在两个不同形状的数组上广播 (Numpy-Python)
Broadcasting over two arrays with different shapes (Numpy-Python)
假设我有以下数组:
first_array = array([[1, 8, 3, 9, 2],
[2, 6, 4, 1, 9],
[4, 2, 12, 8, 16],
[5, 3, 7, 18, 21],
[6, 20, 4, 8, 24]])
所以数组的形状为 (5, 5)
现在我有了第二个数组,它是第一个数组的一部分:
second_array = array([[1, 8, 3, 9, 2],
[2, 6, 4, 1, 9]])
形状为 (2, 5)
的数组。
现在我想用第二个数组的向量减去第一个数组的每个向量(不包括-在第一个数组中-我用来减去的第二个数组的向量),逐个元素。我想对第二个数组的每个向量都这样。
所以我想输出:
subtracted_array = array([[[1, -2, 1, -8, 7],
[3, -6, 9, -1, 14],
[4, -5, 4, 9, 19],
[5, 12, 1, -1, 22]],
[[-1, 2, -1, 8, -7],
[2, -4, 8, 7, 7],
[3, -3, 3, 17, 12],
[4, 14, 0, 7, 15]]])
所以这是一个形状为 (2, 4, 5)
的数组
我们如何使用广播来做到这一点?
此代码是否按照您的意思执行?
欢迎您在您的测试用例上对其进行测试,如果您需要更多帮助,请更新我。
import numpy as np
arr = np.arange(50).reshape(10, 5)
arr_slice = arr[:2, :]
# "outer" tensor subtraction
arr_sub = arr_slice[:, None, :] - arr[None, :, :]
# create an index map of the vector combinations
idx0, idx1 = np.mgrid[:2, :10]
idx0, idx1 = idx0.flatten(), idx1.flatten()
# find the irrelevent combinations and remove them
remove_same = (1 - (idx0 == idx1)).astype(bool)
idx0, idx1 = idx0[remove_same], idx1[remove_same]
arr_sub = arr_sub[idx0, idx1, :].reshape(2, 9, 5)
编辑: 这是一个更有效的方法:
import numpy as np
arr = np.arange(50).reshape(10, 5)
arr_slice = arr[:2, :]
# create an index map of the vector combinations
idx0, idx1 = np.mgrid[:2, :10]
idx0, idx1 = idx0.flatten(), idx1.flatten()
# find the irrelevent combinations and remove them
remove_same = (1 - (idx0 == idx1)).astype(bool)
idx0, idx1 = idx0[remove_same], idx1[remove_same]
arr_sub = arr_slice[idx0, :] - arr[idx1, :]
arr_sub = arr_sub.reshape(2, 9, 5)
输出:
arr =
[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]
[15 16 17 18 19]
[20 21 22 23 24]
[25 26 27 28 29]
[30 31 32 33 34]
[35 36 37 38 39]
[40 41 42 43 44]
[45 46 47 48 49]]
arr_slice =
[[0 1 2 3 4]
[5 6 7 8 9]]
arr_sub =
[[[ -5 -5 -5 -5 -5]
[-10 -10 -10 -10 -10]
[-15 -15 -15 -15 -15]
[-20 -20 -20 -20 -20]
[-25 -25 -25 -25 -25]
[-30 -30 -30 -30 -30]
[-35 -35 -35 -35 -35]
[-40 -40 -40 -40 -40]
[-45 -45 -45 -45 -45]]
[[ 5 5 5 5 5]
[ -5 -5 -5 -5 -5]
[-10 -10 -10 -10 -10]
[-15 -15 -15 -15 -15]
[-20 -20 -20 -20 -20]
[-25 -25 -25 -25 -25]
[-30 -30 -30 -30 -30]
[-35 -35 -35 -35 -35]
[-40 -40 -40 -40 -40]]]
编辑 2: 我以错误的顺序减去数组(整个数组从切片而不是相反)。所以这是一个修复,这次我使用你的例子:
import numpy as np
first_array = np.array(
[[1, 8, 3, 9, 2],
[2, 6, 4, 1, 9],
[4, 2, 12, 8, 16],
[5, 3, 7, 18, 21],
[6, 20, 4, 8, 24]]
)
arr_slice = slice(None, 2)
second_array = first_array[arr_slice]
expected_subtracted_array = np.array(
[[[[1, -2, 1, -8, 7],
[3, -6, 9, -1, 14],
[4, -5, 4, 9, 19],
[5, 12, 1, -1, 22]],
[[-1, 2, -1, 8, -7],
[2, -4, 8, 7, 7],
[3, -3, 3, 17, 12],
[4, 14, 0, 7, 15]]]]
)
# create an index map of the vector combinations
idx0, idx1 = np.mgrid[:second_array.shape[0], :first_array.shape[0]]
idx0, idx1 = idx0.flatten(), idx1.flatten()
# find the irrelevent combinations and remove them
remove_same = (1 - (idx0 == idx1)).astype(bool)
idx0, idx1 = idx0[remove_same], idx1[remove_same]
arr_sub = first_array[idx1, :] - second_array[idx0, :]
arr_sub = arr_sub.reshape(second_array.shape[0], first_array.shape[0]-1, first_array.shape[1])
(arr_sub == expected_subtracted_array).all()
输出:
True
假设我有以下数组:
first_array = array([[1, 8, 3, 9, 2],
[2, 6, 4, 1, 9],
[4, 2, 12, 8, 16],
[5, 3, 7, 18, 21],
[6, 20, 4, 8, 24]])
所以数组的形状为 (5, 5)
现在我有了第二个数组,它是第一个数组的一部分:
second_array = array([[1, 8, 3, 9, 2],
[2, 6, 4, 1, 9]])
形状为 (2, 5)
的数组。
现在我想用第二个数组的向量减去第一个数组的每个向量(不包括-在第一个数组中-我用来减去的第二个数组的向量),逐个元素。我想对第二个数组的每个向量都这样。
所以我想输出:
subtracted_array = array([[[1, -2, 1, -8, 7],
[3, -6, 9, -1, 14],
[4, -5, 4, 9, 19],
[5, 12, 1, -1, 22]],
[[-1, 2, -1, 8, -7],
[2, -4, 8, 7, 7],
[3, -3, 3, 17, 12],
[4, 14, 0, 7, 15]]])
所以这是一个形状为 (2, 4, 5)
我们如何使用广播来做到这一点?
此代码是否按照您的意思执行?
欢迎您在您的测试用例上对其进行测试,如果您需要更多帮助,请更新我。
import numpy as np
arr = np.arange(50).reshape(10, 5)
arr_slice = arr[:2, :]
# "outer" tensor subtraction
arr_sub = arr_slice[:, None, :] - arr[None, :, :]
# create an index map of the vector combinations
idx0, idx1 = np.mgrid[:2, :10]
idx0, idx1 = idx0.flatten(), idx1.flatten()
# find the irrelevent combinations and remove them
remove_same = (1 - (idx0 == idx1)).astype(bool)
idx0, idx1 = idx0[remove_same], idx1[remove_same]
arr_sub = arr_sub[idx0, idx1, :].reshape(2, 9, 5)
编辑: 这是一个更有效的方法:
import numpy as np
arr = np.arange(50).reshape(10, 5)
arr_slice = arr[:2, :]
# create an index map of the vector combinations
idx0, idx1 = np.mgrid[:2, :10]
idx0, idx1 = idx0.flatten(), idx1.flatten()
# find the irrelevent combinations and remove them
remove_same = (1 - (idx0 == idx1)).astype(bool)
idx0, idx1 = idx0[remove_same], idx1[remove_same]
arr_sub = arr_slice[idx0, :] - arr[idx1, :]
arr_sub = arr_sub.reshape(2, 9, 5)
输出:
arr =
[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]
[15 16 17 18 19]
[20 21 22 23 24]
[25 26 27 28 29]
[30 31 32 33 34]
[35 36 37 38 39]
[40 41 42 43 44]
[45 46 47 48 49]]
arr_slice =
[[0 1 2 3 4]
[5 6 7 8 9]]
arr_sub =
[[[ -5 -5 -5 -5 -5]
[-10 -10 -10 -10 -10]
[-15 -15 -15 -15 -15]
[-20 -20 -20 -20 -20]
[-25 -25 -25 -25 -25]
[-30 -30 -30 -30 -30]
[-35 -35 -35 -35 -35]
[-40 -40 -40 -40 -40]
[-45 -45 -45 -45 -45]]
[[ 5 5 5 5 5]
[ -5 -5 -5 -5 -5]
[-10 -10 -10 -10 -10]
[-15 -15 -15 -15 -15]
[-20 -20 -20 -20 -20]
[-25 -25 -25 -25 -25]
[-30 -30 -30 -30 -30]
[-35 -35 -35 -35 -35]
[-40 -40 -40 -40 -40]]]
编辑 2: 我以错误的顺序减去数组(整个数组从切片而不是相反)。所以这是一个修复,这次我使用你的例子:
import numpy as np
first_array = np.array(
[[1, 8, 3, 9, 2],
[2, 6, 4, 1, 9],
[4, 2, 12, 8, 16],
[5, 3, 7, 18, 21],
[6, 20, 4, 8, 24]]
)
arr_slice = slice(None, 2)
second_array = first_array[arr_slice]
expected_subtracted_array = np.array(
[[[[1, -2, 1, -8, 7],
[3, -6, 9, -1, 14],
[4, -5, 4, 9, 19],
[5, 12, 1, -1, 22]],
[[-1, 2, -1, 8, -7],
[2, -4, 8, 7, 7],
[3, -3, 3, 17, 12],
[4, 14, 0, 7, 15]]]]
)
# create an index map of the vector combinations
idx0, idx1 = np.mgrid[:second_array.shape[0], :first_array.shape[0]]
idx0, idx1 = idx0.flatten(), idx1.flatten()
# find the irrelevent combinations and remove them
remove_same = (1 - (idx0 == idx1)).astype(bool)
idx0, idx1 = idx0[remove_same], idx1[remove_same]
arr_sub = first_array[idx1, :] - second_array[idx0, :]
arr_sub = arr_sub.reshape(second_array.shape[0], first_array.shape[0]-1, first_array.shape[1])
(arr_sub == expected_subtracted_array).all()
输出:
True