"for" loop and "if" condition for list creation in python

Question

source=[1,2,3,4,2,3,5,6]

dst=[]
for item in source:
    if item not in dst:
        dst.append(item)

print(dst) # [1,2,3,4,5,6]

Can I simplify code above something like this:

dst=[item for item in [1,2,3,4,2,3,5,6] if item not in 'this array']

Thanks

No, since you do not have a reference to "this" array in advance. Furthermore this uniqueness filter can be improved in case the items are hashable. — Willem Van Onsem, Dec 19 '17 at 10:56
you can just use `set()` for it. you will get all unique items only — Gahan, Dec 19 '17 at 10:57

score 6 · Answer 1 · answered Dec 19 '17 at 10:58

6

No, list comprehensions cannot be self-referential.

You seem to want to remove duplicates from a list. See this and this questions for a boatload of approaches to this problem.

answered Dec 19 '17 at 10:58

timgeb

76,762
20
123
145

grovina · Accepted Answer · 2017-12-19T16:45:14.647

4

A set is probably what you are looking for, since you cannot refer to this array while it's being created:

>>> source = [1,2,3,4,2,3,5,6]
>>> set(source)
{1, 2, 3, 4, 5, 6}

If you do want to keep original order, though, you can keep track of what you have already added to dst with a set (seen):

>>> source = [1,2,3,4,2,3,5,6]
>>> seen = set()
>>> dst = []
>>> for i in source:
>>>     if i not in seen:
>>>         dst.append(i)
>>>         seen.add(i)
>>>
>>> dst
[1, 2, 3, 4, 5, 6]

edited Dec 19 '17 at 16:45

answered Dec 19 '17 at 10:58

grovina

2,999
19
25

1

Not that sets are unordered and may loose your original ordering (actually the fact they _sometimes_ don't loose it is an implementation accident). – bruno desthuilliers Dec 19 '17 at 11:00
@brunodesthuilliers good point. I've complemented the solution with an approach considering original order. – grovina Dec 19 '17 at 11:07
list(set(source)) - quickest way :) – Dec 19 '17 at 11:42
PavelPerminov you don't guarantee original order with `list(set(source))`. – grovina Dec 19 '17 at 16:43

score 3 · Answer 3 · answered Dec 19 '17 at 11:05

You can't reference dst from within the list comprehension, but you can check the current item against the previously iterated items in source by slicing it on each iteration:

source = [1, 2, 3, 4, 2, 3, 5, 6]
dst = [item for i, item in enumerate(source)
       if item not in source[0:i]]

print(dst)  # [1, 2, 3, 4, 5, 6]

score 0 · Answer 4 · answered Dec 19 '17 at 11:03

0

If using if and for is your requirement How about this?

[dst.append(item) for item in source if item not in dst]

answered Dec 19 '17 at 11:03

Sagun Shrestha

1,188
10
23

Gahan · Answer 5 · 2017-12-19T11:12:04.257

Well instead of creating new list you can modify your existing list with list comprehension as shown below:

In [1]: source
Out[1]: [1, 9, 2, 5, 6, 6, 4, 1, 4, 11]

In [2]: [ source.pop(i) for i in range(len(source))[::-1] if source.count(source[i]) > 1 ]
Out[2]: [4, 1, 6]

In [3]: source
Out[3]: [1, 9, 2, 5, 6, 4, 11]

As another approach you can first get unique list with set and then sort it with reference to source index value as follow:

source = [1, 9, 2, 5, 6, 6, 4, 1, 4, 11]
d = list(set(source))
d.sort(key=source.index)
print(d)  # [1, 9, 2, 5, 6, 4, 11]

"for" loop and "if" condition for list creation in python

5 Answers5