CODEWITHSUNDEEP
python
jboyda5
jboyda5

Reputation: 75

How to create a list that results a number from a given list, not the indices?

I am having trouble getting the results to produce the integer in the list, not which index it falls under.

#this function takes, as input, a list of numbers and an option, which is either 0 or 1.
#if the option is 0, it returns a list of all the numbers greater than 5 in the list
#if the option is 1, it returns a list of all the odd numbers in the list
def splitList(myList, option):
    #empty list for both possible options
    oddList = []
    greaterList = []
    #if option is 0, use this for loop
    if int(option) == 0:
        #create for loop the length of myList
        for i in range(0, len(myList)):
            #check if index is greater than 5
            if ((myList[i])>5):
                #if the number is greater than 5, add to greaterList
                greaterList.append(i)
        #return results
        return greaterList
    #if option is 1, use this for loop
    if int(option) == 1:
        #create for loop the length of myList
        for i in range(0, len(myList)):
            #check if index is odd by checking if it is divisible by 2
            if ((myList[i])%2!=0):
                #if index is not divisible by 2, add the oddList
                oddList.append(i)
        #return results
        return oddList

the results I receive are as follows:

>>>splitList([1,2,6,4,5,8,43,5,7,2], 1)
   [0, 4, 6, 7, 8]

I am trying to get the results to be [1, 5, 43, 5, 7]

Upvotes: 0

Views: 122

Answers (4)

Sukrit Kalra
Sukrit Kalra

Reputation: 34493

You're iterating over the range of the index. Iterate over the list instead.

for i in myList:
    #check if index is greater than 5
    if i >5:
        #if the number is greater than 5, add to greaterList
        greaterList.append(i)

So, your code gets rewritten as (with some minor changes)

def splitList(myList, option):
    final_list = []
    if int(option) == 0:
        for i in myList:
            if i > 5:
                final_list.append(i)
    elif int(option) == 1:
        for i in myList:
            if i%2 != 0:
                final_list.append(i)
    return final_list

You could reduce it by doing

def splitList(myList, option):
    if int(option) == 0:
        return [elem for elem in myList if elem > 5]
    elif int(option) == 1:
        return [elem for elem in myList if elem % 2 != 0]

Output - 

>>> splitList([1,2,6,4,5,8,43,5,7,2], 1)
[1, 5, 43, 5, 7]

Upvotes: 2

Grizz
Grizz

Reputation: 320

Take a closer look at your .append() commands... in your compare you're using:

if ((mylList[i])%2!=0)

or 

if ((myList[i])>5)

...but when you put it into the list, you're just using 

greaterList.append(i)

instead of 

greaterList.append(myList[i])

This must be a homework or class somewhere?

Upvotes: 0

Paul Hankin
Paul Hankin

Reputation: 58271

List comprehensions greatly simplify your code.

def split_list(xs, option):
    if option:
        return [x for x in xs if x % 2]
    else:
        return [x for x in xs if x > 5]

Upvotes: 2

Brendan Long
Brendan Long

Reputation: 54242

if ((myList[i])>5):
    #if the number is greater than 5, add to greaterList
    greaterList.append(i)

Instead of adding the index i, add the value (myList[i]):

if ((myList[i])>5):
    #if the number is greater than 5, add to greaterList
    greaterList.append(myList[i])

Same thing for the oddList case.

Note: @Sukrit Kalra's solution is preferable, but I'm leaving this up to show that there's multiple ways of solving this.

Upvotes: 1

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