如何从列表中的字符串中删除前x个字符

Question

So I have a list and a couple of string in it. I just want to remove the first 7 characters from each of the strings. How do I do that?

I've tried:

lst = ["1234567something", "1234567smthelse"]

for i in lst:

    i [7:]

print lst

But I get the same list from the beginning...

Answer 1

When doing i [7:] you are not actually editing the element of the list, you are just computing a new string, without the first 7 characters, and not doing anything with it.

You can do this instead :

>>> lst = [e[7:] for e in lst]
>>> lst
['something', 'smthelse']

This will loop on the elements of your array, and remove the characters from the beginning, as expected.

Answer 2

Try this:

for i in range(0, len(lst)):
   lst[i] = lst[i][7:]

Answer 3

You can do the following:

lst = ["1234567something", "1234567smthelse"]
newlst=[]
    for i in lst:
        newlst.append(i[7:])

print newlst

I hope that helps.

Answer 4

i[7:] is not inplace, it returns a new string which you are doing nothing with.

You can create a new list with the required string:

lst = [string[7:] for string in lst]

Or you can modify the same list:

for idx, string in enumerate(ls):
    ls[idx] = string[7:]

Answer 5

您不会在任何地方保存i[7:]的值……只需创建一个带有修剪后的值的新列表：

lst = [i[7:] for i in lst]

Answer 6

尝试这个：

lst = [s[7:] for s in lst]

Answer 7

You never reassigned the lst in your question, which is why the output of lst in print(lst) does not change. Try reassigning like this:

lst = ["1234567something", "1234567smthelse"]
lst = [i[7:] for i in lst]
print(lst)

returns

['something', 'smthelse']