索引索引数组时设置Numpy数组的值 [英] Setting values of Numpy array when indexing an indexed array
问题描述
我正在尝试索引某些矩阵y，然后使用一些布尔语句对结果重新索引，并将y中的相应元素设置为0.下面显示了我用于测试此索引方案的伪代码.
I'm trying to index some matrix, y, and then reindex that result with some boolean statement and set the corresponding elements in y to 0. The dummy code I'm using to test this indexing scheme is shown below.
x=np.zeros([5,4])+0.1;
y=x;
print(x)
m=np.array([0,2,3]);
y[0:4,m][y[0:4,m]<0.5]=0;
print(y)
我不确定为什么它不起作用.我想要的输出:
I'm not sure why it does not work. The output I want:
[[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]]
[[ 0. 0.1 0. 0. ]
[ 0. 0.1 0. 0. ]
[ 0. 0.1 0. 0. ]
[ 0. 0.1 0. 0. ]
[ 0.1 0.1 0.1 0.1]]
但是我实际上得到的是:
But what I actually get:
[[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]]
[[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]
[ 0.1 0.1 0.1 0.1]]
我确定我缺少一些幕后细节，这些细节解释了为什么这种方法不起作用.有趣的是，如果将m替换为:，则分配有效.由于某些原因，选择列的子集不会让我分配零.
I'm sure I'm missing some under-the-hood details that explains why this does not work. Interestingly, if you replace m with :, then the assignment works. For some reason, selecting a subset of the columns does not let me assign the zeros.
如果有人可以解释正在发生的事情并帮助我找到另一种解决方案(希望该解决方案不涉及生成临时的numpy数组，因为我的实际y将会非常大)，那么我将不胜感激！谢谢！
If someone could explain what's going on and help me find an alternative solution (hopefully one that does not involve generating a temporary numpy array since my actual y will be really huge), I would really appreciate it! Thank you!
y[0:4,:][y[0:4,:]<0.5]=0;
y[0:4,0:3][y[0:4,0:3]<0.5]=0;
等
所有工作都按预期进行.看来问题是当您使用某种类型的列表编制索引时.
all work as expected. It seems the issue is when you index with a list of some kind.
推荐答案
创建一个数组(由于值不同，这是我的最爱之一):
Make an array (this is one of my favorites because the values differ):
In [845]: x=np.arange(12).reshape(3,4)
In [846]: x
Out[846]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
In [847]: m=np.array([0,2,3])
In [848]: x[:,m]
Out[848]:
array([[ 0, 2, 3],
[ 4, 6, 7],
[ 8, 10, 11]])
In [849]: x[:,m][:2,:]=0
In [850]: x
Out[850]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
没有变化.但是，如果我一步一步进行索引编制，它就会改变.
No change. But if I do the indexing in one step, it changes.
In [851]: x[:2,m]=0
In [852]: x
Out[852]:
array([[ 0, 1, 0, 0],
[ 0, 5, 0, 0],
[ 8, 9, 10, 11]])
如果我颠倒顺序，它也可以工作:
it also works if I reverse the order:
In [853]: x[:2,:][:,m]=10
In [854]: x
Out[854]:
array([[10, 1, 10, 10],
[10, 5, 10, 10],
[ 8, 9, 10, 11]])
x[i,j]作为x.__getitem__((i,j))执行. x[i,j]=v作为x.__setitem__((i,j),v).
x[i,j][k,l]=v是x.__getitem__((i,j)).__setitem__((k,l),v).
set适用于get产生的值.如果get返回视图，则更改会影响x.但是，如果生成了副本，则更改不会影响x.
The set applies to the value produced by the get. If the get returns a view, then the change affects x. But if it produces a copy, the change does not affect x.
对于数组m，y[0:4,m]会生成一个副本(我需要证明这一点吗?). y[0:4,:]产生一个视图.
With array m, y[0:4,m] produces a copy (do I need to demonstrate that?). y[0:4,:] produces a view.
因此，简而言之，如果第一个索引生成了一个视图，则第二个索引分配有效.但是，如果产生副本，则第二个无效.
So in short, if the first indexing produces a view the second indexed assignment works. But if produces a copy, the second has no effect.
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