I wanna delete the values of the list
lists = [12,15,15,15,3,15,1,6,4,7,888,56248]
while len(lists) is True:
lists.pop()
print(lists)
But i got this output :
[12, 15, 15, 15, 3, 15, 1, 6, 4, 7, 888, 56248]
2 Answers
The problem here is in your conditional
while len(lists) is True:
is checks for identity, not equality.
[1, 2, 3] == [1, 2, 3] # True
[1, 2, 3] is [1, 2, 3] # False, they are two distinct (but equivalent) lists.
However even equality would be incorrect here, since
42 == True # False
2 == True # False
any_nonzero == True # False
# notably 1 == True
# and 0 == False
# but still (1 is True) == False!
You can coerce an integer into a boolean
bool(42) == True # True
bool(2) == True # True
bool(any_nonzero) == True # True
But it's usually better to just leave the coercion to Python
while lists:
lists.pop()
# or more simply:
# lists = []
4 Comments
Young Mind
yes you are right i forgot . i should put it in bool() .thank you
AJ123
Because
0 means False in python, and anything else is true. This means that the code will keep deleting the values until the statement is false (when there are no values in the list).
Adam Smith
@StefanPochmann either will work, but you're right that
while lists is better. I'll editAdam Smith
@AJ123 not anything else. The empty string (
"") and relevantly empty collections ([], {}, set(), etc) are also Falsey. while lists is definitely the idiomatic approach.If by "delete" you mean removing the element and reducing the list, you can iterate through a copy of lists with a for loop:
for n in lists[:]:
lists.remove(n)
If, instead, you want to have a list of None (or 0) values as long as lists, you can do it like this:
newlists = [0] * len(lists)
As was suggested in the previous comment.
len(lists)is an integer. An integer is never the same object as the Boolean singletonTrue. Perhaps you meant justwhile lists?while lists: lists.pop(). The better solution would belists = []. If it has to be the same list (with the same id) uselists[:] = [].lists.clear()