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])