Python numpy 获取值为 true 的索引
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numpy get index where value is true
提问by change
>>> ex=np.arange(30)
>>> e=np.reshape(ex,[3,10])
>>> e
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, 25, 26, 27, 28, 29]])
>>> e>15
array([[False, False, False, False, False, False, False, False, False,
False],
[False, False, False, False, False, False, True, True, True,
True],
[ True, True, True, True, True, True, True, True, True,
True]], dtype=bool)
I need to find the rows that have true or rows in ewhose value are more than 15. I could iterate using a for loop, however, I would like to know if there is a way numpy could do this more efficiently?
我需要找到具有真e值的行或值大于 15 的行。我可以使用 for 循环进行迭代，但是，我想知道 numpy 是否可以更有效地执行此操作？
采纳答案by Jaime
To get the row numbers where at least one item is larger than 15:
要获取至少一项大于 15 的行号：
>>> np.where(np.any(e>15, axis=1))
(array([1, 2], dtype=int64),)
回答by Hadi
You can use nonzerofunction. it returns the nonzero indices of the given input.
您可以使用非零函数。它返回给定输入的非零索引。
Easy Way
简单的方法
>>> (e > 15).nonzero()
(array([1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]), array([6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]))
to see the indices more cleaner, use transposemethod:
要更清晰地查看索引，请使用transpose方法：
>>> numpy.transpose((e>15).nonzero())
[[1 6]
[1 7]
[1 8]
[1 9]
[2 0]
...
Not Bad Way
不错的方式
>>> numpy.nonzero(e > 15)
(array([1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]), array([6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]))
or the clean way:
或干净的方式：
>>> numpy.transpose(numpy.nonzero(e > 15))
[[1 6]
[1 7]
[1 8]
[1 9]
[2 0]
...
回答by Opt
A simple and clean way:use np.argwhereto group the indices by element, rather than dimension as in np.nonzero(a)(i.e., np.argwherereturns a row for each non-zero element).
一种简单而干净的方法：使用np.argwhere按元素而不是维度对索引进行分组np.nonzero(a)（即，np.argwhere为每个非零元素返回一行）。
>>> a = np.arange(10)
>>> a
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> np.argwhere(a>4)
array([[5],
[6],
[7],
[8],
[9]])
np.argwhere(a)is the same as np.transpose(np.nonzero(a)).
np.argwhere(a)与 相同np.transpose(np.nonzero(a))。
Note:You cannot use a(np.argwhere(a>4))to get the corresponding values in a. The recommended way is to use a[(a>4).astype(bool)]or a[(a>4) != 0]rather than a[np.nonzero(a>4)]as they handle 0-d arrays correctly. See the documentationfor more details. As can be seen in the following example, a[(a>4).astype(bool)]and a[(a>4) != 0]can be simplified to a[a>4].
注意：您不能使用a(np.argwhere(a>4))来获取a. 推荐的方法是使用a[(a>4).astype(bool)]ora[(a>4) != 0]而不是a[np.nonzero(a>4)]因为它们正确处理 0-d 数组。有关更多详细信息，请参阅文档。如在下面的例子中可以看出，a[(a>4).astype(bool)]并且a[(a>4) != 0]可以被简化为a[a>4]。
Another example:
另一个例子：
>>> a = np.array([5,-15,-8,-5,10])
>>> a
array([ 5, -15, -8, -5, 10])
>>> a > 4
array([ True, False, False, False, True])
>>> a[a > 4]
array([ 5, 10])
>>> a = np.add.outer(a,a)
>>> a
array([[ 10, -10, -3, 0, 15],
[-10, -30, -23, -20, -5],
[ -3, -23, -16, -13, 2],
[ 0, -20, -13, -10, 5],
[ 15, -5, 2, 5, 20]])
>>> a = np.argwhere(a>4)
>>> a
array([[0, 0],
[0, 4],
[3, 4],
[4, 0],
[4, 3],
[4, 4]])
>>> [print(i,j) for i,j in a]
0 0
0 4
3 4
4 0
4 3
4 4
