Compare items from two lists in Python

Question

The task is to check whether the last 3 digits from corresponding items from two lists are the same. If the items have a length of less than 3, it checks whether they are the same number.

If the two lists have different lengths it should return false, and if both lists have a length of 0, it should return true.

def corresponding_elements_have_same_end(list1, list2):

if len(list1) == len(list2):

    for i in range(0, len(list1)):

        num1 = str(list1[i])
        num2 = str(list2[i])

        if len(num1) <= 3 and len(num2) <= 3:
            return num1 == num2

        else:
            return num1[-3:] == num2[-3:]

else:
    return False

If I run it through this:

print("1.", corresponding_elements_have_same_end([3452, 43600, 10], [3111452, 600, 10]))
print("2.", corresponding_elements_have_same_end([452, 43600], [52, 600]))
print("3.", corresponding_elements_have_same_end([32, 43032], [32, 32]))
print("4.", corresponding_elements_have_same_end([32, 43132, 300], [32, 56132, 3300]))

It prints out

True

False

True

True

When it should print:

True

False

False

True

You should consider using mod, %, instead of str. num# = list#[i] % 1000. — Claudio Corsi
– Claudio Corsi, Commented May 10, 2016 at 2:08

niemmi · Accepted Answer · 2016-05-10 02:04:20Z

2

The problem is that the function always checks only one item from the lists and returns immediately no matter the length. When [32, 43032], [32, 32] are compared against each other it checks that 32 == 32 and immediately returns True. If you change the order of first list to [42032, 32] you'd get False instead.

In order to fix the problem the loop needs to be modified so that it returns only in case that numbers don't match. If they do then the next pair of numbers should be checked. If the loop completes then you know that all numbers match. Here's an example:

for i in range(0, len(list1)):
    if str(list1[i])[-3:] != str(list2[i])[-3:]:
        return False

return True

Note that you could implement the loop with zip to make indexing unnecessary:

for x, y in zip(list1, list2):
    if str(x)[-3:] != str(y)[-3:]:
        return False

edited May 10, 2016 at 2:04

answered May 10, 2016 at 1:59

niemmi

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Comments

bian · Accepted Answer · 2016-05-10 02:28:59Z

1

def t(m,n):
    if False in map(lambda x,y:str(x)[-3:]==str(y)[-3:],m,n):
        return False
    return True

answered May 10, 2016 at 2:28

bian

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Comments

ml-moron · Accepted Answer · 2016-05-10 02:13:51Z

0

Maybe this is the control flow you are looking for:

def corresponding_elements_have_same_end(list1, list2):
    if len(list1) == len(list2):
        for i in range(len(list1)):
            num1, num2 = str(list1[i]), str(list2[i])
            if len(num1) <= 3 and len(num2) <= 3 and num1 != num2:
                return False
            if num1[-3:] != num2[-3:]:
                return False
        return True
    else:
        return False

answered May 10, 2016 at 2:13

ml-moron

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