[英]tuple as index in multidimensional array together with slicing
I have a 3-dimensional array A
, and I need to access A[:,1,1]
through the tuple x=[1,1]
.我有一个 3 维数组A
,我需要通过元组x=[1,1]
访问A[:,1,1]
] 。 Something like this:像这样的东西:
x = [1,1]
A[:,*x]
However, doing that I get a syntax error.但是,这样做会出现语法错误。 I would love to be able to access the elements of A[:,1,1]
using the variable x
, how can I do that?我希望能够使用变量x
访问A[:,1,1]
的元素,我该怎么做?
Thank you!谢谢!
Second question:第二个问题:
How to do the same but instead of slicing :
do it with a boolean array.如何做同样的事情,而不是切片:
使用 boolean 数组。 For example if t
is an array of booleans, obtain A[t, *x]
例如,如果t
是一个布尔数组,则获得A[t, *x]
You can do the following:您可以执行以下操作:
import numpy as np
A = np.arange(12).reshape((2, 3, 2))
print(A)
x = [1, 1]
print(A[(slice(None), *x)])
You can use slice(None)
instead of :
to build a tuple of slices.您可以使用slice(None)
而不是:
来构建切片元组。 The tuple environment allows for value unpacking with the * operator. tuple 环境允许使用 * 运算符进行值解包。
Output: Output:
[[[ 0 1]
[ 2 3]
[ 4 5]]
[[ 6 7]
[ 8 9]
[10 11]]]
[3 9]
To verify it matches:要验证它是否匹配:
import numpy as np
A = np.arange(12).reshape((2, 3, 2))
x = [1, 1]
s = (slice(None), *x)
print(np.allclose(A[s], A[:, 1, 1])) # True
*This is a modification of answers found here: Slicing a numpy array along a dynamically specified axis *这是对此处找到的答案的修改:沿动态指定的轴切片 numpy 数组
Edit to reflect edit on question and comment:编辑以反映对问题和评论的编辑:
To clarify, you can unpack any iterable you like in the tuple environment.澄清一下,您可以在元组环境中解压缩任何您喜欢的可迭代对象。 The * operator functions normally in within the tuple. * 运算符通常在元组中起作用。 Order your elements however you like.随心所欲地订购您的元素。 Mix in different iterables, types, slice(None)
, how ever you want to build your slices, as long as you end up with a valid sequence of values, it will behave as expected.混合不同的迭代、类型、 slice(None)
,无论你想如何构建你的切片,只要你最终得到一个有效的值序列,它就会按预期运行。
import numpy as np
A = np.arange(12).reshape((2, 3, 2))
t = [True, False]
x = [1, 1]
print(np.allclose(A[(*t, *x)], A[True, False, 1, 1])) # True
You can also add full lists as well in the tuple:您还可以在元组中添加完整列表:
print(np.allclose(A[(t, *x)], A[[True, False], 1, 1])) # True
you can use slice(None)
instead of :
, So你可以使用slice(None)
而不是:
,所以
y = tuple([slice[None]] + x)
A[y]
is what you need.是你需要的。
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