The code is for returning the outlying even or odd number, everything works except for my numbers ar-CodePudding

example of a problem would be [2,4,6,8,9] the code should return 9, instead it returns [9]

def find_outlier(integers):
    even, odd = 0, 0
    outlier = []

    for num in integers:
        if num % 2 == 0:
            even  = 1
        else:
            odd  =1
    if even > odd:
        for num in integers:
            if num % 2 != 0:
                outlier.append(num)
                return outlier
    else:
        if odd > even:
            for num in integers:
                if num % 2 == 0:
                    outlier.append(num)
                    return outlier

CodePudding user response：

You are returning the list and printing the list. That's the reason you can returning the num instead of the outlier list

def find_outlier(integers):
    even, odd = 0, 0
    outlier = []

    for num in integers:
        if num % 2 == 0:
            even  = 1
        else:
            odd  =1
    if even > odd:
        for num in integers:
            if num % 2 != 0:
                outlier.append(num)
                return num
    else:
        if odd > even:
            for num in integers:
                if num % 2 == 0:
                    outlier.append(num)
                    return num
print(find_outlier([2,4,6,8,9]))

But still your code is fully mess. It's not gonna return more than one odd or even. I exactly don't know what you wanted to mean by this code. But this shortened code may help

def CheckEvenOrOdd(integers):
    even, odd= 0,0
    Evenlist=[]
    OddList=[]
    for num in integers:
        if num % 2 == 0:
            Evenlist.append(num)
            even  = 1
        else:
            OddList.append(num)
            odd  =1
    if even > odd:
        return OddList
    else:
        return Evenlist

print(CheckEvenOrOdd([2,4,6,8,9,11]))

It's gonna return list so there is no problem with that

CodePudding user response：

Disclaimer - it's not a direct answer to OP's question, rather to expand it and make it more scalable to include multiple outliers by using Pythonic function from more_itertools.

It uses more_itertools partition. It returns a 2-tuple of iterables derived from the input iterable. The first yields the items that have pred(item) == False. The second yields the items that have pred(item) == True.

from more_itertools import partition   

def find_outliers(iterable):
    ''' find one or more outlier numbers in the iterable list'''
    is_odd = lambda x: x & 1
    odds, evens = partition(is_odd, iterable)

    #  can do the comparison of len(list(odds)) vs. len(list(evens)) ...
    #  ....leave as your exercise..... since you already did it. 

    return list(odds), list(evens)

>>> L = [1, 2, 3, 4, 5, 6, 7, 9]
>>> find_outliers(L)
# ([2, 4, 6], [1, 3, 5, 7, 9])
>>> find_outliers(nums)
# ([2, 4, 6, 8], [9])