How to remove specific element from sets inside a list using list comprehension

Question

I think this may be related to set being mutable .

Basically, I can remove an element from a set using set.discard(element) . However, set.discard(element) itself returns None . But I'd like to get a copy of the updated set. For example, if I have a list of sets, how can I get an updated copy conveniently using list comprehension operations?

Sample code:

test = [{'', 'a'}, {'b', ''}]
print [x.discard('') for x in test]
print test

will return

[None, None]
[set(['a']), set(['b'])]

Answer 1

Whenever you feel constrained by a method that only works in-place, you can use the behavior of or / and to achieve the semantics that you want.

[x.discard('') or x for x in test]

This technique is occasionally useful for achieving things in a lambda (or other situations where you are restricted to a single expression) that are otherwise impossible. Whether it's the most "readable" or "pythonic" is debatable :-)

Answer 2

You can use set difference operator , like this

test, empty = [{'', 'a'}, {'b', ''}], {''}
print [x - empty for x in test]
# [set(['a']), set(['b'])]

Answer 3

>>> s = set( ['a' , 'b', 'c' , 'd' ] )
>>> print(s)
set(['a', 'c', 'b', 'd'])
>>>
>>> s -= {'c'}
>>> print(s)
set(['a', 'b', 'd'])
>>>
>>> s -= {'a'}
>>> print(s)
set(['b', 'd'])

Answer 4

This?(duplicate of @thefourtheye answer)

set subtraction operation returns set data.

test = [{'', 'a'}, {'b', ''}]
print [x - {''} for x in test]
print test

Output:

[set(['a']), set(['b'])]
[set(['a', '']), set(['', 'b'])]