在Python中找到两个列表之间的共同点

Question

Hey guys I need help on this past test question. Basically I am given two list of objects, and I am suppose to find the number of items that appear in the same position of the first list and the second. I have an example that was provided.

>>> commons(['a', 'b', 'c', 'd'], ['a', 'x', 'b', 'd'])
2
>>> commons(['a', 'b', 'c', 'd', 'e'], ['a', 'x', 'b', 'd'])
2

I am having trouble writing out the code. Our class is using python 3. I have no idea where to start writing this from. It is a first year programming course and I never did programming in my life.

Answer 1

I think a more straightforward solution would be:

def commons(L1,L2):
    return len([x for x in zip(L1,L2) if x[0]==x[1]])

Answer 2

This is not a simple problem for a beginner. A more straightforward approach would use functions like sum and zip with a list comprehension like so:

def commons(L1, L2):
    return  sum(el1 == el2 * 1 for el1, el2 in zip(L1, L2))

A more typical but error prone approach taken by beginners is:

def commons(L1, L2):
    count = 0
    for i, elem in enumerate(L2):
        if elem == L1[i]:
            count += 1
    return count

I say this is more error prone because there are more parts to get right.

Without using enumerate you could do:

def commons(L1, L2):
    count = 0
    for i, range(len(L2)):
        if L1[i] == L2[i]:
            count += 1
    return count

but these previous two will work only if len(L2) <= len(L1) . See what I mean by more error prone? To fix this you would need to do:

def commons(L1, L2):
    count = 0
    for i, range(min(len(L2), len(L1))):
        if L1[i] == L2[i]:
            count += 1
    return count

Answer 3

Seems like this would work:

def commons(l1, l2):
   return sum(1 for v1,v2 in map(None, l1,l2) if v1 == v2)

• Note: The degenerate form of map used here results in None being returned for all values in the shorter list (so even if l1 and l2 are not the same length, it'll work.) It assumes that both lists all have values (ie, L1 and L2 do not contain None - since that would end up in false positives if one list was shorter than the other.)