Python List Comprehension and 'not in'

Question

I'm getting started with Python and is currently learning about list comprehensions so this may sound really strange.

Question: Is it possible to use list comprehension to create a list of elements in t that is not found in s?

I tried the following and it gave me an error:

>>> t = [1, 2, 3, 4, 5]
>>> s = [1, 3, 5]
>>>[t for t not in s]

[t for t not in s]
           ^
SyntaxError: invalid syntax

A comprehension can only have two kinds of clauses: `for spam in eggs` and `if bacon`. If you can figure out how to write what you want in terms of a `for` loop or an `if` check, you're set. (In this case, you can.) If not, you either need to factor pieces out into functions, or write an explicit loop block statement instead of a comprehension. — abarnert, Oct 22 '13 at 01:51

Lucas Ribeiro · Accepted Answer · 2013-10-22T02:04:03.787

71

Try this:

[x for x in t if x not in s]

You can nest any for if statements in list comprehensions. Try this identation, to get really long chains of conditionals, with a clearer intuition about what the code is doing.

my_list = [(x,a)
           for x in t
           if x not in s
           if x > 0
           for a in y
           ...]

See?

edited Oct 22 '13 at 02:04

answered Oct 22 '13 at 01:49

Lucas Ribeiro

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score 6 · Answer 2 · answered Oct 22 '13 at 01:51

6

[item  for item  in t if item not in s]

answered Oct 22 '13 at 01:51

Leonardo.Z

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This is identical to [Lucas' answer](http://stackoverflow.com/a/19507719/119527) which was given earlier, and therefore completely redundant. – Jonathon Reinhart Oct 22 '13 at 01:52
6

slow typing is sin for programmers – Leonardo.Z Oct 22 '13 at 02:02
2

That's why I didn't down-vote. Unfortunately, you'll often find that Stack Overflow is a race to the first answer that is correct. – Jonathon Reinhart Oct 22 '13 at 02:04

score 2 · Answer 3 · answered Oct 22 '13 at 01:55

2

I know you're asking about list comprehensions, but I wanted to point out that this specific problem would be better accomplished using sets. The result you want is the difference of set t and s:

>>> t = {1,2,3,4,5}
>>> s = {1,3,5}
>>>
>>> t - s
set([2, 4])
>>>
>>> t.difference(s)
set([2, 4])

Just hoping to expand your knowledge of the tools Python provides to you.

answered Oct 22 '13 at 01:55

Jonathon Reinhart

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This may not be useful if `t` needs to preserve order and any repeated elements... Also... only one needs to be a set here - possibly use the smaller to start with, eg: `set(s).symmetric_difference(t)` – Jon Clements Oct 22 '13 at 01:57
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@JonClements Great insight. I appreciate the feedback. – Jonathon Reinhart Oct 22 '13 at 02:00
Indeed, only one needs to be a set: `"Note, the non-operator versions of union(), intersection(), difference(), and symmetric_difference(), issubset(), and issuperset() methods will accept any iterable as an argument. In contrast, their operator based counterparts require their arguments to be sets."` – Jonathon Reinhart Oct 22 '13 at 02:01

score 1 · Answer 4 · answered Oct 22 '13 at 01:54

1

For better efficiency, use a set:

mySet = set(s)
result = [x for x in t if x not in mySet]

Testing membership of a set is done is O(1), but testing membership in a list is O(n).

answered Oct 22 '13 at 01:54

MangoHands

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This calls `set` multiple times. – falsetru Oct 22 '13 at 01:56
Actually, this result will be more slow, because will create a set of s too. Each time s if checked, i guess – Lucas Ribeiro Oct 22 '13 at 01:57
And, just another way: `list(ifilterfalse(set(s).__contains__, t))` – Jon Clements Oct 22 '13 at 01:59

Python List Comprehension and 'not in'

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