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I need to create an array subtracting two arrays from another array. I know how to subtract one:

am = DeleteCases[amp, Alternatives @@ af]

I can subtract an array each time but there must be a way to subtract more than one array at once. So I want to subtract af and ai from amp. How can I do that at once?

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5 Answers 5

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You can use Select as well :

ar = Range[10];
ar1 = Range[1, 10, 2];
ar2 = Range[1, 10, 3];

Select[ar, Not[MemberQ[Union[ar1, ar2], #]] &]
(* {2, 6, 8} *)

DeleteCases[ar, _?(MemberQ[Union[ar1, ar2], #] &)]
(* {2, 6, 8} *)
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    $\begingroup$ Or, DeleteCases[ar, Alternatives @@ Union[ar1, ar2]]. $\endgroup$
    – VLC
    Feb 25, 2013 at 10:26
  • $\begingroup$ Not@MemberQ is equivalent to FreeQ. $\endgroup$
    – chyanog
    Mar 11, 2013 at 11:18
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The built-in function Complement does exactly what you need:

enter image description here

With

ar = Range[10];
ar1 = Range[1, 10, 2];
ar2 = Range[1, 10, 3];

elements in the first list that are not in any of the subsequent lists are:

Complement[ar, ar1,ar2]
(* {2, 6, 8 *)
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a = {9, 4, 7, 1, 5, 3, 8, 5, 7, 3, 1, 0};
b = {7, 5};
c = {8, 3};

DeleteElements (new in 13.1) keeps duplicates and is equal with the accepted answer

DeleteElements[a, Join[b, c]]

{9, 4, 1, 1, 0}

UniqueElements (also new in 13.1) deletes duplicates

Join @@ UniqueElements[{a, b, c}]

{9, 4, 1, 0}

Compare to other answers:

kglr's solution sorts the elements and deletes duplicates

Complement[a, b, c]

{0, 1, 4, 9}

The accepted answer doesn't sort and keeps duplicates

Select[a, Not[MemberQ[Union[b, c], #]] &]

{9, 4, 1, 1, 0}

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Using eldo's data:

a = {9, 4, 7, 1, 5, 3, 8, 5, 7, 3, 1, 0};
b = {7, 5};
c = {8, 3};

An alternative is to use ReplaceAll:

a /. Thread[Join[b, c] -> Nothing]

(*{9, 4, 1, 1, 0}*)
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Grabbing the data from @eldo we can use UniqueElements which was introduced in v13.1

a = {9, 4, 7, 1, 5, 3, 8, 5, 7, 3, 1, 0};
b = {7, 5};
c = {8, 3};
Flatten[UniqueElements[{a, b, c}]]

{9, 4, 1, 0}

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