python - 基于1D数组中的值更改2D numpy数组中的某些值而无需for循环

我的问题对我来说太基本了，我有点不好意思自己没有解决。尽管咨询了this、this和that，我还是不知道如何在不使用for循环的情况下根据1D numpy数组中的值更改2D numpy数组中的某些值。
具有预期结果的一个例子是：

import numpy as np

# sample data:
a = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
b = np.array([2, 0, 2])
c = np.array([[10, 20, 30], [40, 50, 60], [70, 80, 90]])

# for-loop solution:
for i in range(len(a)):
    a[i][b[i]] = 0.9 * c[i][b[i]]

# desired result:
print(a)
# [[ 1  2 27]
#  [36  5  6]
#  [ 7  8 81]]

编辑1
在对Rafael的答案进行修改之后，我现在得到了期望的结果，而没有for循环。然而，令我惊讶的是，索引解决方案比for循环慢。

import numpy as np
import time

# set seed for reproducibility:
np.random.seed(1)

x = np.random.randint(10, size=(10, 10))
y = np.random.randint(10, size=10)
z = np.random.randint(10, size=(10, 10))

# for-loop solution:
start1 = time.clock()
for i in range(len(x)):
     x[i][y[i]] = 2 * z[i][y[i]]
end1 = time.clock()
print("time loop: " + str(end1 - start1))
# time loop: 0.00045699999999726515
print("result for-loop:")
print(x)
# result for-loop:
# [[ 5  8  9  5  0  0  1  7  6  4]
#  [12  4  5  2  4  2  4  7  7  9]
#  [ 1  7  2  6  9  9  7  6  9  1]
#  [ 2  1  8  8  3  9  8  7  3  6]
#  [ 5  1  9  3  4  8  1 16  0  3]
#  [ 9 14  0  4  9  2  7  7  9  8]
#  [ 6  9  3  7  7  4  5  0  3  6]
#  [ 8  0  2  7  7  9  7  3  0 16]
#  [ 7  7  1  1  3  0  8  6 16  5]
#  [ 6  2  5  7 14  4  4  7  7  4]]

# set seed for reproducibility:
np.random.seed(1)

x = np.random.randint(10, size=(10, 10))
y = np.random.randint(10, size=10)
z = np.random.randint(10, size=(10, 10))

# indexing solution:
start2 = time.clock()
r = x.shape[0]
x[range(r), y] = z[range(r), y] * 2
end2 = time.clock()
print("time indexing: " + str(end2 - start2))
# time indexing: 0.0005479999999948859
print("result indexing:")
print(x)
# result indexing:
# [[ 5  8  9  5  0  0  1  7  6  4]
#  [12  4  5  2  4  2  4  7  7  9]
#  [ 1  7  2  6  9  9  7  6  9  1]
#  [ 2  1  8  8  3  9  8  7  3  6]
#  [ 5  1  9  3  4  8  1 16  0  3]
#  [ 9 14  0  4  9  2  7  7  9  8]
#  [ 6  9  3  7  7  4  5  0  3  6]
#  [ 8  0  2  7  7  9  7  3  0 16]
#  [ 7  7  1  1  3  0  8  6 16  5]
#  [ 6  2  5  7 14  4  4  7  7  4]]

这是什么原因？还有，我怎样才能实现加速呢？

最佳答案

IIUC公司

r = np.arange(a.shape[0]) # same as range(len(a)) here, but faster.
a[r, b] = c[r, b] * 0.9

array([[ 1,  2, 27],
       [36,  5,  6],
       [ 7,  8, 81]])

关于python - 基于1D数组中的值更改2D numpy数组中的某些值而无需for循环，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/51736293/