在递归函数中将字符串转换为列表

Question

I wrote this recursive function to check for a specific string in a list and if the string is found, it should be returned reversed within the list but I can't seem to keep the list as a list. my output is a joined string. can someone please suggest how to maintain the list type?

test_list = ['cat', 'hat', 'bat', 'sat']
test_string = 'bat'

def reverser(some_list, some_string):

    if some_list == []:
        return ''
    elif len(some_list) == 1 and item in some_list == some_string:
        return some_string[::-1] 
    else:
        if some_list[0] == some_string:
            return some_string[::-1] + reverser(some_list[1:], some_string)
        if some_list[0] != some_string:
            return some_list[0] + reverser(some_list[1:], some_string)
reverser(test_list, test_string)

output is:

'cathattabsat'

but i would like it to be:

['cat', 'hat', 'tab', 'sat']

Answer 1

If you want the function to return a list of strings, you need to make sure that in the base case as well as within the loops, you add the appropriate list and not a string.

In addition, in this row:

elif len(some_list) == 1 and item in some_list == some_string:

item is not defined. You can replace this row with:

elif len(some_list) == 1 and some_string in some_list

Finally, this will do the job:

test_list = ['cat', 'hat', 'bat', 'sat']
test_string = 'bat'


def reverser(some_list, some_string):

    if some_list == []:
        return []
    elif len(some_list) == 1 and some_string in some_list:
        return [some_string[::-1]] 
    else:
        if some_list[0] == some_string:
            return [some_string[::-1]] + reverser(some_list[1:], some_string)
        if some_list[0] != some_string:
            return [some_list[0]] + reverser(some_list[1:], some_string)
reverser(test_list, test_string)

This returns

['cat', 'hat', 'tab', 'sat']

Answer 2

Your return statement returns a string always.

return some_string[::-1] + reverser(some_list[1:], some_string)

Note: some_string[::-1] is a string

Wrap it in [] to make it a list of one item.

return [some_string[::-1]] + reverser(some_list[1:], some_string)

However, there's a much simpler way than this weird loop you're doing. You can use map or list comprehensions.

def reverse_string(any, match):
    if any == match: return any[::-1]
    else: return any

def reverser(some_list, some_string):
    return [reverse_string(s, some_string) for s in some_list]

Answer 3

Your recursive function is more complicated than it needs to be:

def reverser(some_list, some_string):
    if some_list == []:
        return []

    word = some_string[::-1] if some_list[0] == some_string else some_list[0]
    return [word] + reverser(some_list[1:], some_string)

But as mentioned by another answerer, this is much better done with a list comprehension:

[word[::-1] if word == some_string else word for word in some_list]

Answer 4

Think about the return type you want, and the expression you are returning:

return some_list[0] + reverser(some_list[1:], some_string)

You have a list of strings. So some_list[0] is going to be a string. You're returning a string, plus whatever.

If you want to return a list, make the string into a list somehow:

return some_list[0:1] + ...
return [some_list[0]] + ...