Issue with array assignment in python

Question

I can't figure out why the program is behaving in a peculiar way. The program is meant to rotate an array cyclically for 'K' times. When the list A is assigned to B in function rotate_array(), the program behaves in an incorrect way as shown in the output. Whereas when it is changed with the line B=[0]*len(A), the problem disappears. Could someone please help ?

def rotate_array(A):
    #B = [0]*len(A)  # This solves the problem
    B = A   # This seems to cause a problem
    print "In rotate_array", A
    for index, item in enumerate(A):
        print "index:item ={}:{}, length of A={}".format(index, item, len(A))
        if index == (len(A) - 1):
            B[0] = A[index]
        else:
            B[index + 1] = item
    print B
    return B


def solution(A, K):
    for index, item in enumerate(A):
        print "in fn soln: index:item ={}:{}, length of A={}".format(index, item, len(A))
    ctr = 0
    while ctr < K:
        A = rotate_array(A)
        ctr += 1
    return A

if __name__ == '__main__':
    A = [1,2,3,4]
    K = 1
    ret_A = solution(A, K)
    print ret_A

Output:

in fn soln: index:item =0:1, length of A=4
in fn soln: index:item =1:2, length of A=4
in fn soln: index:item =2:3, length of A=4
in fn soln: index:item =3:4, length of A=4
In rotate_array [1, 2, 3, 4]
index:item =0:1, length of A=4
index:item =1:1, length of A=4
index:item =2:1, length of A=4
index:item =3:1, length of A=4
[1, 1, 1, 1]
[1, 1, 1, 1]

When function rotate_array(A) is changed to have this line

B = [0]*len(A)  # This solves the problem

instead of

B = A

The output is now correct -

in fn soln: index:item =0:1, length of A=4
in fn soln: index:item =1:2, length of A=4
in fn soln: index:item =2:3, length of A=4
in fn soln: index:item =3:4, length of A=4
In rotate_array [1, 2, 3, 4]
index:item =0:1, length of A=4
index:item =1:2, length of A=4
index:item =2:3, length of A=4
index:item =3:4, length of A=4
[4, 1, 2, 3]
[4, 1, 2, 3]

B=A creates an alias (two references to the same list). If you want a (shallow) copy use B = A[:]. — John Coleman
– John Coleman, Commented May 8, 2018 at 20:14

William Allcock · Accepted Answer · 2018-05-09 15:33:12Z

I see that John answered your question. I dont know all the details of what you are doing, so you may have a reason for looping, but I wanted to suggest an alternative implementation that didn't require looping K times if all you want is the final answer:

def rotate_array(A,K):
    A_length = len(A)
    B = [None] * A_length
    for A_index in range(len(A)):
        B_index = (A_index + K) % A_length
        B[B_index] = A[A_index]
    return(B)

A = list(range(10))
print("A = ", A)
for K in range(10):
    B = rotate_array(A,K)
    print("K = ",K, "B = ", B)

A =  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
K =  0 B =  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
K =  1 B =  [9, 0, 1, 2, 3, 4, 5, 6, 7, 8]
K =  2 B =  [8, 9, 0, 1, 2, 3, 4, 5, 6, 7]
K =  3 B =  [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]
K =  4 B =  [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]
K =  5 B =  [5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
K =  6 B =  [4, 5, 6, 7, 8, 9, 0, 1, 2, 3]
K =  7 B =  [3, 4, 5, 6, 7, 8, 9, 0, 1, 2]
K =  8 B =  [2, 3, 4, 5, 6, 7, 8, 9, 0, 1]
K =  9 B =  [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]

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