
Re: Average the same elements of the list
Posted:
May 23, 2013 3:55 AM


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! > >

