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!!
I suggest that when displaying test data in a post that
you do it like this:

dput(cal)

since then others can simply copy and paste it into
their session.

At any rate, using this test data:

cal <- list(A = data.frame(time = 1:3, X1 = 1:3),
	B = data.frame(1:3, X1 = 3:5))

Pick off the second element of each list component, turn
the result into a data frame and take the row means:

 rowMeans(as.data.frame(lapply(cal, "[", 2)))

Since the times are irregular, you might prefer to represent cal
as a multivariate zoo object using the zoo library:

library(zoo)
cal.zoo <- do.call("merge", lapply(cal, function(x) zoo(x[,2], x[,1])))

Once you have done this and other operations simplify.  For
this one its just:

rowMeans(cal.zoo)

or to plot them all

plot(cal.zoo) # separate plots
plot(call.zoo, plot.type = "single")  # all on one plot
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!!

______________________________________________
R-help at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html

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

Address: Kapucijnenvoer 35, Leuven, Belgium
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.
Thanks in advance !!
Sincerely!!

______________________________________________
R-help at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! 
http://www.R-project.org/posting-guide.html

Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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
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
[[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!!

______________________________________________
R-help at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
Petr Pikal
petr.pikal at precheza.cz