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I'm simply looking for the {}+{} function for sublist elements of a list: (Parallel sums of elements)

c = {{1, 2, 3, 4}, {8, 7, 6, 5}};
Plus[c]
Mean[c]  
desired={9, 9, 9, 9}

Where Mean does directly what it's supposed to, unfortunately Plus doesn't. I tried Take and Part but they both do not seem to work here . Additionally, my c consists of a varying number of elements. I tried to work with Length[c] and Range .. ({}+{} is no solution as list element number varies) thank you for help!!

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4 Answers 4

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Use either 

Plus@@c

or

Total[c]
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    $\begingroup$ For completeness: if one treats the list of lists as a matrix, Total[c] and Total[c, {1}] sums columns, Total[c, {2}] sums rows, and Total[c, 2] sums all elements. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Feb 15, 2012 at 23:29
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    $\begingroup$ @J.M. Good points. For once, I did not try to be complete :) $\endgroup$
    – Leonid Shifrin
    Commented Feb 15, 2012 at 23:35
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    $\begingroup$ Leonid, are you ill? :o) $\endgroup$
    – Mr.Wizard
    Commented Feb 16, 2012 at 5:10
  • $\begingroup$ @Mr.Wizard Looks like that :) $\endgroup$
    – Leonid Shifrin
    Commented Feb 16, 2012 at 13:36
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Before version 5, I used Tr a lot for summing vectors as Total wasn't available:

c = {{1, 2, 3, 4}, {8, 7, 6, 5}};
Tr /@ Transpose[c]

(*
==> {9, 9, 9, 9}
*)
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7
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Use

Apply[Plus, {{1, 2, 3, 4}, {8, 7, 6, 5}}]
(*
==> {9, 9, 9, 9}
*)

or its shortcut 

Plus @@ {{1, 2, 3, 4}, {8, 7, 6, 5}}

Another way is 

MapThread[Plus, {{1, 2, 3, 4}, {8, 7, 6, 5}}]
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4
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Here is another method that I hope will be instructive.

Sum[i, {i, c}]

{9, 9, 9, 9}

In TraditionalForm:

Mathematica graphics

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