Finding substrings that contain each other in two different lists in Pytyhon-CodePudding

Really basic question, but I couldn't find the way to properly achieve this.

I have two lists:

vowels = ['a', 'e', 'i', 'o', 'u']

and

usernames = ['example1', 'zzzzz23', 'eeeee43', 'llllll5', 'pppapp1', 'wwsd0']

I want, to extract from usernames all elements that don't have vowels in them.

I tried:

usernames_without_vowels = []
for username in usernames:
  if str(vowels) not in username:
    usernames_without_vowels.append(username)
  else: pass
print(usernames_without_vowels)

OUTPUT 1:

>> ['example1', 'zzzzz23', 'eeeee43', 'llllll5', 'pppapp1', 'wwsd0']

As you can see, it printed the whole usernames list, as it seems to not look for substrings too.

I then tried zipping both lists as it follows:

usernames_without_vowels = []
for username,vowel in zip(usernames,vowels):
  if str(vowels) not in str(username):
    usernames_without_vowels.append(username)
  else: pass
print(usernames_without_vowels)

but, then again: OUTPUT 2:

>> ['example1', 'zzzzz23', 'eeeee43', 'llllll5', 'pppapp1']

it printed the whole usernames list EXCEPT the last value: wwsd0.

Also tried:

usernames_without_vowels = [username for username in usernames if str(vowels) not in username]
print(usernames_without_vowels)

and OUTPUT 3: >> ['example1', 'zzzzz23', 'eeeee43', 'llllll5', 'pppapp1', 'wwsd0']

I want to get all usernames in which EACH STRING of vowels is NOT present, but can't find a way.

EXPECTED OUTPUT:

>> ['zzzzz23', 'llllll5', 'wwsd0']

SOLUTION

Following @Helios solution, it was accomplished by:

username_without_vowels = [u for u in usernames if not any([v in u for v in vowels])]

Simple, and (very) effective.

Thank you for all the help!

CodePudding user response：

There are many ways to skin this cat, one way is to use set intersection and filter out all results that have results of such an intersection.

vowels = set(vowels)

[user for user in usernames if len(vowels.intersection(user)) == 0]

# OR

[user for user in usernames if not vowels.intersection(user)]

Both yield

> ['zzzzz23', 'llllll5', 'wwsd0']

CodePudding user response：

[u for u in usernames if not any(v in u for v in vowels)]

EDIT:

Updated to include @cglacet comment