I need help comparing two lists and returning the indices that they don't match.
a = [0, 1, 1, 0, 0, 0, 1, 0, 1]
b = [0, 1, 1, 0, 1, 0, 1, 0, 0]
indices 4 and 8 don't match and i need to return that as a list [4,8]
I've tried a few methods but they haven't worked for me.
Use zip to iterate over both lists at the same time and enumerate to get the indices during iteration, and write a list comprehension that filters out the indices where the list values don't match:
>>> [i for i, (x, y) in enumerate(zip(a, b)) if x != y]
[4, 8]
[4, 8] as it should. Really, there must be a bug somewhere in your code.You could also just use a simple loop which scans the lists, item by item:
a = [0, 1, 1, 0, 0, 0, 1, 0, 1]
b = [0, 1, 1, 0, 1, 0, 1, 0, 0]
diff=[]
for i in range(0,len(a)):
if a[i]!=b[i]:
diff.append(i)
print diff
A list comprehension could also do the same thing:
diff=[i for i in range(len(a)) if a[i]!=b[i]]
print diff
range(0,len(a)) could be simplified to just range(len(a))If you are happy to use a 3rd party library, numpy provides one way:
import numpy as np
a = np.array([0, 1, 1, 0, 0, 0, 1, 0, 1])
b = np.array([0, 1, 1, 0, 1, 0, 1, 0, 0])
res = np.where(a != b)[0]
# array([4, 8], dtype=int64)
Relevant: Why NumPy instead of Python lists?
You can use zip :
a = [0, 1, 1, 0, 0, 0, 1, 0, 1]
b = [0, 1, 1, 0, 1, 0, 1, 0, 0]
count=0
indices=[]
for i in zip(a,b):
if i[0]!=i[1]:
indices.append(count)
count+=1
print(indices)
output:
[4, 8]
