[英]Index multiple dimensions of a multi-dimensional array with another - NumPy/ Python
可以说我有以下形式的张量:
import numpy as np
a = np.array([ [[1,2],
[3,4]],
[[5,6],
[7,3]]
])
# a.shape : (2,2,2) is a tensor containing 2x2 matrices
indices = np.argmax(a, axis=2)
#print indices
for mat in a:
max_i = np.argmax(mat,axis=1)
# Not really working I would like to
# change 4 in the first matrix to -1
# and 3 in the last to -1
mat[max_i] = -1
print a
现在我想做的是使用索引作为a上的掩码,用-1替换每个max元素。 有这样一种笨拙的方式吗? 到目前为止,我所知道的是使用for循环。
这是在3D
使用linear indexing
的一种方法 -
m,n,r = a.shape
offset = n*r*np.arange(m)[:,None] + r*np.arange(n)
np.put(a,indices + offset,-1)
样品运行 -
In [92]: a
Out[92]:
array([[[28, 59, 26, 70],
[57, 28, 71, 49],
[33, 6, 10, 90]],
[[24, 16, 83, 67],
[96, 16, 72, 56],
[74, 4, 71, 81]]])
In [93]: indices = np.argmax(a, axis=2)
In [94]: m,n,r = a.shape
...: offset = n*r*np.arange(m)[:,None] + r*np.arange(n)
...: np.put(a,indices + offset,-1)
...:
In [95]: a
Out[95]:
array([[[28, 59, 26, -1],
[57, 28, -1, 49],
[33, 6, 10, -1]],
[[24, 16, -1, 67],
[-1, 16, 72, 56],
[74, 4, 71, -1]]])
这是另一种使用linear indexing
的方法,但在2D
-
m,n,r = a.shape
a.reshape(-1,r)[np.arange(m*n),indices.ravel()] = -1
运行时测试并验证输出 -
In [156]: def vectorized_app1(a,indices): # 3D linear indexing
...: m,n,r = a.shape
...: offset = n*r*np.arange(m)[:,None] + r*np.arange(n)
...: np.put(a,indices + offset,-1)
...:
...: def vectorized_app2(a,indices): # 2D linear indexing
...: m,n,r = a.shape
...: a.reshape(-1,r)[np.arange(m*n),indices.ravel()] = -1
...:
In [157]: # Generate random 3D array and the corresponding indices array
...: a = np.random.randint(0,99,(100,100,100))
...: indices = np.argmax(a, axis=2)
...:
...: # Make copies for feeding into functions
...: ac1 = a.copy()
...: ac2 = a.copy()
...:
In [158]: vectorized_app1(ac1,indices)
In [159]: vectorized_app2(ac2,indices)
In [160]: np.allclose(ac1,ac2)
Out[160]: True
In [161]: # Make copies for feeding into functions
...: ac1 = a.copy()
...: ac2 = a.copy()
...:
In [162]: %timeit vectorized_app1(ac1,indices)
1000 loops, best of 3: 311 µs per loop
In [163]: %timeit vectorized_app2(ac2,indices)
10000 loops, best of 3: 145 µs per loop
您可以使用indices
来索引到的最后一个维度a
前提是,你还可以指定数组索引到前两个维度,以及:
import numpy as np
a = np.array([[[1, 2],
[3, 4]],
[[5, 6],
[7, 3]]])
indices = np.argmax(a, axis=2)
print(repr(a[range(a.shape[0]), range(a.shape[1]), indices]))
# array([[2, 3],
# [2, 7]])
a[range(a.shape[0]), range(a.shape[1]), indices] = -1
print(repr(a))
# array([[[ 1, -1],
# [ 3, 4]],
# [[ 5, 6],
# [-1, -1]]])
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