').split(','))
[1234, 4567, 19234, 786222]

In Python 3 map returns a map object which only lazily evaluates. To create the list we need to be explicit. This is a good thing, more efficient and general:

>>> list(map(int, ''.join(list_1).strip('[]').split(',')))
[1234, 4567, 19234, 786222]

Answer 4

You can use translate to remove your unwanted characters.

>>> [int(s.translate(None,"[],")) for s in list_1]
[1234, 4567, 19234, 786222]

Hope this helps.

Answer 5

If you don't know the set of "bad" characters up front, you can also filter all digits from each string and then convert the result to an integer:

>>> list_1 =  [ '[1234,', '4567,', '19234,', '786222]' ]
>>> [int(filter(lambda x: x.isdigit(), s)) for s in list_1]
[1234, 4567, 19234, 786222]