# Re: [R] How to simplify

From: Petr Pikal <petr.pikal_at_precheza.cz>
Date: Thu 08 Dec 2005 - 02:29:08 EST

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@gmail.com>
To:             	r-help@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
> [[1]]
> 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
>
> [[2]]
> 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
>
> [[3]]
> 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
>
> [[4]]
> .......
>
> [[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[[1]][1]+a[[2]][1]+a[[3]][1]+.....)/n
> >mean.b<-(b[[1]][1]+b[[2]][1]+b[[3]][1]+.....)/n
> >mean.c<-(c[[1]][1]+c[[2]][1]+c[[3]][1]+.....)/n
> >mean.d<-(d[[1]][1]+d[[2]][1]+d[[3]][1]+.....)/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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