Re: Pure function in a pure function (again)

*To*: mathgroup at smc.vnet.net*Subject*: [mg69818] Re: [mg69797] Pure function in a pure function (again)*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Sat, 23 Sep 2006 23:45:19 -0400 (EDT)*Reply-to*: hanlonr at cox.net

data=Array[a,{15,5}]; Table[data[[n,1]]=Random[Integer,{1,5}],{n,Length[data]}]; Needs["Statistics`DataManipulation`"]; f[x_] := Plus@@Column[Select[data, #[[1]] == x &], 2] ; Given your description, presumably you meant Plus@@f/@Range[3] a[1, 2] + a[2, 2] + a[3, 2] + a[6, 2] + a[9, 2] + a[14, 2] This can be done without a helper function (and without the add-on package) as Plus@@Cases[data,x_?(MemberQ[Range[3],#[[1]]]&):>x[[2]]] a[1, 2] + a[2, 2] + a[3, 2] + a[6, 2] + a[9, 2] + a[14, 2] %==%% True Bob Hanlon ---- Skirmantas Janusonis <janusonis at psych.ucsb.edu> wrote: > Hello, > > I've already asked this question, but I thought there was a better way to state it. If Dat is a 2D-table, the following picks out the rows whose first element is 1,2 or 3 and adds up their the second elements: > > f[x_] := Plus @@ Column[Select[Dat, #[[1]] == x &], 2] > f /@ Range[3] > > Instead of f/@Range[3], I'd like to use a second pure function, something like this: > > Map[Plus @@ Column[Select[Dat, #1[[1]] == #2 &], 2]&,Range[3]] > > This line is incorrect, however. How do I fix it? > > Thanks! > > Skirmantas >