Detect where elements of two lists differ from one another according to their index-CodePudding

given these two lists

l1=[0,0,0,0,1,1,1,1]
l2=[0,0,0,1,1,1,1,1]

they are identical except for the 4th element. I need a code that detects the difference between these two sets and prints out the location of the detected difference. In this case it would be = 4. The intersection and union command wouldn't work as they don't take the index into consideration.

I have tried this code, but it does not print out anything:

l1=[0,0,0,0,1,1,1,1]
l2=[0,0,0,1,1,1,1,1]

for i in l1:
    if i != l2[l1.index(i)]:
        print(l1.index(i),i)

CodePudding user response：

Your code does not work because list.index(value [, pos]) only reports the first occurence of a value in that list [after pos].

This would report the differences:

l1=[0,0,0,0,1,1,1,1]
l2=[0,0,0,1,1,1,1,1]
print(*(p for p,v in enumerate(zip(l1,l2)) if v[0]^v[1]))

Output:

The zip(..) pairs up values positionally in to tuples, enumerate(..) gets the index, tuple value and v[0]^v[1] is a logical xor that is only true if the values differ at least in 1 bit.

See:

The simpler version of this works without zip:

for index,number in enumerate(l1): # get index & number of one list
    if l2[index] != number:        # compare to other number
        print(f"On {index}: {number} != {l2[index]}")

CodePudding user response：

output = [idx for idx,(i,j) in enumerate(zip(l1,l2), start=1) if i!=j]
print(output) # [4]

Or you can do:

for i in range(len(l1)):
    if l1[i] != l2[i]:
        print(i 1)

which is closer to what you wrote.

CodePudding user response：

You can use zip

l1=[0,0,0,0,1,1,1,1]
l2=[0,0,0,1,1,1,1,1]

diff_idx = [idx for idx,(x,y) in enumerate(zip(l1,l2)) if x != y]
print(diff_idx)

output

[3]

CodePudding user response：

l1=[0,0,0,0,1,1,1,1]
l2=[0,0,0,1,1,1,1,1]
print(l1.index(l1 != l2))

The third line returns the index position of l1 which does not satisfy the condition l1 not equal to l2.