2
$\begingroup$

Suppose I have the following two lists: 

l1 = {{"apple", "banana", "cream"}, {"beef", "pepper", 
   "salt"}, {"wine", "beer", "beef", "olive"}}
l2 = {"beef:salt:pepper", "apple:cream", "wine:beer:beef:olive"}

For each element in l2 I want to go through l1 and select the element in l2 if it exists in l1, such that the output would be: 

{"beef:salt:pepper", "wine:beer:beef:olive"}

I know generally one should use Intersection[l1, l2] yet the format of two lists are different so it turns an empty list. I wonder how one go around this?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

You can reformat l2, so that you have it in the same format as l1, using StringSplit:

reformatted = StringSplit[#, ":"] & /@ l2

{{"beef", "salt", "pepper"}, {"apple", "cream"}, {"wine", "beer",
"beef", "olive"}}

Intersection[l1, reformatted]

{{"wine", "beer", "beef", "olive"}}

You'll notice that {"beef", "pepper", "salt"} was left out, because order matters. If you don't want to consider the order in which elements appear, you can fix it like this:

Intersection[l1, reformatted, SameTest -> (Sort[#1] === Sort[#2] &)]

{{"beef", "pepper", "salt"}, {"wine", "beer", "beef", "olive"}}

Improve this answer
$\endgroup$
1
  • $\begingroup$ One could also include the reformatting directly in the SameTest, which would allow to work directly with the original lists $\endgroup$
    – Lukas Lang
    Jul 31, 2019 at 11:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.