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Topic: Average the same elements of the list
Replies: 4   Last Post: May 24, 2013 5:24 AM

 Messages: [ Previous | Next ]
 Bob Hanlon Posts: 906 Registered: 10/29/11
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!
>
>

Date Subject Author
5/23/13 Bob Hanlon
5/23/13 Tomas Garza Hernandez
5/23/13 Murray Eisenberg
5/24/13 Daniel