CODEWITHSUNDEEP
pythonlistsorting
BattleDrum
BattleDrum

Reputation: 818

Sorting a list based on whether item is odd or even

I am trying to sort a list of numbers depending on whether it is odd or even (even gains higher priority). Example: 

a=[1,2,3,4] 
a.sort(key=org(a)) sorted will produce [2,4,1,3]. I want to use the sort method 

def org(a):
    for i in range(len(a)):
        if a[i]%2==0:
            b.append(a[i])
            b.sort()
    else:
        c.append(a[i])
        c.sort()
    print(b+c)

I got this error from running a.sort(key=org(a))

Traceback (most recent call last):
File "<pyshell#80>", line 1, in <module>
a.sort(key=org(a))
TypeError: 'list' object is not callable

I realize that sorting it every time makes it slow. Which other way can I do this without having to sort after every loop?

Upvotes: 2

Views: 2396

Answers (3)

John La Rooy
John La Rooy

Reputation: 304137

To sort by "evenness" and then by magnitude, you can use a function that returns a tuple

>>> a=[1,2,3,4] 
>>> a.sort(key=lambda x: (x % 2, x))
>>> a
[2, 4, 1, 3]

To sort the odd entries first, you can simply negate the value of the modulus. This is a useful trick for reversing the sort of numeric fields in general.

>>> a.sort(key=lambda x:(-(x % 2), x))
>>> a
[1, 3, 2, 4]

Upvotes: 7

LetzerWille
LetzerWille

Reputation: 5658

one can use helper function from intertools 

def partition_by_pred(pred, iterable):
    from itertools import tee, filterfalse
    t1, t2 = tee(iterable)
    return filterfalse(pred, t1), filter(pred, t2)

l, r = partition_by_pred(lambda x: False if x % 2 == 0  else True ,[1,2,3,4]  )

print(*l,*r)

2 4 1 3

Upvotes: 0

rofls
rofls

Reputation: 5115

It looks like you were on the right track:

a = [1,2,3,4]
>>> sorted(a,key=lambda x: x%2)
[2, 4, 1, 3]

Here sorted is a function that returns a sorted list. You can also use the sort method, as John said. Apparently if your original list is not sorted, sorted will not work, so you then you would have to do:

a = [3,2,1,4]
a.sort()
newList = sorted(a,key=lambda x: x%2)

at which point John's code is probably cleaner. His can also be implemented with the sorted function though, by doing sorted(a,key=lambda x: (x%2,x)).

Upvotes: 2

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