Getting the count of items under diagonal in numpy - numpy

Question

I have a correlation matrix, and I want to get the count of number of items below the diagonal. Preferably in numpy.
[[1, 0, 0, 0, 0],
[.35, 1, 0, 0, 0],
[.42, .31, 1, 0, 0],
[.25, .38, .41, 1, 0],
[.21, .36, .46, .31, 1]]
I want it to return 10. Or, to return the mean of all numbers under the diagonal.

Answer 1

Setup
a = np.array([[1. , 0. , 0. , 0. , 0. ],
[0.35, 1. , 0. , 0. , 0. ],
[0.42, 0.31, 1. , 0. , 0. ],
[0.25, 0.38, 0.41, 1. , 0. ],
[0.21, 0.36, 0.46, 0.31, 1. ]])
numpy.tril_indices will give the indices of all elements under the diagonal (if you provide an offset of -1), and from there, it becomes as simple as indexing and calling mean and size
n, m = a.shape
m = np.tril_indices(n=n, k=-1, m=m)
a[m]
# array([0.35, 0.42, 0.31, 0.25, 0.38, 0.41, 0.21, 0.36, 0.46, 0.31])
a[m].mean()
# 0.346
a[m].size
# 10

Answer 2

A more primitive and bulky answer since numpy provides np.tril_indices as user3483203 mentioned, but what you want per row iteration i is the following (in terms of [row,col] indices):
(i=0)
[1,0] (i=1)
[2,0] [2,1] (i=2)
[3,0] [3,1] [3,2] (i=3)
...
This is essentially the zip of list [i,i,i,...] = [i]*i (i repetitions of i) with [0,1,...,i-1] = range(i). So iterating over the rows of the table, you can actually get the indices per iteration and perform the operator of your choice.
Example setup:
test = np.array(
[[1, 0, 0, 0, 0],
[.35, 1, 0, 0, 0],
[.42, .31, 1, 0, 0],
[.25, .38, .41, 1, 0],
[.21, .36, .46, .31, 1]])
Function definition:
def countdiag(myarray):
numvals = 0
totsum = 0
for i in range(myarray.shape[0]): # row iteration
colc = np.array(range(i)) # calculate column indices
rowc = np.array([i]*i) # calculate row indices
if any(rowc):
print(np.sum(myarray[rowc,colc]))
print(len(myarray[rowc,colc]))
numvals += len(myarray[rowc,colc])
totsum += np.sum(myarray[rowc,colc])
print(list(zip([i]*i, np.arange(i))))
mean = totsum / numvals
return mean, numvals
Test:
[165]: countdiag(test)
[]
0.35
1
[(1, 0)]
0.73
2
[(2, 0), (2, 1)]
1.04
3
[(3, 0), (3, 1), (3, 2)]
1.34
4
[(4, 0), (4, 1), (4, 2), (4, 3)]
0.346
Out[165]:
(0.346, 10)