Trying to understanding lstrip('0x') behaviour when remaining string should be '0'

Question

Here is the sample code:

test = '0x0'
test.lstrip('0x')

I expect '0' but I see ''.

Is this some bug or expected behavior? If the latter, what is the best solution to obtain expected output?

It is expected. Use `test[2:] if test.startswith('0x') or test` or something similar. — InSync, Jun 13 '23 at 23:29
`test.lstrip('0x')` means that every `0` and `x` is striped from the left side. So with `0x0` you end with empty string. — Andrej Kesely, Jun 13 '23 at 23:29

score 1 · Accepted Answer · answered Jun 13 '23 at 23:34

The misunderstanding is thinking that str.lstrip strips that exact literal string from the left side of a string. Rather it strips each of the characters in that string.

>>> 'foo bar'.lstrip('ofb ')
'ar'

If you need to remove an exact substring from the beginning of a string, you may wish to use an approach like the following.

>>> s = 'foo bar'
>>> if s.startswith('fo'): s = s[len('fo'):]
... 
>>> s
'o bar'

Trying to understanding lstrip('0x') behaviour when remaining string should be '0'

1 Answers1