# Creating sums of elements from a list

I have a list $(d_1, d_2, .. d_k)$ and I want to create all sums that I get for adding only two elements for my list $(d_1+d_2, d_1+d_3,...d_{k-1}+d_k)$. The RotateLeft function gives me only some of my sums and I need all of them.

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Welcome to Mathematica.SE! I see that you have been given a lot of answers to your 3 questions, but you did not accept any as most helpful. Please, when you see good Q&A, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. ALSO, remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign - thanks! – Vitaliy Kaurov Sep 4 '12 at 15:09

l = {a, b, c, d};
Plus @@@ Subsets[l, {2}]
(*
{a + b, a + c, a + d, b + c, b + d, c + d}
*)


edit

Some timings

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Come downvote @Verde

l = {a, b, c, d}

l~Subsets~{2}~Total~{2}

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(deleted comment) – Dr. belisarius Sep 4 '12 at 12:41
This one is six times faster than the Plus @@@ solution, twice as fast as the Outer solution, and three times faster than the Table solution. – whuber Sep 4 '12 at 12:50
And ten times slower than Total /@ Subsets[l, {2}] – Dr. belisarius Sep 4 '12 at 13:39
@whuber l = RandomReal[1, 10^3]; {(Timing@(l~Subsets~{2}~Total~{2}))[[1]], (Timing[ Plus @@@ Subsets[l, {2}]])[[1]]} -> {2.89, 0.547} ... – Dr. belisarius Sep 4 '12 at 13:43
@Verde You're right: that's astounding, given how similar the Total /@ and ~Total~2 constructs are!. – whuber Sep 4 '12 at 14:34

Just to show that there's more than one way to do things in Mathematica:

test = {a, b, c, d, e};
Total /@ (Join @@ MapIndexed[Drop[#1, First[#2]] &,
Outer[List, test, test]])
{a + b, a + c, a + d, a + e, b + c, b + d, b + e, c + d, c + e, d + e}


Of course, Oleksandr's and Verde's suggestions are the more compact way of going about it.

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Something like :

data = {a, b, c, d};

Flatten[Table[data[[i]] + data[[j]], {i, 1, Length[data] - 1}, {j, i + 1, Length[data]}],1]

(* {a + b, a + c, a + d, b + c, b + d, c + d} *)


Alternatively (plus suggestion from @Oleksandr R.) :

Total /@ Subsets[data, {2}]


And just because RotateLeft was mentioned :

Union[Flatten[Total /@ Subsets[NestList[RotateLeft[#] &, data, Length[data] - 1], {2}], 1]]

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The third one seems a memory hog :) – Dr. belisarius Sep 4 '12 at 14:52
l = {a, b, c, d};


Let's make use of pattern matching ( even though there are faster methods especially for list manipulations) :

ReplaceList[ l, {___, x_, ___, y_, ___} -> x + y]

{a + b, a + c, a + d, b + c, b + d, c + d}


Typically, efficiency of pattern matching solutions is worse than that of functional approach, nevertheless we point out a remarkable feature of the result of ReplaceList: it is identical with other (functional) methods, e.g. (taking a longer list) we have:

ls = {a, b, c, d, e, f, g, h, i, j, k, l, , m, n, o, p, q, r, s};

ReplaceList[ls, {___, x_, ___, y_, ___} -> x + y] ==
Plus @@@ Subsets[ls, {2}] == ls~Subsets~{2}~Total~{2}

True

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Since somebody mentioned timings...

Module[{x = Outer[Plus, l, l]},
Flatten[x[[#, # + 1 ;;]] & /@ Range[Length@x - 1]]]

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