bit-flipping decoding

[divon]

Hello everyone, especially to @jefsummers
I will try to explain the algorithm:

1. I have M = [0 1 1 0 1 1] and the value will be taken individually later.
   
2. The explanation of B_j is the same as in my previous reply.
   
3. In the loop of j in range 4 and i in range 6, I will do the equation E[j, i] = XOR of all M[i'].
   For example: when j = 0, i = 0, B_0 = 0, 1, 3, i' = 0, 1, 3, E[0, 0] = M[1] ^ M[3] (M[0] is not included because i' = i).
   Same thing also happened in i = 1 at E[0, 1] = M[0] ^ M[3] (M[1] is not included because i' = i)
   But when j = 0, i = 2, E[0, 2] = M[0] ^ M[1] ^ M[3]. How to implement this condition in python?
   

Here is the update of my code:

import numpy as np

#declaration of parity-check matrix
H = np.array([[1, 1, 0, 1, 0, 0], [0, 1, 1, 0, 1, 0],
            [1, 0, 0, 0, 1, 1], [0, 0, 1, 1, 0, 1]])

#setting the input and output of the LDPC
c = np.array([0, 0, 1, 0, 1, 1])
y = np.array([0, 1, 1, 0, 1, 1])

#initialization of bit flipping decoding
w_r = 2
w_c = 3
m = np.size(H,0)
N = np.size(H,1)
E = np.zeros((m, N))
#Code for obtain each value of matrix M
M = np.hsplit(y[::].copy(), N)

#initializing value of B
B_j = np.where(H == 1)
B = B_j[1].reshape(m, w_c)

for j in range (m):
    for i in range (N):
        for i_prime in B[j]:
            print (i_prime)
            #this part is supposed to be for line 11 of the algorithm and I am stuck at it
            if i_prime != i:
                E[j, i] = M[i_prime != i] ^ M[i_prime != i]