Please I need help with debugging this code or writing it entirely

Question

def my_sort (numbers):
If numbers == 0:
return "invalid"
elif len(numbers []) ==1:
return numbers
else:
a = set(list(filter(lambda x:x%2!=0, numbers )))
b =set (list(filter(lambda x:x%2==0,numbers)))
c=a|b
return c

print (my_sort([1,2,3,4,6]))

Answer 1

Thanks Bogdan.... However I tweaked and arrived at this

def my_sort (numbers):
if len(numbers) == 0:
return "invalid"
elif len(numbers) ==1:
return numbers
else:
a = list(filter(lambda x:x%2!=0, numbers ))
a.sort()
b =list(filter(lambda x:x%2==0,numbers))
b.sort()

c=0
max_index = len(b)- 1
while c <= max_index:
a.append(b[c])
c = c +1
return a

print (my_sort([1,20,3,40,5,6,7]))

Answer 2

Edited

Answer 3

The objective is to sort the list such that the odd numbers come first and even numbers follow

Answer 4

@Obumneme - your code definitely does _not_ sort the list - you are returning a set, which is unsorted by definition :)

Since you are already getting your lists correctly, I see no reason for converting them to sets.

Try this:

a = list(filter(lambda x:x%2!=0,numbers))
b = list(filter(lambda x:x%2==0,numbers))
a.extend(b)
return a

Answer 5

That's great - you basically implemented "extend" in your own code.

There are always multiple ways of solving the same problem - so as long as your code works, it's a valid solution! :D

Answer 6

def my_sort (numbers):
if len(numbers) == 0:
return "invalid"
elif len(numbers) ==1:
return numbers
else:
a = set(list(filter(lambda x:x%2!=0, numbers )))
b =set (list(filter(lambda x:x%2==0,numbers)))
c=a|b
return c

print (my_sort([1,2,3,4,5,6,7]))