Car*_*osH 4 python numpy vectorization argmax
我有一个N维数组(名为A)。对于A的第一轴的每一行,我想获得沿A的其他轴的最大值的坐标。然后,我将返回一个二维数组,其中包含第一轴的每一行的最大值的坐标的A。
我已经使用循环解决了我的问题,但是我想知道是否有更有效的方法可以做到这一点。我当前的解决方案(对于示例数组A)如下:
import numpy as np
A=np.reshape(np.concatenate((np.arange(0,12),np.arange(0,-4,-1))),(4,2,2))
maxpos=np.empty(shape=(4,2))
for n in range(0, 4):
maxpos[n,:]=np.unravel_index(np.argmax(A[n,:,:]), A[n,:,:].shape)
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在这里,我们将有:
A:
[[[ 0 1]
[ 2 3]]
[[ 4 5]
[ 6 7]]
[[ 8 9]
[10 11]]
[[ 0 -1]
[-2 -3]]]
maxpos:
[[ 1. 1.]
[ 1. 1.]
[ 1. 1.]
[ 0. 0.]]
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如果有多个最大化器,我不介意选择哪个。
我尝试使用np.apply_over_axes,但是我没有设法使它返回我想要的结果。
你可以做这样的事情-
# Reshape input array to a 2D array with rows being kept as with original array.
# Then, get idnices of max values along the columns.
max_idx = A.reshape(A.shape[0],-1).argmax(1)
# Get unravel indices corresponding to original shape of A
maxpos_vect = np.column_stack(np.unravel_index(max_idx, A[0,:,:].shape))
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样品运行-
In [214]: # Input array
...: A = np.random.rand(5,4,3,7,8)
In [215]: # Setup output array and use original loopy code
...: maxpos=np.empty(shape=(5,4)) # 4 because ndims in A is 5
...: for n in range(0, 5):
...: maxpos[n,:]=np.unravel_index(np.argmax(A[n,:,:,:,:]), A[n,:,:,:,:].shape)
...:
In [216]: # Proposed approach
...: max_idx = A.reshape(A.shape[0],-1).argmax(1)
...: maxpos_vect = np.column_stack(np.unravel_index(max_idx, A[0,:,:].shape))
...:
In [219]: # Verify results
...: np.array_equal(maxpos.astype(int),maxpos_vect)
Out[219]: True
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