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I have pairs of numbers {a1, b1}, {a2, b2}, {a3, b3}, ..., {aN, bN}

What is the shortest way to compute and plot {a1, b1}, {a2, b1+b2}, {a3, b1+b2+b3},....?

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    $\begingroup$ {as, bs} = Transpose[pairs]; result = Transpose[{as, Accumulate[bs]}] $\endgroup$ – user484 Nov 14 '15 at 5:08
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    $\begingroup$ Or, Transpose[MapAt[Accumulate, Transpose[pairs], 2]]. $\endgroup$ – J. M.'s technical difficulties Nov 14 '15 at 7:20
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Accumulate@Array[b, {3}]
(* {b[1], b[1] + b[2], b[1] + b[2] + b[3]} *)

therefore:

{a, b} = Transpose[list];
Transpose[{a, Accumulate[b]}]

Also this will do the job:

Rest@FoldList[{#2[[1]], #1[[2]] + #2[[2]]} &, {0, 0}, list]

or even easier

list[[All,2]]=Accumulate@list[[All,2]]; list
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FoldList[{0, 1} # + #2 &, list]
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thanks to Mike Honeychurch, without pure function:

list = {{a1, b1}, {a2, b2}, {a3, b3}};
Thread[{Thread[list][[1]], Rest@FoldList[Plus, 0, Thread[list][[2]]]}]
(* {{a1, b1}, {a2, b1 + b2}, {a3, b1 + b2 + b3}} *)

and shorter (one expression):

Thread[{list[[All, 1]], Accumulate[list][[All, 2]]}]
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  • $\begingroup$ Could you please explain a little bit more how Rest@FoldList structure works? $\endgroup$ – Maria Nov 14 '15 at 17:30
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    $\begingroup$ @Maria. Rest just removes first element in the list, i.e. Rest[{1, 2, 3}] gives {2, 3}. @ is a shorthand for head application. So Rest@FoldList[...] is the same as Rest[FoldList[...]]. Finally, FoldList[Plus, 0, {a, b, c}] gives {0, a, a + b, a + b + c}. Now you may see why we need Rest - just to get rid of 0. $\endgroup$ – garej Nov 14 '15 at 17:55

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