Count all that satisfy a condition in Python

Question

Is there a way to add all that satisfy a condition in a shorthand nested loop? My following attempt was unsuccessful:

count += 1 if n == fresh for n in buckets['actual'][e] else 0

score 5 · Accepted Answer · edited May 23 '17 at 11:50

5

Use sum with a generator expression:

sum(n == fresh for n in buckets['actual'][e])

as True == 1 and False == 0, so else is not required.

edited May 23 '17 at 11:50

Community

answered Oct 14 '13 at 19:47

Ashwini Chaudhary

If you feel uncomfortable implicitly using the fact that `True == 1` and `False == 0`, you can make this fact explicit by doing `int(n == fresh)`. This is only useful to make the code more obvious... the language doesn't care. – SethMMorton Oct 14 '13 at 19:52
Sure, but there is also the line [Explicit is better than implicit](http://www.python.org/dev/peps/pep-0020/). I personally don't care, but it was in case the OP has a gut reaction against implicitly using `bool`s as `int`s. To be clear, I'm not saying you are wrong... my comment was an addendum intended for the OP in case they wanted to make a more self-documenting code. – SethMMorton Oct 14 '13 at 19:59

score 1 · Answer 2 · answered Oct 14 '13 at 19:47

1

Using sum() function:

sum(1 if n == fresh else 0  for n in buckets['actual'][e])

or:

sum(1 for n in buckets['actual'][e] if n == fresh)

answered Oct 14 '13 at 19:47

Rohit Jain

2 Answers2