Can anyone help me to optimize this piece of code?

Question

I want to store even number and odd number in a separate list. But, here I am facing a unique problem. I am able to store it in sets but not in lists. Is there a way wherein I can store these in a List without repetition.

I have tried this in Jupyter notebook

list_loop=[1,2,3,4,5,6,7,8,9,10,11,12,13,1,4,1,51,6,17,]
for i in list_loop:
    if i % 2 == 0 :
        list_even = list_even + [i]
    else:
        list_odd = list_odd + [i]
print(set(list_even))
print(set(list_odd))

Expected output:

[2,4,6,8,10,12]
[1,3,5,7,9,11,13,17,51]

You haven't said what the actual output is, or why you don't like it. — Ned Batchelder, Apr 07 '19 at 11:31
See also: https://stackoverflow.com/questions/11233355/generating-a-list-of-even-numbers-in-python — Jens, Apr 07 '19 at 11:34

Lante Dellarovere · Answer 1 · 2019-04-07T12:58:22.457

0

You can apply the list function to your set object in order to convert it to a list.

list_from_set = list(set(list_even))
>>> print(list_from_set)
[2, 4, 6, 8, 10, 12]

edited Apr 07 '19 at 12:58

answered Apr 07 '19 at 11:34

Lante Dellarovere

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jdowner · Answer 2 · 2019-04-07T11:44:52.067

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There are a couple of ways you could do this. You could use the OrderedDict in the collections library, or you could just sort the set and get a list,

...
print(sorted(set(list_even)))
print(sorted(set(list_odd)))

Also, I would personally create those lists using a set comprehension

list_even = sorted({x for x in list_loop if x % 2 == 0})
list_odd = sorted({x for x in list_loop if x % 2 == 1})

edited Apr 07 '19 at 11:44

answered Apr 07 '19 at 11:35

jdowner

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Jens · Answer 3 · 2019-04-07T11:43:22.587

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Use a comprehension

>>> list_loop=[1,2,3,4,5,6,7,8,9,10,11,12,13,1,4,1,51,6,17,]
>>> print(list(set(_ for _ in list_loop if _ % 2)))
[1, 3, 5, 7, 9, 11, 13, 17, 51]

Similarly for even numbers.

edited Apr 07 '19 at 11:43

answered Apr 07 '19 at 11:36

Jens

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I guess he does not want repetitions – Lante Dellarovere Apr 07 '19 at 11:39
I guess he should specify that, then – Jens Apr 07 '19 at 11:40
he did. "Is there a way wherein I can store these in a List without repetition." – Lante Dellarovere Apr 07 '19 at 11:41

glhr · Answer 4 · 2019-04-07T11:41:22.080

0

Define list_odd and list_even as lists and don't convert them to sets before printing. Note that you can use list comprehension to fill list_odd and list_even:

list_odd = []
list_even = []

list_loop=[1,2,3,4,5,6,7,8,9,10,11,12,13,1,4,1,51,6,17,]
list_odd = [elem for elem in list_loop if elem % 2 != 0]
list_even = [elem for elem in list_loop if elem % 2 == 0]

print(list_even)
print(list_odd)

Output:

[2, 4, 6, 8, 10, 12, 4, 6]
[1, 3, 5, 7, 9, 11, 13, 1, 1, 51, 17]

Edit: for uniqueness, turn list_loop into a set:

list_loop=set([1,2,3,4,5,6,7,8,9,10,11,12,13,1,4,1,51,6,17,])

Output:

[2, 4, 6, 8, 10, 12]
[1, 3, 5, 7, 9, 11, 13, 17, 51]

edited Apr 07 '19 at 11:41

answered Apr 07 '19 at 11:37

glhr

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This makes no sense. Why use a list comprehension, throw it away while building a new list inside of that same comprehension? – Jens Apr 07 '19 at 11:38
Not sure what I was thinking there. Edited. – glhr Apr 07 '19 at 11:40
2

The expected output indicates that the numbers in the lists should be unique. – jdowner Apr 07 '19 at 11:40

score 0 · Answer 5 · answered Apr 07 '19 at 11:49

You can solve this using a list comprehension with a filter condition - but you then iterate your list twice.

By using a simple for loop you only need to touch any number once at it will conserve the original order - what putting your numbers through a set might not do - order in a set is not guaranteed:

Keep a set of seen numbers, only add anything if your current number was not yet seen.

list_loop = [1,2,3,4,5,6,7,8,9,10,11,12,13,1,4,1,51,6,17,]

list_even = []
list_odd = [] 
seen = set()

trick = [list_even, list_odd]  # even list is at index 0, odd list at index 1

for i in list_loop:
    if i in seen: 
        continue
    else:
        seen.add(i)
                                 # the trick eliminates the need for an if-clause
        trick[i%2].append(i)     # you use i%2 to get either the even or odd index


print(list_even)
print(list_odd)

Output:

[2, 4, 6, 8, 10, 12]
[1, 3, 5, 7, 9, 11, 13, 51, 17]

Can anyone help me to optimize this piece of code?

5 Answers5