CODEWITHSUNDEEP
python
Kashif
Kashif

Reputation: 3327

every element of list is True boolean

I know that

all(map(compare,new_subjects.values()))==True

would tell me if every element of the list is True. However, how do I tell whether every element except for one of them is True?

Upvotes: 21

Views: 28568

Answers (4)

Paddy3118
Paddy3118

Reputation: 4772

Assuming the compare function returns a boolean vaue, and knowing that True/False become 1/0 in an integer context you can do:

values = new_subjects.values()
sum(compare(v) for v in values) == len(values) -1

Upvotes: 0

dting
dting

Reputation: 39287

If you mean is actually True and not evaluates to True, you can just count them?

>>> L1 = [True]*5
>>> L1
[True, True, True, True, True]
>>> L2 = [True]*5 + [False]*2
>>> L2
[True, True, True, True, True, False, False]
>>> L1.count(False)
0
>>> L2.count(False)
2
>>>

checking for only a single False:

>>> def there_can_be_only_one(L):
...     return L.count(False) == 1
... 
>>> there_can_be_only_one(L1)
False
>>> there_can_be_only_one(L2)
False
>>> L3 = [ True, True, False ]
>>> there_can_be_only_one(L3)
True
>>>

edit: This actually answer your question better:

>>> def there_must_be_only_one(L):
...     return L.count(True) == len(L)-1
... 
>>> there_must_be_only_one(L3)
True
>>> there_must_be_only_one(L2)
False
>>> there_must_be_only_one(L1)
False

Upvotes: 7

Jochen Ritzel
Jochen Ritzel

Reputation: 107608

Count how many are not True:

values = (compare(val) for val in new_subjects.itervalues())
if sum(1 for x in values if not x) == 1: # just one
    ...

Upvotes: 4

Rafe Kettler
Rafe Kettler

Reputation: 76955

values = map(compare, new_subjects.values())
len([x for x in values if x]) == len(values) - 1

Basically, you filter the list for true values and compare the length of that list to the original to see if it's one less.

Upvotes: 12

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