如何一个讨厌的单线?
一,数据;数组的形状与您的形状相同,但我使用整数使示例更易于阅读。
In [81]: A
Out[81]:
array([[ 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]])
In [82]: B
Out[82]:
array([[ 0, 100, 200, 300, 400],
[ 500, 600, 700, 800, 900],
[1000, 1100, 1200, 1300, 1400],
[1500, 1600, 1700, 1800, 1900],
[2000, 2100, 2200, 2300, 2400]])
In [83]: C
Out[83]: array([1, 3, 2, 4, 0])
而这里的肮脏的一行:
In [84]: np.insert(A.ravel(), np.ravel_multi_index((range(A.shape[0]), C), A.shape) + 1, B[range(B.shape[0]), C]).reshape(A.shape[0], A.shape[1]+1)
Out[84]:
array([[ 0, 1, 100, 2, 3, 4],
[ 5, 6, 7, 8, 800, 9],
[ 10, 11, 12, 1200, 13, 14],
[ 15, 16, 17, 18, 19, 1900],
[ 20, 2000, 21, 22, 23, 24]])
这里是破旧的版本:
A.ravel()
变平A
成1-d阵列,我会打电话给F
:
In [87]: F = A.ravel()
In [88]: F
Out[88]:
array([ 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])
(编辑:原来这第一步 - 展平A
- 没有必要。正如@hpaulj在他的回答中指出的那样,np.insert
会默认将该阵列展平。)
np.ravel_multi_index
用于将所需的二维位置转换为折叠数组中的索引。
In [89]: insert_indices = np.ravel_multi_index((range(A.shape[0]), C), A.shape) + 1
In [90]: insert_indices
Out[90]: array([ 2, 9, 13, 20, 21])
B[range(B.shape[0]), C]
拉期望值出B
:
In [91]: values = B[range(B.shape[0]), C]
In [92]: values
Out[92]: array([ 100, 800, 1200, 1900, 2000])
np.insert
进行实际的+ 1
末,因为你想后插入的元素在C
给出的指标是必要的插入并创建一个新阵列:
In [93]: np.insert(F, insert_indices, values)
Out[93]:
array([ 0, 1, 100, 2, 3, 4, 5, 6, 7, 8, 800,
9, 10, 11, 12, 1200, 13, 14, 15, 16, 17, 18,
19, 1900, 20, 2000, 21, 22, 23, 24])
现在只是重塑,要得到最终的结果是:
In [94]: np.insert(F, insert_indices, values).reshape(A.shape[0], A.shape[1]+1)
Out[94]:
array([[ 0, 1, 100, 2, 3, 4],
[ 5, 6, 7, 8, 800, 9],
[ 10, 11, 12, 1200, 13, 14],
[ 15, 16, 17, 18, 19, 1900],
[ 20, 2000, 21, 22, 23, 24]])
你是一个错误的例子结果(你不能将一个6元素数组赋给'A [i,:]',它只有5个元素的空间)。 – 2014-10-07 03:57:43
@WarrenWeckesser对不起,你是对的,我写下了示例矩阵,并没有多想太多。我正在修复:) – ProGM 2014-10-07 04:49:36
我调整了你的迭代工作。 – hpaulj 2014-10-07 06:56:27