np.array(
[[0,13,0,2,0,0,0,0,0,0,0,0],
[0,0,15,0,9,0,0,0,0,0,0,0],
[0,0,0,0,0,18,0,0,0,0,0,0],
[0,0,0,0,27,0,20,0,0,0,0,0],
[0,0,0,0,0,20,0,10,0,0,0,0],
[0,0,0,0,0,0,0,0,8,0,0,0],
[0,0,0,0,0,0,0,14,0,14,0,0],
[0,0,0,0,0,0,0,0,12,0,25,0],
[0,0,0,0,0,0,0,0,0,0,0,11],
[0,0,0,0,0,0,0,0,0,0,15,0],
[0,0,0,0,0,0,0,0,0,0,0,7],
[0,0,0,0,0,0,0,0,0,0,0,0]])
我正在尝试找到如何像上面这样的 numpy 数组,然后在一个高性能操作中用我想要清零的元素索引屏蔽它
[0,1][1,4][4,7][7,8][8,11]
所以我剩下的就是
np.array(
[[0,0,0,2,0,0,0,0,0,0,0,0],
[0,0,15,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,18,0,0,0,0,0,0],
[0,0,0,0,27,0,20,0,0,0,0,0],
[0,0,0,0,0,20,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,8,0,0,0],
[0,0,0,0,0,0,0,14,0,14,0,0],
[0,0,0,0,0,0,0,0,0,0,25,0],
[0,0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,15,0],
[0,0,0,0,0,0,0,0,0,0,0,7],
[0,0,0,0,0,0,0,0,0,0,0,0]])
类似于 np.in1d 的功能,但对于 2d 数组?我可以遍历每个元素,但数组可以真正庞大,因此矢量单操作掩码是最好的。可能吗?如果这是一个愚蠢的问题,我相信我会被告知!
您可以通过以下方式直接访问这些索引
indexes = [[0,1], [1,4], [4,7], [7,8], [8,11]]
indexes =zip(*indexes)
>>[(0, 1, 4, 7, 8), (1, 4, 7, 8, 11)]
a[indexes[0], indexes[1]]=0
>>
[[ 0 0 0 2 0 0 0 0 0 0 0 0]
[ 0 0 15 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 18 0 0 0 0 0 0]
[ 0 0 0 0 27 0 20 0 0 0 0 0]
[ 0 0 0 0 0 20 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 8 0 0 0]
[ 0 0 0 0 0 0 0 14 0 14 0 0]
[ 0 0 0 0 0 0 0 0 0 0 25 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 15 0]
[ 0 0 0 0 0 0 0 0 0 0 0 7]
[ 0 0 0 0 0 0 0 0 0 0 0 0]]
我想你在寻找这个
a = np.array(
[[0,13,0,2,0,0,0,0,0,0,0,0],
[0,0,15,0,9,0,0,0,0,0,0,0],
[0,0,0,0,0,18,0,0,0,0,0,0],
[0,0,0,0,27,0,20,0,0,0,0,0],
[0,0,0,0,0,20,0,10,0,0,0,0],
[0,0,0,0,0,0,0,0,8,0,0,0],
[0,0,0,0,0,0,0,14,0,14,0,0],
[0,0,0,0,0,0,0,0,12,0,25,0],
[0,0,0,0,0,0,0,0,0,0,0,11],
[0,0,0,0,0,0,0,0,0,0,15,0],
[0,0,0,0,0,0,0,0,0,0,0,7],
[0,0,0,0,0,0,0,0,0,0,0,0]])
b = np.array([[0,1],[1,4],[4,7],[7,8],[8,11]])
# get x coordinates in an array
c1 = b[:,0]
# get y coordinates in an array
c2 = b[:,1]
a[c1[:,None],c2] = 0
a
array([[ 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 15, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 18, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 27, 0, 20, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 20, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 14, 0, 14, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 25, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])