With a
as the data array and idx
as the array of indices such that each row corresponds to one element to be set in the data array, you could do -
a[tuple(idx.T)] = 5
Sample run -
In [94]: a = np.zeros((2,2,3),dtype=int)
In [95]: idx = np.array([[0,0,0],[1,1,0],[0,1,2]])
In [96]: a[tuple(idx.T)] = 5
In [97]: a
Out[97]:
array([[[5, 0, 0],
[0, 0, 5]],
[[0, 0, 0],
[5, 0, 0]]])
In [98]: a[tuple(idx.T)] = [5,10,15] # or set different values
In [99]: a
Out[99]:
array([[[ 5, 0, 0],
[ 0, 0, 15]],
[[ 0, 0, 0],
[10, 0, 0]]])
Alternatively, we could compute the linear indices with np.ravel_multi_index
and then perform the assignment with np.put
, like so -
np.put(a,np.ravel_multi_index(idx.T,a.shape),5)
If you are dealing with three dimensional arrays, we could slice the three dimensional indices and assign to have another method, like so -
a[idx[:,0],idx[:,1],idx[:,2]] = 5
If it's just one element needed to be set, just do -
a[tuple(idx)] = 5
Sample run -
In [118]: a = np.zeros((2,2,3),dtype=int)
In [119]: idx = np.array([0,0,0])
In [120]: a[tuple(idx)] = 5
In [121]: a
Out[121]:
array([[[5, 0, 0],
[0, 0, 0]],
[[0, 0, 0],
[0, 0, 0]]])