Date: May 23, 2013 3:55 AM
Author: Bob Hanlon
Subject: Re: Average the same elements of the list

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

On Wed, May 22, 2013 at 2:18 AM, BBabic <bipsich101@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!

>

>