the task here is :

A. match_ends Given a list of strings, return the count of the number of strings where the string length is 2 or more and the first and last chars of the string are the same. Note: python does not have a operator, but = works.

def match_ends(words):
    for i in words:
        if (len(i) >= 2) & (i[0] == i[-1]):
            return [i]


print(match_ends(['aba', 'xyz', 'aa', 'x', 'bbb']))

and the output is :

['aba']

CodePudding user response：

the problem is that returns the first and it doesn't keep going

the return keyword stops the function, so you have to store it in an array-like to return it

here is the code optimized:

def match_ends(*words):
    result = []
    for i in list(words):
        if len(i) >= 2 and i[0] == i[-1]:
            result.append([i])
    return result


print(match_ends('aba', 'xyz', 'aa', 'x', 'bbb'))

CodePudding user response：

if (len(i) >= 2) & (i[0] == i[-1]):
     return [i]

You return when your if condition is true, and this will stop your function.

You should then save your data in a variable if you want to get all the occurence of it as res.append(i) (save i in the list res)

Then return it outside when you have iterate through all your list.

def match_ends(words):
    for i in words:
        if (len(i) >= 2) & (i[0] == i[-1]):
            res.append(i)
    return res

CodePudding user response：

This is happening because your first item matches your if conditions and as you are using return, your code exits with the result as the first item. I am not sure what you want to do with this function.