# matrix elements counting

Hello,
I have the following matrix of 1s a,d 0s
5x6:                                   5x5
r1 1 1 0 0 1 1                     4 2 3 2 2
r2 1 0 0 0 1 0                     2 2 2 0 1
r3 1 0 1 0 1 1     =>            3 2 4 2 3
r4 0 1 1 1 0 1                     2 0 2 4 3
r5 1 0 1 0 0 1                    2 1 3 3 3

What i want to do as un can see in the example is to compare:
1 with 1            2 with 1           etc.....
1 with 2            2 with 2
1 with 3           2 with  3
1 with 4           2 with  4
1 with 5          2 with  5

when i compre for example r1 with all the rows: If row === 1 i incremement the matrix(5x5) by 1 so M[1,1] = M[1,1] + 1
and see where all the element of row1 are 1 with the same elements in row2 are 1:
as u can see:
r1 1 1 0 0 1 1
r2 1 0 0 0 1 0

row1 and row2 are only similar in 2 places the M[1,2] will be equal to 2 as u can see in the 5x5 matrix/

Hope that it's clear, i'll explain it again if there is a problem.

Thank you so much
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Author Commented:
I solved the problem on python:
But can u find an easier way?

``````for i in range(0,data.shape[0]):
M1 = []
for j in range(0,data.shape[1]):
M1.append(data[i,j])

for k in range(0,data.shape[0]):
count = 0
M2 = []
for l in range(0,data.shape[1]):
M2.append(data[k,l])

for c in range(0,len(M1)):
if (M1[c] and M2[c] == 1.0):
count = count + 1

count = count + 1
distance[i,k] = count
``````
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