官網的api說明如下：
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
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)
解釋：
給定一個矩陣，shape=（6,7,8），即3維的矩陣，求第n個元素的下標是什麼？矩陣各維的下標從0開始
如果indices參數是一個標量，那麼返回的是一個向量，維數=矩陣的維數，向量的值其實就是在矩陣中對應的下標。如6*7*8*9的矩陣，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。所以返回的向量爲array(3,1,4,1)
如果indices參數是一個向量的，那麼通過該向量中值求出對應的下標。下標的個數就是矩陣的維數，每一維下標組成一個向量，所以返回的向量的個數=矩陣維數。如7*6的矩陣，第22個元素是 3*6+4，所以對應的下標是(3,4)，那麼返回的值是 array([3]),array([4])