Fastest method: Split string and get last element-CodePudding

I have a list of urls:

urls = ['https://www.website.com/407161a', 'https://www.website.com/359426a', 'https://www.website.com/441885a', 'https://www.website.com/331791a']

I'm trying to split each url string on / and get the last element, for the following output: ['407161a', '359426a', '441885a', '331791a', '403123a'].

I use this code to get the above: [u.rpartition('/')[2] for u in urls]. The problem: it's somewhat slow on larger lists. It takes ~0.75 seconds on a list with 2 million urls on my machine. I'm trying to find a faster method since I'll be running this multiple times on lists containing 10 million elements.

Is there a way to make this code faster?

CodePudding user response：

You can use str.rsplit with maxsplit parameter as 1 (using rsplit with maxsplit parameter will just avoid unnecessary splits), then take the last element. You can use list-comprehension for each of the items.

>>> [i.rsplit('/',1)[-1] for i in urls]
['407161a', '359426a', '441885a', '331791a']

CodePudding user response：

You can use str.rindex and find index of '/' from last then return all chars from that index to last string.

Try this:

[u[u.rindex('/') 1:] for u in URLs]
# ['407161a', '359426a', '441885a', '331791a']