numpy.ma.indices¶

numpy.ma.
indices
(dimensions, dtype=<class 'int'>, sparse=False)[source]¶ Return an array representing the indices of a grid.
Compute an array where the subarrays contain index values 0, 1, … varying only along the corresponding axis.
Parameters:  dimensions : sequence of ints
The shape of the grid.
 dtype : dtype, optional
Data type of the result.
 sparse : boolean, optional
Return a sparse representation of the grid instead of a dense representation. Default is False.
New in version 1.17.
Returns:  grid : one ndarray or tuple of ndarrays
 If sparse is False:
Returns one array of grid indices,
grid.shape = (len(dimensions),) + tuple(dimensions)
. If sparse is True:
Returns a tuple of arrays, with
grid[i].shape = (1, ..., 1, dimensions[i], 1, ..., 1)
with dimensions[i] in the ith place
See also
mgrid
,ogrid
,meshgrid
Notes
The output shape in the dense case is obtained by prepending the number of dimensions in front of the tuple of dimensions, i.e. if dimensions is a tuple
(r0, ..., rN1)
of lengthN
, the output shape is(N, r0, ..., rN1)
.The subarrays
grid[k]
contains the ND array of indices along thekth
axis. Explicitly:grid[k, i0, i1, ..., iN1] = ik
Examples
>>> grid = np.indices((2, 3)) >>> grid.shape (2, 2, 3) >>> grid[0] # row indices array([[0, 0, 0], [1, 1, 1]]) >>> grid[1] # column indices array([[0, 1, 2], [0, 1, 2]])
The indices can be used as an index into an array.
>>> x = np.arange(20).reshape(5, 4) >>> row, col = np.indices((2, 3)) >>> x[row, col] array([[0, 1, 2], [4, 5, 6]])
Note that it would be more straightforward in the above example to extract the required elements directly with
x[:2, :3]
.If sparse is set to true, the grid will be returned in a sparse representation.
>>> i, j = np.indices((2, 3), sparse=True) >>> i.shape (2, 1) >>> j.shape (1, 3) >>> i # row indices array([[0], [1]]) >>> j # column indices array([[0, 1, 2]])