I have a list of integers and I need to count how many of them are > 0.
I'm currently doing it with a list comprehension that looks like this:
sum([1 for x in frequencies if x > 0])
It seems like a decent comprehension but I don't really like the "1"; it seems like a bit of a magic number. Is there a more Pythonish way to do this?
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
>>> isinstance(True,int) True
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.