Writing odd numbers to array in Python

Question

I am trying to write even numbers upto a limit into a list in Python using the following code

def odd_count(n):
    arr = list(50)
    for i in range(0,n,2):
        arr[i] = i
    return arr

I am gettting an error sayng TypeError: 'int' object is not iterable. What is wrong in this code? How can I fix it?

Your title says you want odd numbers, your body says even. Which is it? — ShadowRanger, Nov 11 '17 at 02:26

Anonta · Answer 1 · 2017-11-11T02:22:13.717

This line raises the above mentioned exception

arr = list(50)

list expects an iterable object (something that has a __iter__() method). You are passing an int instead.

To create a list of size 50 you can use the * operator on a list:

arr = [None] * 50

This will create a list with all 50 elements set to None.

A more pythonic way to solve your problem is to pass the range object directly to the list constructor like:

def odd_count(n):
    return list(range(1,n+1, 2))

print(odd_count(50))

score 0 · Answer 2 · edited Nov 11 '17 at 02:22

0

You can initialize the list with brackets. In the for loop, use .append(). The final code should be

def odd_count(n):
    arr = []
    for i in range(0,n,2):
        arr.append(i)
    return arr

edited Nov 11 '17 at 02:22

eyllanesc

answered Nov 11 '17 at 02:17

Anish Shenoy

1

identation, sir – grovina Nov 11 '17 at 02:20
1

@PatrickHaugh: The question says odd in the title, even in the body. – ShadowRanger Nov 11 '17 at 02:27

score 0 · Answer 3 · answered Nov 11 '17 at 02:24

0

You can also use Python's list comprehension if you want to be Pythonic:

def even_numbers(n):
    return [i for i in range(0,n,2)]

answered Nov 11 '17 at 02:24

Justin

No-op listcomps aren't really Pythonic; if it's the identity transform with no filter, you'd just do `list(range(0, n, 2))`, which is faster (for all but the smallest inputs), shorter, and equivalent. – ShadowRanger Nov 29 '17 at 21:38

3 Answers3