Delete elements of a list that appear in another list

Question

Basically I need to do a set substraction operation A-B=C between two lists. For example, I have two lists A and B:

A = {{1,2,3,4},{3,4,3,4},{1,1,1,5},{1,2,1,2},{4,3,2,1}} 

B = {{1,2,1,2}, {1,2,3,4}}

and the result would be all the elements of A that do not appear in B and keeping the order of A:

C  = {{3,4,3,4},{1,1,1,5},{4,3,2,1}}

It is important to know that list B could be empty in some cases

I am also avoiding loops or recursion. Thank you all <3.

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Vitaliy Kaurov · Accepted Answer · 2017-10-02 19:44:39Z

8

As @Szabolcs said with no order:

Complement[A, B]

{{1, 1, 1, 5}, {3, 4, 3, 4}, {4, 3, 2, 1}}

but to keep order you can try:

DeleteCases[A, Alternatives @@ B]

{{3, 4, 3, 4}, {1, 1, 1, 5}, {4, 3, 2, 1}}

You might need too look up levelspec option for DeleteCases depending on a situation. Some other ways to keep order

Delete[A, Position[A, Alternatives @@ B]]

Select[A, ! MemberQ[B, #] &]

with perhaps a few more others.

answered Oct 2, 2017 at 19:26

Vitaliy Kaurov

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$\begingroup$ What about if list B is empty? $\endgroup$
– Gabriela
Commented Oct 2, 2017 at 20:53
1

$\begingroup$ @Gabriela did you check if these methods work with empty list? $\endgroup$
– Vitaliy Kaurov
Commented Oct 2, 2017 at 21:38
1

$\begingroup$ @Gabriela, the nice thing about Mathematica is that as long as you know the function to use, questions on how it would behave for a particular case can be easily tried out by doing the experiments yourself. $\endgroup$
– J. M.'s missing motivation ♦
Commented Oct 2, 2017 at 23:41
$\begingroup$ @VitaliyKaurov yeah it works :D It did not work at first but was a mistake of mine, thank you so much. $\endgroup$
– Gabriela
Commented Oct 3, 2017 at 2:10

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eldo · Accepted Answer · 2024-01-15 00:25:42Z

2

a = {{1, 2, 3, 4}, {3, 4, 3, 4}, {1, 1, 1, 5}, {1, 2, 1, 2}, {4, 3, 2, 1}};

b = {{1, 2, 1, 2}, {1, 2, 3, 4}};

Using DeleteElements and UniqueElements (both new in 13.1)

DeleteElements[a, b]

{{3, 4, 3, 4}, {1, 1, 1, 5}, {4, 3, 2, 1}}

First @ UniqueElements[{a, b}]

{{3, 4, 3, 4}, {1, 1, 1, 5}, {4, 3, 2, 1}}

"Both are much slower than Complement, but keep the sort order.

answered Jan 15 at 0:25

eldo

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2 Answers 2