# [R] How to simplify

Dimitris Rizopoulos dimitris.rizopoulos at med.kuleuven.be
Wed Dec 7 16:10:39 CET 2005

```assuming that you have the same number of measurements in each
sub-data.frame, you could use something like:

cal <- lapply(1:10, function(x) data.frame(time = c(0, 0.5, 1, 2, 4,
6, 8, 10), X1 = rnorm(8, 10:3)))
##############
rowMeans(as.data.frame(lapply(cal, "[", "X1")))

I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://www.med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm

----- Original Message -----
From: "Rhett Eckstein" <glaxowell at gmail.com>
To: <r-help at stat.math.ethz.ch>
Sent: Wednesday, December 07, 2005 3:38 PM
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.
> Sincerely!!
>
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