以最pythonian方式编写ndarray的sub-ndarray.的Python 2 [英] Writting in sub-ndarray of a ndarray in the most pythonian way. Python 2

问题描述

我有一个像这样的ndarray:

I have a ndarray like this one:

number_of_rows = 3
number_of_columns = 3
a = np.arange(number_of_rows*number_of_columns).reshape(number_of_rows,number_of_columns)
a

array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])

但是我想要这样的东西:

But I want something like this:

array([[0, 100, 101],
       [3, 102, 103],
       [6, 7, 8]])

要做到这一点，我想避免一一做，我宁愿在数组或矩阵中做，因为以后我想扩展代码. 不，我已经更改了初始矩阵的子矩阵(在数学上，就本例ndarray而言).在示例中，考虑的列为[1,2]，行为[0,1].

To do that I want to avoid to do it one by one, I rather prefer to do it in arrays or matrices, because later I want to extend the code. Nothe I have change a submatrix of the initial matrix (in mathematical terms, in terms of this example ndarray). In the example the columns considered are [1,2] and the rows [0,1].

columns_to_keep = [1,2] 
rows_to_keep = [0,1]

我的第一次尝试是做

a[rows_to_keep,:][:,columns_to_keep] = np.asarray([[100,101],[102,103]])

但是这个不会修改首字母a，我没有任何错误，所以a =

array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])

所以我实现了一段代码来完成这项工作:

So I have implemented a piece of code that goes do the job:

b = [[100, 101],[102, 103]]

for i in range(len(rows_to_keep)):
    a[i,columns_to_keep] = b[i]

我以为前几行做的很好，我想知道如何以更快的速度进行切片.也可以通过以下方式实现:

Al thought the previous lines do the job I am wondering how to do it slicing and in a faster fashion. Also in a way that with:

columns_to_keep = [0,2] 
rows_to_keep = [0,2]

所需的输出是

array([[100, 1, 101],
       [3, 4, 5],
       [102, 7, 103]]).

非常感谢！

推荐答案

使用像[1,2]这样的列表进行索引称为高级索引.它本身会产生一个副本，而不是视图.您必须使用一个索引表达式，而不是两个索引表达式来分配或更改值.即a[[1,2],:]是副本，a[[1,2],:][:,[1,2]] += 100修改该副本，而不是原始的a.

Indexing with lists like [1,2] is called advanced indexing. By itself it produces a copy, not a view. You have to use one indexing expression, not two to assign or change values. That is a[[1,2],:] is a copy, a[[1,2],:][:,[1,2]] += 100 modifies that copy, not the original a.

In [68]: arr = np.arange(12).reshape(3,4)

使用切片索引；这是基本的索引编制:

Indexing with slices; this is basic indexing:

In [69]: arr[1:,2:]
Out[69]: 
array([[ 6,  7],
       [10, 11]])

In [70]: arr[1:,2:] += 100

In [71]: arr
Out[71]: 
array([[  0,   1,   2,   3],
       [  4,   5, 106, 107],
       [  8,   9, 110, 111]])

对列表进行相同的索引编制需要相互广播"的数组. ix_是生成这些的便捷方法:

Doing the same indexing with lists requires arrays that 'broadcast' against each other. ix_ is a handy way of generating these:

In [73]: arr[np.ix_([1,2],[2,3])]
Out[73]: 
array([[106, 107],
       [110, 111]])

In [74]: arr[np.ix_([1,2],[2,3])] -= 100

In [75]: arr
Out[75]: 
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])

这是ix_生成的-一个数组元组，一个数组的形状为(2,1)，另一个数组的形状为(1,2).他们一起索引了一个(2,2)块:

Here's what ix_ produces - a tuple of arrays, one is (2,1) in shape, the other (1,2). Together they index a (2,2) block:

In [76]: np.ix_([1,2],[2,3])
Out[76]: 
(array([[1],
        [2]]), array([[2, 3]]))

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