python 将元组/数组/列表解包为 Numpy 数组的索引
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2444923/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Unpacking tuples/arrays/lists as indices for Numpy Arrays
提问by ntimes
I would love to be able to do
我希望能够做到
>>> A = numpy.array(((1,2),(3,4)))
>>> idx = (0,0)
>>> A[*idx]
and get
并得到
1
however this is not valid syntax. Is there a way of doing this without explicitly writing out
但是，这不是有效的语法。有没有办法在不明确写出的情况下做到这一点
>>> A[idx[0], idx[1]]
?
?
EDIT: Thanks for the replies. In my program I was indexing with a Numpy array rather than a tuple and getting strange results. Converting to a tuple as Alok suggests does the trick.
编辑：感谢您的答复。在我的程序中，我使用 Numpy 数组而不是元组进行索引并得到奇怪的结果。正如 Alok 建议的那样转换为元组可以解决问题。
采纳答案by Vicki Laidler
It's easier than you think:
这比你想象的要容易：
>>> import numpy
>>> A = numpy.array(((1,2),(3,4)))
>>> idx = (0,0)
>>> A[idx]
1
回答by Alok Singhal
Try
尝试
A[tuple(idx)]
Unless you have a more complex use case that's not as simple as this example, the above should work for all arrays.
除非你有一个更复杂的用例，不像这个例子那么简单，否则以上应该适用于所有数组。
回答by lunixbochs
Indexing an object calls:
索引对象调用：
object.__getitem__(index)
When you do A[1, 2], it's the equivalent of:
当你做 A[1, 2] 时，它相当于：
A.__getitem__((1, 2))
So when you do:
所以当你这样做时：
b = (1, 2)
A[1, 2] == A[b]
A[1, 2] == A[(1, 2)]
Both statements will evaluate to True.
这两个语句都将评估为 True。
If you happen to index with a list, it mightnot index the same, as [1, 2] != (1, 2)
如果您碰巧使用列表进行索引，则它的索引可能与 [1, 2] != (1, 2) 不同
回答by Mike Graham
No unpacking is necessary—when you have a comma between [and ], you are making a tuple, not passing arguments. foo[bar, baz]is equivalent to foo[(bar, baz)]. So if you have a tuple t = bar, bazyou would simply say foo[t].
不需要解包——当你在[and之间有逗号时]，你正在创建一个元组，而不是传递参数。foo[bar, baz]相当于foo[(bar, baz)]。因此，如果您有一个元组，t = bar, baz您只需说foo[t].
