[1, 1, 1, 2, 2, 3].count(True)
>>> 3
Why does this return 3 instead of 6, if bool(i) returns True for all values i not equal to 0?
Add a comment
|
2 Answers
In [33]: True == 1
Out[33]: True
In [34]: True == 2
Out[34]: False
In [35]: True == 3
Out[35]: False
True and False are instances of bool, and bool is a subclass of int.
From the docs:
[Booleans] represent the truth values False and True. The two objects representing the values False and True are the only Boolean objects. The Boolean type is a subtype of plain integers, and Boolean values behave like the values 0 and 1, respectively, in almost all contexts, the exception being that when converted to a string, the strings "False" or "True" are returned, respectively.
Comments
This is better done with a comprehension:
>>> sum(1 for i in [1,1,1,2,2,3,0] if i)
6
or
sum(bool(i) for i in [1,1,1,2,2,3,0])
Or count the opposite way, since there is no ambiguity about False is something other than 0
>>> li=[1, 1, 1, 2, 2, 3, 0]
>>> len(li) - li.count(False)
6
Better still:
sum(map(bool,li))
5 Comments
FreeAsInGimme
I should have clarified that it wasn't code I was actually intending to use--I was just experimenting with the interpreter, finding edge cases and such.
FreeAsInGimme
No, no, I appreciate the input. The more I can learn, the better.
PaulMcG
Don't do comparisons of '== True' or '!= True', just treat the values as booleans:
sum(1 for i in seq if i), or simpler, sum(bool(i) for i in seq), or simplest sum(map(bool,seq)).PaulMcG
Here's a surprise:
sum(not not i for i in seq) is twice as fast as sum(bool(i) for i in seq) (using timeit module).
the wolf
Function calls are very expensive in Python... While you are timing, what about
map(bool,li)?