在python/numpy
中,如何将多维数组子集在一起,其中另一个具有相同形状的数据沿某个轴(例如第一个)最大?
假设我有两个3 * 2 * 4阵列,a
和b
。我想在b
沿第一轴有最大值的位置获得一个包含a
值的2 * 4数组。
import numpy as np
np.random.seed(7)
a = np.random.rand(3*2*4).reshape((3,2,4))
b = np.random.rand(3*2*4).reshape((3,2,4))
print a
#[[[ 0.07630829 0.77991879 0.43840923 0.72346518]
# [ 0.97798951 0.53849587 0.50112046 0.07205113]]
#
# [[ 0.26843898 0.4998825 0.67923 0.80373904]
# [ 0.38094113 0.06593635 0.2881456 0.90959353]]
#
# [[ 0.21338535 0.45212396 0.93120602 0.02489923]
# [ 0.60054892 0.9501295 0.23030288 0.54848992]]]
print a.argmax(axis=0) #(I would like b at these locations along axis0)
#[[1 0 2 1]
# [0 2 0 1]]
我可以做这个非常难看的手动子集:
index = zip(a.argmax(axis=0).flatten(),
[0]*a.shape[2]+[1]*a.shape[2], # a.shape[2] = 4 here
range(a.shape[2])+range(a.shape[2]))
# [(1, 0, 0), (0, 0, 1), (2, 0, 2), (1, 0, 3),
# (0, 1, 0), (2, 1, 1), (0, 1, 2), (1, 1, 3)]
这将使我获得我想要的结果:
b_where_a_is_max_along0 = np.array([b[i] for i in index]).reshape(2,4)
# For verification:
print a.max(axis=0) == np.array([a[i] for i in index]).reshape(2,4)
#[[ True True True True]
# [ True True True True]]
实现这一目标的智能numpy
方法是什么?谢谢:))
答案 0 :(得分:1)
m,n = a.shape[1:]
b_out = b[a.argmax(0),np.arange(m)[:,None],np.arange(n)]
示例运行 -
设置输入数组a
并沿第一轴获取argmax
-
In [185]: a = np.random.randint(11,99,(3,2,4))
In [186]: idx = a.argmax(0)
In [187]: idx
Out[187]:
array([[0, 2, 1, 2],
[0, 1, 2, 0]])
In [188]: a
Out[188]:
array([[[49*, 58, 13, 69], # * are the max positions
[94*, 28, 55, 86*]],
[[34, 17, 57*, 50],
[48, 73*, 22, 80]],
[[19, 89*, 42, 71*],
[24, 12, 66*, 82]]])
使用b
-
In [193]: b
Out[193]:
array([[[18*, 72, 35, 51], # Mark * at the same positions in b
[74*, 57, 50, 84*]], # and verify
[[58, 92, 53*, 65],
[51, 95*, 43, 94]],
[[85, 23*, 13, 17*],
[17, 64, 35*, 91]]])
In [194]: b[a.argmax(0),np.arange(2)[:,None],np.arange(4)]
Out[194]:
array([[18, 23, 53, 17],
[74, 95, 35, 84]])
答案 1 :(得分:1)
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ogrid