ValueError The truth value of an array with more than one element is ambiguous Use a any or a all

So I’m trying to create an array of the same size as [A]. What I want is a forloop that checks if the 1st value of the ith element in the Array == 0, However it keeps telling me that there is more than one element in the truth value of an array when there shouldnt be as I indexed the 1st value of the ith element in my array. Here is my code:

n = 4
N = [i for i in range(1,n+1)]
V = [0] + N
K = N + [5]
M = [0,1]
A = np.array([(i,j,k) for i in V for j in K for k in M if i!=j])

C=[]

for i in A:
    if A[i][0]==0:
        C.append([0.7])
    elif abs(A[i][0]-A[i][1])<=1:
        C.append([1])
    else:
        C.append([0])

>Solution :

When you go through your for loop, i is already each list in A, you can check this with the below:

for i in A:
    print(i)

Which returns:

[0 1 0]
[0 1 1]
[0 2 0]
[0 2 1]...

So then calling A[i][0] gives an array each time rather than an integer, so the comparison A[i][0] == 0 is not possible. To fix your problem, either do the below, which will change your i to get an index for every element in A:

for i in range(len(A)):
    if A[i][0]==0:
        C.append([0.7])
    elif abs(A[i][0]-A[i][1])<=1:
        C.append([1])
    else:
        C.append([0])

Or change all instances of A[i][x] to i[x], and use the each list element of A that way, as follows:

for i in A:
    if i[0]==0:
        C.append([0.7])
    elif abs(i[0]-i[1])<=1:
        C.append([1])
    else:
        C.append([0])