6
$\begingroup$
{
 a -> 1,
 a -> 2,
 b -> 3,
 b -> 4
 }

You can see that both a and b have two values, so how do I add these two vaule?Like this

{a->3,b->7}
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5 Answers 5

12
$\begingroup$

Merge[Total]@{a -> 1, a -> 2, b -> 3, b -> 4}

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1
  • 6
    $\begingroup$ Or preface with Normal@ to convert to rules. $\endgroup$
    – Bob Hanlon
    Jan 6, 2023 at 6:25
7
$\begingroup$

Another suggested solution is suggested below.

The logic is to transform the list of rules into a list of lists, sort it and then sum over.

With

lst = {a -> 1, a -> 2, b -> 3, b -> 4};

we do

Rule @@@ ({#[[1, 1]], Total[#[[All, 2]]]} & /@ 
   GatherBy[Sort[List @@@ lst], First])

having borrowed from Syed

The above returns

res

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7
$\begingroup$

Just another way to do this using GroupBy:

Map[Part[#, 1] -> Total@Part[#, 2, All, 2] &, Normal@GroupBy[lst, Keys]]
(*{a -> 3, b -> 7}*)

Or in a compact way using the third argument of GroupBy:

Normal@GroupBy[lst, Keys,Total@#[[All, 2]] &]
(*{a -> 3, b -> 7}*)
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5
$\begingroup$
Reap[MapApply[Sow[#2,#1]&]@rules,_,#1->Total@#2&][[2]]

(* {a -> 3, b -> 7} *) 

Original Answer

Reap[Sow[Last@#,First@#]&/@rules,_,#1->Total[#2]&][[2]]
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2
$\begingroup$
list = {a -> 1, a -> 2, b -> 3, b -> 4};

A variant of E. Chan-López GroupBy-solution

GroupBy[list, First -> Last, Total]

<|a -> 3, b -> 7|>

KeyValueMap[List][%]

{{a, 3}, {b, 7}}

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