Sort list of strings with lambda and ignore prefix

Question

I'm trying to sort a list but ignoring a prefix. This question has been answered here. It should be straight forward, only it's not working. Here's what I have:

def sort_it(lst, ignore):

  return lst.sort(key=lambda x: x.strip(ignore))

myList = ["cheesewhiz", "www.cheese.com", "www.wagons.com", "www.apples.com", "www.bananas.com"]

ignoreThis = "www."
sort_it(myList, ignoreThis)
print myList

Only the sorting is getting mixed up as the first item doesn't have anything to ignore as part of the string. I'm not sure if adding a check to see if the string contains the ignore string is the Pythonic approach with Lambda.

I expect the results to be in alphabetic order ignoring the www.

www.apples.com
www.bananas.com
www.cheese.com
cheesewhiz
www.wagons.com

Change from sort to sorted. The sort function changes the list in place; `sorted leaves the original list alone, and returns a copy, sorted. — Prune, Nov 20 '17 at 22:03
Your key function is wrong. `str.strip` isn't suited to remove a prefix. Use `lambda x: x[4:] if x.startswith('www.') else x`. — Aran-Fey, Nov 20 '17 at 22:04
The problem is that the strip function is stripping the values only when comparing. It is not affecting actual values in the list. — Swakeert Jain, Nov 20 '17 at 22:07
@SwakeertJain The problem is that `"www.wagons.com".strip("www.")` returns `'agons.com'` — juanpa.arrivillaga, Nov 20 '17 at 22:09
@juanpa.arrivillaga That is also an issue, but I believe a different one than the OP is facing. They want the output to be without the "www." part! — Swakeert Jain, Nov 20 '17 at 22:13
@SwakeertJain no, they very clearly do not. Look at their example of desired output, and it includes the strings with the `www.` prefix. — juanpa.arrivillaga, Nov 20 '17 at 22:13
Sorry! my bad. I misunderstood the last statement in the question — Swakeert Jain, Nov 20 '17 at 22:27

Jean-François Fabre · Accepted Answer · 2017-11-21T20:04:00.110

strip doesn't work that way. It will try to strip every single character of the passed argument, so possibly more than the string you passed. Also, you're sorting in-place, no need to return None (or use sorted which will sort a copy of your parameter, maybe less a surprise for callers)

You probably want str.replace instead to get rid of www., or re.sub("^www.","",x)

def sort_it(lst, ignore):
  lst.sort(key=lambda x: x.replace(ignore,""))

myList = ["cheesewhiz", "www.cheese.com", "www.wagons.com", "www.apples.com", "www.bananas.com"]

ignoreThis = "www."
sort_it(myList, ignoreThis)
print(myList)

result:

['www.apples.com', 'www.bananas.com', 'www.cheese.com', 'cheesewhiz', 'www.wagons.com']

More accurately, if you want to remove www. from the key only if it starts with www. you could go with regex (although you'd need to escape the text):

import re

def sort_it(lst, ignore):
  lst.sort(key=lambda x: re.sub("^"+re.escape(ignore),"",x))

or without regex, maybe the best solution, with a ternary expression and startswith since we don't need regular expressions:

def sort_it(lst, ignore):
  lst.sort(key=lambda x: x[len(ignore):] if x.startswith(ignore) else x)

Thank you for your comprehensive answer. Updated my code in the question to use the ignore parameter. — Ghoul Fool, Nov 21 '17 at 19:40

Sort list of strings with lambda and ignore prefix

1 Answers1