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,...].