Counting positive integer elements in a list with Python list comprehensions

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==1True

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])   # sth100 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.

A slightly more Pythonic way would be to use a generator instead:

sum(1 for x in frequencies if x > 0)

This avoids generating the whole list before calling sum().

You could use len() on the filtered list:

len([x for x in frequencies if x > 0])