# [R] How to simplify

Petr Pikal petr.pikal at precheza.cz
Wed Dec 7 16:29:08 CET 2005

```Hi

changing list to matrix and making summary could do the trick

lll <- list(a=cbind(1:10, rnorm(10)), b=cbind(1:10, rnorm(10)))
mat <- do.call("rbind", lll)
tapply(mat[,2], mat[,1],  mean)

BTW I found a suitable thread with similar question in CRAN search

list summary mean

HTH
Petr

On 7 Dec 2005 at 22:38, Rhett Eckstein wrote:

Date sent:      	Wed, 7 Dec 2005 22:38:56 +0800
From:           	Rhett Eckstein <glaxowell at gmail.com>
To:             	r-help at stat.math.ethz.ch
Subject:        	[R] How to simplify

> Dear list,
> I have a list containing parameters (time and X1),  and have "n"
> similar data set like
> the following:
> > cal
> []
>   time        X1
> 1  0.0 10.006306
> 2  0.5  9.433443
> 3  1.0  8.893405
> 4  2.0  7.904274
> 5  4.0  6.243807
> 6  6.0  4.932158
> 7  8.0  3.896049
> 8 10.0  3.077604
>
> []
>   time        X1
> 1  0.0 10.015972
> 2  0.5  9.460064
> 3  1.0  8.935039
> 4  2.0  7.970755
> 5  4.0  6.343151
> 6  6.0  5.047900
> 7  8.0  4.017131
> 8 10.0  3.196856
>
> []
>   time       X1
> 1  0.0 9.985741
> 2  0.5 9.552583
> 3  1.0 9.138239
> 4  2.0 8.362664
> 5  4.0 7.003394
> 6  6.0 5.865057
> 7  8.0 4.911747
> 8 10.0 4.113382
>
> []
> .......
>
> [[n]]
> .......
>
> And I would like to put all  X1( when time=0) together, time=0.5,1...
> are the same. then calculate the mean value. > a<-list() > b<-list() >
> c<-list() > d<-list() > e<-list() ....... > for(i in 1:n){ +
> a[[i]]<-cal[[i]][1,2] +   b[[i]]<-cal[[i]][2,2] +
> c[[i]]<-cal[[i]][3,2] +   d[[i]]<-cal[[i]][4,2] +
> e[[i]]<-cal[[i]][5,2] +   ......... }
> >mean.a<-(a[]+a[]+a[]+.....)/n
> >mean.b<-(b[]+b[]+b[]+.....)/n
> >mean.c<-(c[]+c[]+c[]+.....)/n
> >mean.d<-(d[]+d[]+d[]+.....)/n >.............
> >xy<-c(mean.a,mean.b,mean.c,mean.d,........) But the way I use seem
> not very smart. So please give me some hints to the simplify this.
> Thanks in advance !! Sincerely!!
>
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