Numpy掩码计算满足条件的元素数量

Question

如何使用Numpy对此for循环进行向量化？

count=0
arr1 = np.random.rand(184,184)
for i in range(arr1.size[0]):
    for j in range(arr1.size[1]):
        if arr1[i,j] > 0.6:
            count += 1
print count

我试过了:

count=0
arr1 = np.random.rand(184,184)
mask = (arr1>0.6)
indices = np.where(mask)
print indices , len(indices) 

我期望len(指数)给予计数,但事实并非如此.请给我任何建议.

Answer 1

获取一个布尔掩码并计算"True":

(arr1 > 0.6).sum()

Answer 2

np.count_nonzero 应该比总和快一点：

np.count_nonzero(arr1 > 0.6)

实际上，速度是它的三倍

>>> from timeit import repeat
>>> kwds = dict(globals=globals(), number=10000)
>>> 
>>> arr1 = np.random.rand(184,184)
>>> 
>>> repeat('np.count_nonzero(arr1 > 0.6)', **kwds)
[0.15281831508036703, 0.1485864429268986, 0.1477385900216177]
>>> repeat('(arr1 > 0.6).sum()', **kwds)
[0.5286932559683919, 0.5260644309455529, 0.5260107989888638]