Check to see if a string has the same amount of 'x's and 'o's. The method must return a boolean and be case insensitive. The string can contain any char.

Examples input/output:

XO("xooxx") => false
XO("ooxXm") => true
XO("zpzpzpp") => true // when no 'x' and 'o' is present should return true
XO("zzoo") => false

So I tried to make new lists, and then append for each 'x' and each 'o' but its not quite working. Here is my code so far:

def xo(s):
    ex = []
    oh = []
    for letter in s.split():
        if letter.islower() == 'x':
            ex = ex.append(letter)
        elif letter.islower() == 'o':
            oh = oh.append(letter)
    if len(ex) == len(oh):
        return True
    else:
        return False

It only returns True on the first ones but not on the last third. It seems to me that the code is not appending properly.

CodePudding user response：

There are (at least) three problems in your code:

• s.split() splits s on every space and returns a list (so assuming you have no spaces, this gives you a list with a single item). Instead, just use for letter in s: which iterates over each letter.
• letter.islower() checks if the letter is lowercased, returning True or False. What you want is letter.lower() which returns the lowercased version of a character.
• Also (thanks Mark) append doesn't return a value, so when you run oh = oh.append(letter) you are assigning a value of None to the oh variable. You should instead just use oh.append(letter) (no assignment).

CodePudding user response：

Just use collections.Counter.

>>> from collections import Counter
>>> c = Counter('xooxXo'.lower())
>>> c['x'] == c['o']
True

or

>>> c = Counter('xooxx'.lower())
>>> c['x'] == c['o']
False

Wrap it in your own function if you want to.

CodePudding user response：

David's answer covers some potential fixes to your code, but if you're looking for something more concise, I would suggest using .lower() to lowercase the string upfront, and then use .count() to count the number of Xs and Os rather than using a for loop:

def xo(s):
    return s.lower().count('x') == s.lower().count('o')

This outputs:

False