Count non zero elements in list python

>>> np.count_nonzero[np.eye[4]]
4
>>> a = np.array[[[0, 1, 7, 0],
...               [3, 0, 2, 19]]]
>>> np.count_nonzero[a]
5
>>> np.count_nonzero[a, axis=0]
array[[1, 1, 2, 1]]
>>> np.count_nonzero[a, axis=1]
array[[2, 3]]
>>> np.count_nonzero[a, axis=1, keepdims=True]
array[[[2],
       [3]]]

If you want to reduce the amount of memory, you can avoid generating a temporary list by using a generator:

sum[x > 0 for x in frequencies]

This works because bool is a subclass of int:

>>> isinstance[True,int]
True

and True's value is 1:

>>> True==1
True

However, as Joe Golton points out in the comments, this solution is not very fast. If you have enough memory to use a intermediate temporary list, then sth's solution may be faster. Here are some timings comparing various solutions:

>>> frequencies = [random.randint[0,2] for i in range[10**5]]

>>> %timeit len[[x for x in frequencies if x > 0]]   # sth
100 loops, best of 3: 3.93 ms per loop

>>> %timeit sum[[1 for x in frequencies if x > 0]]
100 loops, best of 3: 4.45 ms per loop

>>> %timeit sum[1 for x in frequencies if x > 0]
100 loops, best of 3: 6.17 ms per loop

>>> %timeit sum[x > 0 for x in frequencies]
100 loops, best of 3: 8.57 ms per loop

Beware that timeit results may vary depending on version of Python, OS, or hardware.

Of course, if you are doing math on a large list of numbers, you should probably be using NumPy:

>>> frequencies = np.random.randint[3, size=10**5]
>>> %timeit [frequencies > 0].sum[]
1000 loops, best of 3: 669 us per loop

The NumPy array requires less memory than the equivalent Python list, and the calculation can be performed much faster than any pure Python solution.

How do you count nonzero entries in NumPy array?

Is there a NumPy count?

How do you count values in an array in Python?

How do you count the number of zeros in an array in Python?