计算所有满足条件的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

Answer 1

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.

---

Related reads: Is it Pythonic to use bools as ints? , Is False == 0 and True == 1 in Python an implementation detail or is it guaranteed by the language?

Answer 2

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)