Re: Average the same elements of the list
- To: mathgroup at smc.vnet.net
- Subject: [mg130888] Re: Average the same elements of the list
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Thu, 23 May 2013 04:10:37 -0400 (EDT)
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- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-outx@smc.vnet.net
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- References: <20130522061810.01F4F6A31@smc.vnet.net>
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 at 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! > >
- References:
- Average the same elements of the list
- From: BBabic <bipsich101@gmail.com>
- Average the same elements of the list