我想使用 numpy 有效地将 2D 数组中的每个元素与 1D 数组相乘,以便返回 3D 数组。
基本上,代码应该执行以下操作:
import numpy as np
#create dummy data
arr1=np.arange(0,9).reshape((3,3))
arr2=np.arange(0,9)
#create output container
out = []
#loop over every increment in arr1
for col in arr1:
row = []
for i in col:
#perform calculation
row.append(i*arr2)
out.append(row)
#convert output to array
out = np.array(out)
没有形状 (3, 3, 9),因此等于
array([[[ 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8],
[ 0, 2, 4, 6, 8, 10, 12, 14, 16]],
[[ 0, 3, 6, 9, 12, 15, 18, 21, 24],
[ 0, 4, 8, 12, 16, 20, 24, 28, 32],
[ 0, 5, 10, 15, 20, 25, 30, 35, 40]],
[[ 0, 6, 12, 18, 24, 30, 36, 42, 48],
[ 0, 7, 14, 21, 28, 35, 42, 49, 56],
[ 0, 8, 16, 24, 32, 40, 48, 56, 64]]])
提前非常感谢您!
最佳答案
使用numpy.outer
:
np.outer(arr2,arr1).reshape(3,3,9)
获取:
array([[[ 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 1, 2, 3, 4, 5, 6, 7, 8],
[ 0, 2, 4, 6, 8, 10, 12, 14, 16]],
[[ 0, 3, 6, 9, 12, 15, 18, 21, 24],
[ 0, 4, 8, 12, 16, 20, 24, 28, 32],
[ 0, 5, 10, 15, 20, 25, 30, 35, 40]],
[[ 0, 6, 12, 18, 24, 30, 36, 42, 48],
[ 0, 7, 14, 21, 28, 35, 42, 49, 56],
[ 0, 8, 16, 24, 32, 40, 48, 56, 64]]])
关于python - 如何在 Numpy 中有效地将二维数组中的每个元素乘以一维数组?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59878166/