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bundles / numpy latest / numpy / indices

function

numpy:indices

source: /dev/numpy/build-install/usr/lib/python3.14/site-packages/numpy/_core/numeric.py :1777

Signature

def   indices ( dimensions dtype = <class 'int'> sparse = False )

Summary

Return an array representing the indices of a grid.

Extended Summary

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.

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

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, ..., rN-1) of length N, the output shape is (N, r0, ..., rN-1).

The subarrays grid[k] contains the N-D array of indices along the k-th axis. Explicitly

grid[k, i0, i1, ..., iN-1] = ik

Examples

import numpy as np
grid = np.indices((2, 3))
grid.shape
grid[0]        # row indices
grid[1]        # column indices
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]
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
j.shape
i        # row indices
j        # column indices

See also

meshgrid
mgrid
ogrid

Aliases

  • numpy.indices

Referenced by