Excellent, thank you to both of you!!!
Gail
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Thanks Elk,
That was the missing link.
#or
lapply(list6,function(x)max(x$target))
[[1]]
[1] 6
[[2]]
[1] 18
[[3]]
[1] 9
max(unlist(lapply(list6,function(x) max(x$target
A.K.
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Hi Arun,
if you name the data.frame in the list, e.g.
mango <- list( y=c(4,5,3,2,1,8), bw=c(4,18,12,3,4,9),
nn=paste0("a",1:6),target=data.frame(coconut=1:6))
banana <- list(target=data.frame(coconut=c(1,2,18,16,15)), y=c(4,5,9,2,1),
bw=c(4,18,22,3,4), nn=paste0("a",1:5))
pineapple <- list(target=d
Hi Elk,
I tried to test with another case where coconut is in different position in
the sublist.
mango <- list( y=c(4,5,3,2,1,8), bw=c(4,18,12,3,4,9),
nn=paste0("a",1:6),data.frame(coconut=1:6))
banana <- list(data.frame(coconut=c(1,2,18,16,15)), y=c(4,5,9,2,1),
bw=c(4,18,22,3,4), nn=paste0("a",1:
Hello,
I was in a hurry at the time, when I replied.
You could do one thing: extract sublist using:
list4 <- list(mango, banana, pineapple)
b1<-list()
for(i in 1:3){
b1[[i]]<-list()
b1[[i]]<-list4[[i]][1]
}
b1
b<-list()
for(i in 1:3){
b[[i]]<-list()
b[[i]]<-lapply(b1[[i]][[1]],FUN=functio
Hi Elk,
Thanks for the input. Somehow, I got entangled with which.max. I should
have used max.
lapply(list4,function(x) max(x[[1]]$coconut))
[[1]]
[1] 6
[[2]]
[1] 18
[[3]]
[1] 9
max(unlist(lapply(list4,function(x) max(x[[1]]$coconut
#[1] 18
A.K.
I still try to figure out, what you are finally asking for, but maybe
res<-sapply(list4,function(x)max(x[[1]]$coconut))
names(res)<-c("mango", "banana", "pineapple")
res
max(res)
is worth a try?
Especially i did not get the point of doing something like
x[which.max(x)] because this is in any cas
Hello,
If I understand it now, can't you do this:
Individual cases:
list4<-list(mango[[1]],banana[[1]],pineapple[[1]])
b<-list()
for(i in 1:3){
b[[i]]<-list()
b[[i]]<-lapply(list4[[i]],FUN=function(x)x[which.max(x)])
}
b1<-data.frame(do.call(rbind,b))
row.names(b1)<-c("mango","banana","pineap
Hi,
my dataset is the result of the function "density" on another set of data.
It refers to that data in its variable "call", though since all the results
are actually reproduced (except that I've removed all rows bar 10), I am not
sure why R still needs it. But I've understood now why your code d
Michael, here is the data I am working with. I've shortened it without, I
trust, changing its nature. I am not using any package other than what is
pre-installed. Many thanks, Gail
> dput(list.example)
list
(structure(list(
x = c(379.455895016957, 380.116796110287,
380.777697203618, 381.438598
Well but the whole point was to be able to plug in a single command and get
the result for all sublists. If I have to create a new list and type in each
sublist and the column I'm interested in, that completely defeats the
purpose, I might as well just ask for max(sublist$coconut) for each sublist,
Arun, I see you've defined dat5 in your later message; however the same
applies as to the above: the code doesn't work if the list contains
non-numerical elements.
Gail
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Hi,
I am not able to read your dataset.
Getting errors.
source("list1.txt")
Error in paste(locuteur, i, sep = "") : object 'locuteur' not found
Please check.
A.K.
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Hello Gail,
The dat5 was in earlier post. I forgot to paste it.
Now, your question regarding individual cases,
This is a case that includes nonnumeric columns
list1<-list(data.frame(x=1:6,y=c(4,5,3,2,1,8),bw=c(4,18,12,3,4,9),nn=paste0("a",1:6)))
list2<-list(data.frame(x=c(1,2,18,16,15),y=c(4,5,9,
HI,
I tested the code for another set of data with added complexity. Seems to
be working fine.
list1<-list(data.frame(x=1:6,y=c(4,5,3,2,1,8),bw=c(4,18,12,3,4,9)))
list2<-list(data.frame(x=c(1,2,18,16,15),y=c(4,5,9,2,1),bw=c(4,18,22,3,4)))
list3<-list(data.frame(x=c(4,6,9),y=c(8,24,12),bw=c(14,31,3
I'm a little confused on your data structure -- can you use dput() as
described here [1] to give a small reproducible example?
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
Michael
On Wed, Aug 1, 2012 at 10:38 AM, gail wrote:
> Thank you!!! But I realise
On Wed, Aug 1, 2012 at 8:43 AM, gail wrote:
> Hello. Please forgive me if this problem has already been posted (and solved)
> by someone else ... I can't find it anywhere though it seems so very basic.
> Here it is:
>
> I have a list comprised of several matrices, each of which has two columns.
>
Thank you!!! But I realise I've simplified my data to the point that your
solution doesn't actually work -- not your fault, mine! My list is actually
more complicated than what I presented it to be; it's not composed of
numerical matrices but of lists, each being composed of 7 columns, the first
tw
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