numpy.unravel_index 说明
官网的api说明如下：
numpy.
unravel_index
(
indices
,
dims
,
order='C'
)
Converts a flat index or array of flat indices into a tuple of coordinate arrays.
| Parameters: | indices : array_like An integer array whose elements are indices into the flattened version of an array of dimensions dims. Before version 1.6.0, this function accepted just one index value. dims : tuple of ints The shape of the array to use for unraveling indices. order : {‘C’, ‘F’}, optional Determines whether the indices should be viewed as indexing in row-major (C-style) or column-major (Fortran-style) order. New in version 1.6.0. |
| Returns: | unraveled_coords : tuple of ndarray Each array in the tuple has the same shape as the indices array. |
See also
ravel_multi_index
Examples
>>>
>>>
np
.
unravel_index([
22
,
41
,
37
], (
7
,
6
))
(array([3, 6, 6]), array([4, 5, 1]))
>>>
np
.
unravel_index([
31
,
41
,
13
], (
7
,
6
), order
=
'F'
)
(array([3, 6, 6]), array([4, 5, 1]))
>>>
>>>
np
.
unravel_index(
1621
, (
6
,
7
,
8
,
9
))(3, 1, 4, 1)
例子：
Consider a (6,7,8) shape array, what is the index (x,y,z) of the 100th element ?
>>> print
np.unravel_index(
100
,(
6
,
7
,
8
))
(
1
,
5
,
4
)
解释：
给定一个矩阵，shape=（6,7,8），即3维的矩阵，求第n个元素的下标是什么？
矩阵各维的下标从0开始
如果indices参数是一个标量，那么返回的是一个向量，维数=矩阵的维数，向量的值其实就是在矩阵中对应的下标。如6*7*8*9的矩阵，1621/(7*8*9)=3，(1621-3*7*8*9)/(8*9)=1，(1621-3*7*8*9-1*8*9)/9=4，(1621-3*7*8*9-1*8*9-4*9)=1。所以返回的向量为array(3,1,4,1)
如果indices参数是一个向量的，那么通过该向量中值求出对应的下标。下标的个数就是矩阵的维数，每一维下标组成一个向量，所以返回的向量的个数=矩阵维数。如7*6的矩阵，第22个元素是 3*6+4，所以对应的下标是(3,4)，那么返回的值是 array([3]),array([4])