Python 有效地计算numpy数组中的零元素？
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Efficiently count zero elements in numpy array?
提问by Gabriel
I need to count the number of zero elements in numpyarrays. I'm aware of the numpy.count_nonzerofunction, but there appears to be no analog for counting zero elements.
我需要计算numpy数组中零元素的数量。我知道numpy.count_nonzero函数，但似乎没有用于计算零元素的模拟。
My arrays are not very large (typically less than 1E5 elements) but the operation is performed several millions of times.
我的数组不是很大（通常少于 1E5 个元素），但该操作执行了数百万次。
Of course I could use len(arr) - np.count_nonzero(arr), but I wonder if there's a more efficient way to do it.
当然我可以使用len(arr) - np.count_nonzero(arr)，但我想知道是否有更有效的方法来做到这一点。
Here's a MWE of how I do it currently:
这是我目前如何做的 MWE：
import numpy as np
import timeit
arrs = []
for _ in range(1000):
arrs.append(np.random.randint(-5, 5, 10000))
def func1():
for arr in arrs:
zero_els = len(arr) - np.count_nonzero(arr)
print(timeit.timeit(func1, number=10))
回答by kmario23
A 2xfaster approach would be to just use np.count_nonzero()but with the conditionas needed.
一种快2倍的方法是仅使用np.count_nonzero()但根据需要使用条件。
In [3]: arr
Out[3]:
array([[1, 2, 0, 3],
[3, 9, 0, 4]])
In [4]: np.count_nonzero(arr==0)
Out[4]: 2
In [5]:def func_cnt():
for arr in arrs:
zero_els = np.count_nonzero(arr==0)
# here, it counts the frequency of zeroes actually
You can also use np.where()but it's slower than np.count_nonzero()
您也可以使用，np.where()但它比np.count_nonzero()
In [6]: np.where( arr == 0)
Out[6]: (array([0, 1]), array([2, 2]))
In [7]: len(np.where( arr == 0))
Out[7]: 2
Efficiency:(in descending order)
效率：（降序排列）
In [8]: %timeit func_cnt()
10 loops, best of 3: 29.2 ms per loop
In [9]: %timeit func1()
10 loops, best of 3: 46.5 ms per loop
In [10]: %timeit func_where()
10 loops, best of 3: 61.2 ms per loop
