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MathGroup Archive 2009

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Re: compress lists with mean for equal elements

  • To: mathgroup at smc.vnet.net
  • Subject: [mg101005] Re: compress lists with mean for equal elements
  • From: Ray Koopman <koopman at sfu.ca>
  • Date: Sat, 20 Jun 2009 04:01:47 -0400 (EDT)
  • References: <h1hbfr$n4$1@smc.vnet.net>

On Jun 19, 5:45 pm, vJD <epho... at gmail.com> wrote:
> I have a nested list of this form:
>
> {{{1, 13}, 97.6493}, {{1, 13}, 97.9511}, {{1, 14}, 99.3002}, {{1, 14},
> 99.4602}, {{1, 14}, 99.6193}, {{1, 15}, 100.513}, {{1, 15}, 101.149},
> {{1, 15}, 101.483}, {{1, 15}, 101.494}, {{1, 16}, 101.51}, {{1, 16},
> 101.895}}
>
> I want to shrink the list in the way that I calculate the mean over
> the second part for all elements where the first part is equal. so
> e.g. the first element of the new list will be {{1,13}, 97.8002},
> because they both have {1,13} and the mean of 97.6493 and 97.9511 is
> 97.8002.
>
> The problem is that there is no fixed number for the equal elements
> (so e.g. two elements with {1,13}, three elements with {1,14} and so
> on). The number is random.
>
> Can anybody please provide me with a statement or hint how to
> accomplish that task.
>
> Thank you very much
>
> Regards,
> Holger

data = {{{1,13},97.6493}, {{1,13},97.9511}, {{1,14},99.3002},
        {{1,14},99.4602}, {{1,14},99.6193}, {{1,15},100.513},
        {{1,15},101.149}, {{1,15},101.483}, {{1,15},101.494},
        {{1,16},101.51}, {{1,16},101.895}};

{#[[1,1]],Mean[#[[All,2]]]}& /@ Split[data,#1[[1]]==#2[[1]]&]

{{{1,13},97.8002},{{1,14},99.4599},{{1,15},101.16},{{1,16},101.703}}

Note: if 'data' is not already sorted then use Split[Sort@data,...].


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