Python :: remove all occurrences until first space

Question

how do I remove all occurrences in a string up to the first space, so that:

strings = ["1234 zoocore", "4356 00's punk"]

becomes: ["zoocore", "00's punk"] ?

I have tried regex:

for s in strings:
    new_s = re.sub(r'\d+','', s)

but that erases 00' as well, which I don't want.

Answer 1

You can use str.split with maxsplit parameter:

>>> strings = ["1234 zoocore", "4356 00's punk"]
>>> [s.split(None, 1)[1] for s in strings]
['zoocore', "00's punk"]

If you have strings that don't contain space you can use -1 as index:

>>> strings = ["1234 zoocore", "4356 00's punk", "rock"]
>>> [s.split(None, 1)[-1] for s in strings]
['zoocore', "00's punk", 'rock']

Answer 2

Your mistake is that you didn't anchor the match at the start of the string and that you didn't include the space. Here's a fixed version:

>>> [re.sub(r'^\d* ', '', s) for s in strings]
['zoocore', "00's punk"]

And just another way, finding the space (if any) and then slicing the string.

>>> strings = ["1234 zoocore", "4356 00's punk", "rock"]
>>> [s[s.find(' ') + 1:] for s in strings]
['zoocore', "00's punk", 'rock']