5
$\begingroup$

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.

| improve this question | | | | |
$\endgroup$
8
$\begingroup$
l = {a, b, c, d};
Plus @@@ Subsets[l, {2}]
(*
  {a + b, a + c, a + d, b + c, b + d, c + d}
*)

edit

Some timings

Mathematica graphics

| improve this answer | | | | |
$\endgroup$
6
$\begingroup$

Come downvote @Verde

l = {a, b, c, d}

l~Subsets~{2}~Total~{2}
| improve this answer | | | | |
$\endgroup$
  • $\begingroup$ (deleted comment) $\endgroup$ – Dr. belisarius Sep 4 '12 at 12:41
  • 1
    $\begingroup$ 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. $\endgroup$ – whuber Sep 4 '12 at 12:50
  • $\begingroup$ And ten times slower than Total /@ Subsets[l, {2}] $\endgroup$ – Dr. belisarius Sep 4 '12 at 13:39
  • 2
    $\begingroup$ @whuber l = RandomReal[1, 10^3]; {(Timing@(l~Subsets~{2}~Total~{2}))[[1]], (Timing[ Plus @@@ Subsets[l, {2}]])[[1]]} -> {2.89, 0.547} ... $\endgroup$ – Dr. belisarius Sep 4 '12 at 13:43
  • $\begingroup$ @Verde You're right: that's astounding, given how similar the Total /@ and ~Total~2 constructs are!. $\endgroup$ – whuber Sep 4 '12 at 14:34
3
$\begingroup$

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.

| improve this answer | | | | |
$\endgroup$
3
$\begingroup$

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]]
| improve this answer | | | | |
$\endgroup$
  • $\begingroup$ The third one seems a memory hog :) $\endgroup$ – Dr. belisarius Sep 4 '12 at 14:52
3
$\begingroup$
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
| improve this answer | | | | |
$\endgroup$
2
$\begingroup$

Since somebody mentioned timings...

Module[{x = Outer[Plus, l, l]},
 Flatten[x[[#, # + 1 ;;]] & /@ Range[Length@x - 1]]]
| improve this answer | | | | |
$\endgroup$
0
$\begingroup$
Subsets[Plus @@ l, {2}]
(* or Subsets[Total@l, {2}] *)

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

| improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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