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Sort odd numbers in a list

How can I sort the ascending odd numbers in a list of integers, but leaving the even numbers in their original places?

Example:

sortArray([5, 3, 2, 8, 1, 4]) == [1, 3, 2, 8, 5, 4]

My code:

def sort_array(source_array):     
    odd_numbers = [n for n in source_array if n%2!=0]
    odd_numbers = sorted(odd_numbers)

How to switch the indices of odd numbers in source_array

with those of odd_numbers

?

+3


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4 answers


It looks like you're almost there - you can make your sorted odd numbers iterable and rebuild the original list with either the original even number or the next sorted odd number, like this:



>>> data = [5, 3, 2, 8, 1, 4]
>>> odds = iter(sorted(el for el in data if el % 2))
>>> [next(odds) if el % 2 else el for el in data]
[1, 3, 2, 8, 5, 4]
+10


source


Different lines, but also more readable



a = [5, 3, 2, 8, 1, 4]
b = sorted([item for item in a if item%2 != 0])
odd_int = 0
for i in range(len(a)):
    if a[i] %2 != 0:
        a[i] = b[odd_int]
        odd_int += 1
+2


source


If you are open to using numpy, you can get the indices of odd numbers using np.where

, sort the odd numbers, and update the array using the previously obtained indices assigning the sorted array of odd numbers:

import numpy as np

a = np.array([5, 3, 2, 8, 1, 4])
ind = np.where(a%2)                 # get indices of odd items
a[ind] = np.sort(a[ind])            # update items at indices using sorted array
print(a)
# array([1, 3, 2, 8, 5, 4])
0


source 


def sort_array(source_array):
    odd_numbers = sorted([n for n in source_array if n%2!=0])
    c = 0
    res = []
    for i in source_array:
        if i %2!=0:
            res.append(odd_numbers[c])
            c += 1
        else:
            res.append(i)
    return res
0


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