How do you return only odd numbers in python?

My issue here is that the code filters out the even numbers correctly which is what I want, however it stops at seven and doesn't display number 9 which is what I would expect it to do. I've tried going over my code but I can't seem to find the issue

def remove_even[numbers] :
    new_list = []
    for i in range[0,len[numbers]-1] :
        if i % 2 != 0 :
            new_list.append[i]
    return new_list
l = [1,2,3,4,5,6,7,8,9,10]
print[remove_even[l]]

asked Jan 16, 2017 at 12:14

7

You should just directly loop through your values instead of indices

for i in numbers:

Otherwise if you wanted to use range you would have to index into your list

for i in range[0, len[numbers]]:
    if numbers[i] % 2 != 0 :
        new_list.append[numbers[i]]

For brevity, list comprehensions are well-suited for this type of task

>>> new_list = [num for num in l if num % 2 == 1]
>>> new_list
[1, 3, 5, 7, 9]

answered Jan 16, 2017 at 12:15

Cory KramerCory Kramer

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0

[k for k in l if k %2]

Is a simple list comprehension that returns

[1, 3, 5, 7, 9]

PM 2Ring

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answered Jan 16, 2017 at 12:16

e4c5e4c5

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0

You can do this more easily with the filter built-in function:

list[filter[lambda x: x % 2, l]]

[1, 3, 5, 7, 9]

It will only retain elements for which the condition evaluates to True

answered Oct 14, 2020 at 18:00

Nicolas GervaisNicolas Gervais

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This is because you are starting your range[] function of your for loop from 0 and ending at len[numbers]-1 [which is 9 in your case], python range[] already will run till end-1:

for eg:

for i in range[0,9]:
    print[i]

will print no's: 0 1 2 3 4 5 6 7 8

and that's why your 9 is not here in the output.

you don't have to start your loop from 0. If you are starting from 0 you can arrange your for loop like this:

1]

for i in range[0, len[numbers]+1]

2]Or you can code like more pythonic way.

def remove_even[numbers] :
    new_list = []
    for i in numbers :
        if i % 2 != 0 :
            new_list.append[i]
    return new_list

answered Jan 16, 2017 at 13:06

Ajay RawatAjay Rawat

1422 silver badges12 bronze badges

1

#simpliest way of doing it
mylist = [1,2,3,4,5,6,7,8,9,10,11]
for x in mylist:
    if x % 2 == 1: #this displays odd numbers
        print[x]

answered Jul 3, 2018 at 14:55

numbers = [1,2,3,4,5,6,7,8,9,10]

odds = [i for i in numbers if i%2!=0]

answered Jan 16, 2017 at 12:16

def odd_numbers[n]:return [x for x in range[1,n+1] if x%2==1]print[odd_numbers[5]] # Should print [1, 3, 5]

answered Jul 18, 2020 at 10:23

1

def remove_even[numbers] :
    new_list = []
    for i in range[0,len[numbers]-1] :
        if i % 2 != 0 :
            new_list.append[i]
    return new_list 

l = [1,2,3,4,5,6,7,8,9,10] 
print[remove_even[l]]

you got the algorithm right in the money. All you need to change is -1 to +1 instead and you will get your desired result.

why change the logic algorithm if only one item is to be changed.

answered Mar 14, 2021 at 22:17

1

It's simple using list comprehension:

def odd_no[n]:
      return [x for x in range[n+1] if n%2==1]

answered Oct 14, 2020 at 17:30

1

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