Average the same elements of the list

[BBabic]

Hello,
I have list which is something like
data={
{{a1,b1,c1},d1},{{a2,b2,c2},d2}}
I would like to get new list which gets average of the second elements if the first elements in the sublists are all the same.
Namely if a1=a2,b1=b2,c1=c2
new list would look like
datanew={{a1,b1,c1},Mean[{d1,d2}]
Is there an elegant way to do this ?
Thanks!

[BBabic]

[Tomas Garza]

Elegance is often a matter of individual taste (it lies in the eyes of the
beholder?), but a possible way is
datanew:=If[data[[1,1]]==data[[2,1]],{data[[1,1]],Mean[{data[[1,2]],data[[2,2]]}]},data]
data={{{1,2,3},4},{{1,2,3},5}}{{{1,2,3},4},{{1,2,3},5}}
datanew{{1,2,3},9/2}
data={{{1,1,3},4},{{1,2,3},5}}{{{1,1,3},4},{{1,2,3},5}}
datanew{{{1,1,3},4},{{1,2,3},5}}
-Tomas
> From: bipsi...@gmail.com
> Subject: Average the same elements of the list
> To: math...@smc.vnet.net
> Date: Wed, 22 May 2013 02:18:09 -0400

[Ray Koopman]

datanew = {data[[1,1]], Mean@Pick[data[[All,2]],data[[All,1]],data[[1,1]]]}

[Murray Eisenberg]

You didn't specify what to return in case the first elements of the
sublists are NOT all the same; let's return the entire original list in
this case. Then perhaps something like the following will do.

op[lis_] :=
If[And @@ MapThread[Equal, First /@ lis],
{First@First@lis, Mean[Last /@ lis]},
lis]

op[{ {{2, 5, -1}, d1}, {{2, 5, -1}, d2}}]
(* { {2, 5, -1}, (d1 + d2)/2 } *)

op[{{{2, 5, -1}, d1}, {{2, 9, -1}, d2}}]
(* {{{2, 5, -1}, d1}, {{2, 9, -1}, d2}} *)

Undoubtedly the code can be shortened.

On May 22, 2013, at 2:18 AM, BBabic <bipsi...@gmail.com> wrote:

> Hello,
> I have list which is something like
> data={
> {{a1,b1,c1},d1},{{a2,b2,c2},d2}}
> I would like to get new list which gets average of the second elements if the first elements in the sublists are all the same.
> Namely if a1=a2,b1=b2,c1=c2
> new list would look like
> datanew={{a1,b1,c1},Mean[{d1,d2}]
> Is there an elegant way to do this ?
> Thanks!

---
Murray Eisenberg mur...@math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2838 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305

[Bill Rowe]

On 5/22/13 at 2:18 AM, bipsi...@gmail.com (BBabic) wrote:

>Hello, I have list which is something like data={
>{{a1,b1,c1},d1},{{a2,b2,c2},d2}} I would like to get new list which
>gets average of the second elements if the first elements in the
>sublists are all the same. Namely if a1=a2,b1=b2,c1=c2 new list
>would look like datanew={{a1,b1,c1},Mean[{d1,d2}] Is there an
>elegant way to do this ? Thanks!

Something like

{#[[1,1]],Mean[#[[All,-1]]]}/@GatherBy[data/.{a__,d_}->{{a},d}]

should work

[Bob Hanlon]

data = {{{a1, b1, c1}, d1}, {{a2, b2, c2}, d2},
{{a1, b1, c1}, d3}, {{a2, b2, c2}, d4},
{{a1, b1, c1}, d5}, {{a3, b3, c3}, d7}};


For v7 and later


sol = Plus @@ #/Length[#] & /@
GatherBy[data, First]


{{{a1, b1, c1}, (1/3)*(d1 + d3 + d5)},
{{a2, b2, c2}, (d2 + d4)/2}, {{a3, b3, c3}, d7}}


or (probably faster for longer lists)


sol == ({#[[1, 1]], Mean[#[[All, 2]]]} & /@
GatherBy[data, First])


True


For v3 or later


sol == (Plus @@ #/Length[#] & /@
Split[data // Sort, #1[[1]] == #2[[1]] &])


True



Bob Hanlon

[Daniel]

data={{{a1,b1,c1},d1},{{a2,b2,c2},d2}};
{#[[1, 1]], #[[All, 2]] // Mean} & /@ SplitBy[data, First]

output:

{{{a1, b1, c1}, d1}, {{a2, b2, c2}, d2}}

data={{{a1,b1,c1},d1},{{a1,b1,c1},d2}};
{#[[1, 1]], #[[All, 2]] // Mean} & /@ SplitBy[data, First]

output:
{{{a1, b1, c1}, (d1 + d2)/2}}