15

I have two lists, l1 and l2. I need items from l1 which are not in l2.

l1 = [2, 3, 4, 5]
l2 = [0, 1, 2, 3]

I want to get only [4,5] - only new values in l1.

[i for i in l1 if not i in l2 ]

Can I do that without iteration?

Peter Mortensen
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Pol
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    Sounds like premature optimization. Why do you say 'no iteration' if this is fundamentally an iterative problem? – the wolf Mar 15 '11 at 05:21

10 Answers10

17

Short answer, yes: list(set(l1) - set(l2)), though this will not keep order.

Long answer, no, since internally the CPU will always iterate. Though if you use set() that iteration will be done highly optimized and will be much faster then your list comprehension (not to mention that checking for membership value in list is much faster with sets then lists).

orlp
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  • Someone told me that inline iteration is highly optimized as well. And its different from what it perfor when there is: for i in l1: do() – Pol Mar 14 '11 at 22:15
  • List comprehensions (not inline iteration) are indeed more optimized than a for loop, but set difference is probably done in high speed C which will be even better. And because you are using sets the `in` keyword (or it's C cousin) is blazing fast thus increasing speed even more. – orlp Mar 14 '11 at 22:18
12

You can't do it without iteration. Even if you call a single method, internally that will iterate.

Your approach is fine for a small list, but you could use this approach instead for larger lists:

s2 = set(l2)
result = [i for i in l1 if not i in s2 ]

This will be fast and will also preserve the original order of the elements in l1.

Peter Mortensen
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Mark Byers
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3

If you don't care about the order of the elements, you can use sets:

l1 = set([2, 3, 4, 5])
l2 = set([0, 1, 2, 3])
print l1 - l2

prints

set([4, 5])
Sven Marnach
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  • Yep.. I was trying doing something similar. I did not wark for me. Now it's ok. Thanks – Pol Mar 14 '11 at 22:12
2

The conversion to sets is great when your list elements can be converted to sets. Otherwise you'll need something like Mark Byers' solution. If you have large lists to compare you might not want to pay the memory allocation overhead and simplify his line to:

[l1.remove(m) for m in l1 if m in l2]
Community
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user960249
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  • Welcome to SO. Maybe you should reconsider your answer, because the OP seems to want an iteration-free solution. I guess he means no for-loop even not as a list-comprehension. So maybe a map or filter solution would be more appropriate. Also i don't think he wants to alter the original list. – Don Question Nov 08 '12 at 22:31
1

You can use use set_1.difference_update(set_2) for in place difference:

>>sl1 = set([2, 3, 4, 5])
>>sl2 = set([0, 1, 2, 3])
>>sl1.difference_update(sl2)
>>sl1
set([4, 5])
Peter Mortensen
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joaquin
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1

You can do this simply as follows:

list( set(l1) - set(l2) )

This should do the trick.

Sam
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0

Using the built-in module set

>>> a = set([1,2,3,4,5])
>>> b = set([1,3,5])
>>> a.difference(b)
set([2, 4])

Another approach

>>> a = set([1,2,3,4,5])
>>> b = [1,3,5]
>>> a.difference(b)
set([2, 4])
Superpaul
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0

Convert them to sets and use the difference operator:

l1=[2,3,4,5]
l2=[0,1,2,3]

answer = set(l1) - set(l2)
Peter Mortensen
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koblas
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0

Simply as it is programming, a simple task can be done in a variety of ways. We can use list comprehension methods like this for exactly the same problem

fruits = ["apple", "banana", "cherry", "kiwi", "mango"]
ddd = ["apple", "banana", "mango"]

newlist = [x for x in fruits if x not in ddd]

print(newlist)
Muhammad Abdullah Nabeel
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-1

Example:

>>a = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
>>b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

We can concatenate as well to get the complete difference:

>>list (set(a) -set(b)) +  list (set(b) -set(a))
>>[89, 34, 21, 55, 4, 6, 7, 9, 10, 11, 12]
Peter Mortensen
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Ashwin
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