我有这个人。 numpy数组:
inds = []
for index,item in enumerate(arr):
if item == 0:
inds.append(index)
这就是我获取数组中所有0的索引的方法:
image
是否有一个numpy函数来做同样的事情?
答案 0 :(得分:2)
>>> arr = np.array([0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,0,0,0,0,0,0,0,0,0,1])
>>> (arr==0).nonzero()[0]
array([ 0, 1, 2, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25])
答案 1 :(得分:2)
您可以使用numpy.argwhere
作为评论中指出的@chappers:
arr = np.array([0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,0,0,0,0,0,0,0,0,0,1])
In [34]: np.argwhere(arr == 0).flatten()
Out[34]:
array([ 0, 1, 2, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25], dtype=int32)
或与astype(bool)
的倒数:
In [63]: (~arr.astype(bool)).nonzero()[0]
Out[63]:
array([ 0, 1, 2, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25], dtype=int32)