对于循环，如果元素不等于值，则用空字符串替换

Question

my_list = ['1 ab ac bbba','23 abcba a aabb ab','345 ccc ab aaaaa']

I'm trying to get rid of the numbers and the spaces, basically everything that's not an 'a','b', or 'c'

I tried this but it didn't work and I'm not sure why:

for str in my_list:
    for i in str:
        if i != 'a' or 'b' or 'c':
            i = ''
        else:
            pass

I want to eventually get:

my_list2 = ['abacbbba','abcbaaaabbab','cccabaaaaa']

Answer 1

You're misunderstanding how or works:

if i != 'a' or 'b' or 'c':

is equivalent to

if (i != 'a') or ('b') or ('c'):

and will therefore always be True (because b evaluates to True ).

You probably meant to write

if i != 'a' and i != 'b' and i != 'c':

which can also be written as

if i not in ('a', 'b', 'c'):

or even (since a string can iterate over its characters)

if i not in 'abc':

But even then, you're not doing anything with that information; a string is immutable, and by assigning '' to i , you're not changing the string at all. So if you want to do it without a regex, the correct way would be

>>> my_list = ['1 ab ac bbba','23 abcba a aabb ab','345 ccc ab aaaaa']
>>> new_list = [''.join(c for c in s if c in 'abc') for s in my_list]
>>> new_list
['abacbbba', 'abcbaaaabbab', 'cccabaaaaa']

Answer 2

Use re.sub to replace everything that is not a , b , or c , ie, [^abc] , with an empty string:

import re
my_list2 = []
for str in my_list:
    my_list2.append(re.sub("[^abc]", "", str))

DEMO .

Answer 3

  m = ['1 ab ac bbba','23 abcba a aabb ab','345 ccc ab aaaaa']
  n=[m[x][m[x].index(" "):] for x in range(len(m))]
  n=[x.replace(" ","") for x in n]