CODEWITHSUNDEEP
python
user15404163
user15404163

Reputation:

How to convert every even number in a list to odd

one of my revision questions was to create a func that takes a list of integer values as a list, and convert every even number into an odd number.

Expected Output:
integer_list = [1, 2, 3 , 4, 5, 6, 7]
Before: [1, 2, 3, 4, 5, 6, 7]
After: [1, 3, 3, 5, 5, 7, 7]

integer_list = [1, -2, 3 , -4, 8, 16, 21, -17]
Before: [1, -2, 3, -4, 8, 16, 21, -17]
After: [1, -3, 3, -5, 9, 17, 21, -17]

Code thus far: 

def make_all_odd(integer_list):
    for number in integer_list:
        if (number % 2) == 0:
            number += 1
            integer_list.append(number)

So far, I've only figured out that I need to iterate through every element in the list and check if it is even using a for loop, but I'm not sure how to modify the original string to change the identified even numbers into odd. I know my code just adds the recently converted even to odd numbers to the end of the list, but in my case I have to replace, not add. I first thought of using find() but that doesn't work for lists

Upvotes: 0

Views: 3170

Answers (3)

KetZoomer
KetZoomer

Reputation: 2914

You can modify a list using this syntax: list[index] = whatever you want here. Indexes start at 0. Enumerate just allows you to get the index while looping, instead of doing list.index(element).

Code:

Using Enumerate:

integer_list = [1, 2, 3 , 4, 5, 6, 7]
def make_all_odd(integer_list):
    for index, number in enumerate(integer_list):
        if (number % 2) == 0:
            integer_list[index] += 1

make_all_odd(integer_list)
print(integer_list)

Without Using Enumerate (doesn't take into account duplicate value, so the even numbers all have to be different):

integer_list = [1, 2, 3 , 4, 5, 6, 7]
def make_all_odd(integer_list):
    for number in integer_list:
        index = integer_list.index(number)
        if (number % 2) == 0:
            integer_list[index] += 1

make_all_odd(integer_list)
print(integer_list)

Here I use a variable to keep track of the index instead of the .index method, and this fixes not being able to use duplicates.

integer_list = [1, 2, 3 , 4, 5, 6, 7]
index = 0
def make_all_odd(integer_list):
    for number in integer_list:
        index = integer_list.index(number)
        if (number % 2) == 0:
            integer_list[index] += 1
    index+=1

make_all_odd(integer_list)
print(integer_list)

You could also do this return a new list (if you don't want the original list to be modified):

Returning New List Using Enumerate:

integer_list = [1, 2, 3 , 4, 5, 6, 7]
def make_all_odd(integer_list):
    integer_list_copy = integer_list.copy() # remember to use .copy(), look at https://stackoverflow.com/questions/2612802/list-changes-unexpectedly-after-assignment-why-is-this-and-how-to-prevent-it 
    for index, number in enumerate(integer_list):
        if (number % 2) == 0:
            integer_list_copy[index] = number + 1
    return integer_list_copy

make_all_odd(integer_list)
print(integer_list)

Also, don't change a list when iterating through it. It just gives everyone headaches.

Upvotes: -1

Q-man
Q-man

Reputation: 2209

Keep it simple using the True/False array:

import numpy as np
L = [1, 2, 3, 4, 5, 6, 7]
L + (np.mod(L,2)==0)

which returns:

array([1, 3, 3, 5, 5, 7, 7])

Upvotes: -1

Jonathon Reinhart
Jonathon Reinhart

Reputation: 137398

You're making a critical mistake: You're modifying a list while you are iterating it. Don't do that. Sometimes the language or implementation will save you, but often times you'll end up getting strange and incorrect results.

Try instead iterating over the list by index:

for i in range(len(integer_list)):
   # You can read an item from the list:
   val = integer_list[i]

   # And you can overwrite an item in the list:
   integer_list[i] = new_val

(The exact solution is up to you to come up with.)

Upvotes: 3

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