Fill matrix diagonal with different values for each python numpy

Question

I saw a function numpy.fill_diagonal which assigns same value for diagonal elements. But I want to assign different random values for each diagonal elements. Ho

Answer 1

You can use np.diag_indices to get those indices and then simply index into the array with those and assign values.

Here's a sample run to illustrate it -

In [86]: arr          # Input array
Out[86]: 
array([[13, 69, 35, 98, 16],
       [93, 42, 72, 51, 65],
       [51, 33, 96, 43, 53],
       [15, 26, 16, 17, 52],
       [31, 54, 29, 95, 80]])

# Get row, col indices
In [87]: row,col = np.diag_indices(arr.shape[0])

# Assign values, let's say from an array to illustrate
In [88]: arr[row,col] = np.array([100,200,300,400,500])

In [89]: arr
Out[89]: 
array([[100,  69,  35,  98,  16],
       [ 93, 200,  72,  51,  65],
       [ 51,  33, 300,  43,  53],
       [ 15,  26,  16, 400,  52],
       [ 31,  54,  29,  95, 500]])

You can also use np.diag_indices_from and probably would be more idomatic, like so -

row, col = np.diag_indices_from(arr)

Note : The tried function would work just fine. This is discussed in a previous Q&A - Numpy modify ndarray diagonal too.

Answer 2

That the docs call the fill val a scalar is an existing documentation bug. In fact, any value that can be broadcasted here is OK.

Fill diagonal works fine with array-likes:

>>> a = np.arange(1,10).reshape(3,3)
>>> a
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])
>>> np.fill_diagonal(a, [99, 42, 69])
>>> a
array([[99,  2,  3],
       [ 4, 42,  6],
       [ 7,  8, 69]])

It's a stride trick, since the diagonal elements are regularly spaced by the array's width + 1.

From the docstring, that's a better implementation than using np.diag_indices too:

Notes
-----
.. versionadded:: 1.4.0

This functionality can be obtained via `diag_indices`, but internally
this version uses a much faster implementation that never constructs the
indices and uses simple slicing.