How to set single element of multi dimensional Numpy Array using another Numpy array?

Question

If we have a numpy array like:

Array = np.zeros((2, 10, 10))

and we want to set one element of it, given by another

index         


        

Answer 1

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]]])