Very elegant! Thank you Eric.
(You omitted library(plyr), so I had to search for arrange())
-Michael
On 12/13/2013 3:01 PM, rmail...@justemail.net wrote:
Perhaps this?
library(reshape2)
library(stringr)
GeisslerLong <- melt (Geissler, id.vars = c("boys"))
GeisslerLong <- transform ( Geissler
Thanks Hervé Pagès and A. K.. It works.
Thank you!
David
2013/12/14 Hervé Pagès
> Hi David,
>
>
> On 12/14/2013 01:06 PM, Marino David wrote:
>
>> Hi all:
>>
>> Assume that I have variables, say v1, v2,...,v100 and I want to use one
>> variable in each roop. How can I do this? See below
>>
>>
Hi
have a look at the latticeExtra package
library(latticeExtra)
? useOuterStrips
or
xyplot(y~x| paste(grouping1, grouping2), data=na.omit(mydata), layout =
c(...
HTH
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home: mac...@northnet
or for a different view
library(lattice)
xyplot(B+C ~A, data = s,
outer = T,
scales = list(relation = "free"),
pch = as.numeric(rownames(s)),
col = as.numeric(rownames(s)))
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of
Hi David,
On 12/14/2013 01:06 PM, Marino David wrote:
Hi all:
Assume that I have variables, say v1, v2,...,v100 and I want to use one
variable in each roop. How can I do this? See below
for (i in 1:100){
f(vi)
}
for (i in 1:100){
f(get(paste0("v", i)))
}
Cheers,
H.
Thanks
David
Hi,
If the variables described are the columns in a data.frame,
set.seed(24)
dat1 <- as.data.frame(matrix(sample(100,100*10,replace=TRUE),ncol=100))
f1 <- function(x) mean(x,na.rm=TRUE)
sapply(1:100,function(i) f1(dat1[,i]))
#or
sapply(colnames(dat1),function(x) f1(dat1[,x]))
#IF these ar
Hi all:
Assume that I have variables, say v1, v2,...,v100 and I want to use one
variable in each roop. How can I do this? See below
for (i in 1:100){
f(vi)
}
Thanks
David
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
Thanks both Uwe and Daniel for the great help!
On Sat, Dec 14, 2013 at 3:30 PM, Dániel Kehl wrote:
> Dear Gang,
>
> this seem to solve your problem.
>
>
> http://stackoverflow.com/questions/1195826/dropping-factor-levels-in-a-subsetted-data-frame-in-r
>
>
> best
> daniel
> _
Dear Gang,
this seem to solve your problem.
http://stackoverflow.com/questions/1195826/dropping-factor-levels-in-a-subsetted-data-frame-in-r
best
daniel
Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ;
meghatalmazó: Gang Chen [gangc
On 14.12.2013 21:09, Gang Chen wrote:
Suppose I have a dataframe 'd' defined as
L3 <- LETTERS[1:3]
d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace
= TRUE))
(d <- d0[d0$fac %in% c('A', 'B'),])
x y fac
2 1 2 B
3 1 3 A
4 1 4 A
5 1 5 A
6 1 6 B
8
Suppose I have a dataframe 'd' defined as
L3 <- LETTERS[1:3]
d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace
= TRUE))
(d <- d0[d0$fac %in% c('A', 'B'),])
x y fac
2 1 2 B
3 1 3 A
4 1 4 A
5 1 5 A
6 1 6 B
8 1 8 A
Even though factor 'fac' in 'd' onl
Hi Eliza,
Try:
ggplot(s,aes(x=A,y=B))+geom_point(colour="white",shape=21,size=4,aes(fill=factor(C)))+theme_bw()+theme(legend.position="none")
A.K.
On Saturday, December 14, 2013 12:29 PM, eliza botto
wrote:
Dear Arun and david,
Thanks for your reply. If instead of text i want to add poin
Hi,
Try:
dat1 <- read.table(text="x1 x2 x3
1 5 7
4 8 9
8 6 12
4 8 13",sep="",header=TRUE)
vec1 <- dat1[,1]
vec1
#[1] 1 4 8 4
A.K.
On Saturday, December 14, 2013 12:55 PM, Patty Haaem
wrote:
Hi every one,
I have a simple question. I want to make a vector from one of the c
On Dec 14, 2013, at 6:34 AM, Patty Haaem wrote:
> Hi every one,
> I have a simple question. I want to make a vector from one of the colomns in
> a data set. for example I have this data:
> x1 x2 x3
> 1 5 7
> 4 8 9
> 86 12
> 4 8 13
> I want to convert x1 to a vector such as:
>
Dear Arun and david,Thanks for your reply. If instead of text i want to add
points, what change in code should occur?
eliza
> Date: Sat, 14 Dec 2013 07:48:52 -0800
> From: smartpink...@yahoo.com
> Subject: Re: [R] plot two columns against one
> To: r-help@r-project.org
> CC: eliza_bo...@hotmail.c
This is the R-help mailing list. If your problem persists when using R
from the command line or with the GUI shipped with R on your
(unspecified) platform post back here. Otherwise the RStudio support
forum is at https://support.rstudio.com
Best,
Ista
On Sat, Dec 14, 2013 at 12:22 AM, yuanzhi wr
On 14.12.2013 17:50, David Winsemius wrote:
On Dec 14, 2013, at 7:53 AM, Uwe Ligges wrote:
David,
how should R interpret "110+"? It cannot be numeric, perhaps you have not recognized the
"+" there?
I specifically included the fragment ofthe much longer file that was throwing
the error.
On 13.12.2013 20:11, J Karon wrote:
I get an invalid connection method error message when trying to write an R
object from a user-defined function to my hard drive (running Windows 7)
using write.csv. I have previously not had this problem with the same
user-defined function. The error messag
Hi every one,
I have a simple question. I want to make a vector from one of the colomns in a
data set. for example I have this data:
x1 x2 x3
1 5 7
4 8 9
8 6 12
4 8 13
I want to convert x1 to a vector such as:
x1= c(1,4,8,4)
How I can do it?
Thanks
[[alternative HTML vers
On Dec 14, 2013, at 7:53 AM, Uwe Ligges wrote:
> David,
>
> how should R interpret "110+"? It cannot be numeric, perhaps you have not
> recognized the "+" there?
>
I specifically included the fragment ofthe much longer file that was throwing
the error. If this behavior doesn't appear flawed
On Dec 13, 2013, at 1:34 PM, Trevor Davies wrote:
> Is there a quick function that can convert minutes (seconds) after midnight
> to a time?
>
> i.e 670.93 (minutes after midnight) --> 11:10:56.**
>
> I know it can be done by hand but I thought there must be a function for
> this already.
form
Hi,
Try:
x[-1]+x[-length(x)]
A.K.
On Saturday, December 14, 2013 5:46 AM, 水静流深 <1248283...@qq.com> wrote:
x<-c(1,4,9,20,3,7)
i want to get a serie c(5,13,29,23,10).
y <- c()
for (i in 2:length(x)){
y[i-1] <- x[i-1]+x[i]}
is there more simple way to get?
[[alternative HTML version del
David,
how should R interpret "110+"? It cannot be numeric, perhaps you have
not recognized the "+" there?
Uwe
On 14.12.2013 01:35, David Winsemius wrote:
I thought that setting colClasses to numeric would coerce errant data to NA.
Instead read.table is throwing
errors. This is not what
HI,
Using ?ggplot()
s <-read.table(text="A B C
0.451 0.333 1134
0.491 0.270 1433
0.389 0.249 7784
0.425 0.819 6677
0.457 0.429 99053
0.436 0.524 111049
0.423 0.270 121093
0.463 0.315 131019",sep="",header=TRUE)
library(ggplot2)
ggplot(s,aes(x=A
Please do not send emails using html. Please use dput() to send
your data to the list
> st <- read.table(text=s, header=TRUE)
> plot(B~A, st, type="n")
> with(st, text(A, B, C))
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station,
Dear users of R,
How can i plot the values in column "C" with "A" on x-axis and "B" on y-axis?s
<-
"A BC 0.451 0.333 1134
0.491 0.270 1433
0.389 0.249 7784
0.425 0.819 6677
0.457 0.429 99053
0.436 0.524 111049 0.423
See ?source
Uwe Ligges
On 14.12.2013 12:22, 水静流深 wrote:
there is a file which contain such lines:
test <-function(x,f){
n<- length(x);
sum(f*(x[1:(n-1)]+x[2:n])/2)/sum(f) -> result;
return(result);
}
i can read it into terminal , cat test.R
how can i do it in R?
readLines("
there is a file which contain such lines:
test <-function(x,f){
n<- length(x);
sum(f*(x[1:(n-1)]+x[2:n])/2)/sum(f) -> result;
return(result);
}
i can read it into terminal , cat test.R
how can i do it in R?
> readLines("c:/test.R",n=-1)
[1] "test <-function(x,f){"
On 14-Dec-2013 10:46:10 Ë®¾²Á÷Éî wrote:
> x<-c(1,4,9,20,3,7)
> i want to get a serie c(5,13,29,23,10).
> y <- c()
> for (i in 2:length(x)){
> y[i-1] <- x[i-1]+x[i]}
>
> is there more simple way to get?
x <- c(1,4,9,20,3,7)
N <- length(x)
x[1:(N-1)] + x[2:N]
# [1] 5 13 29 23 10
B
> On Fri, Dec 13, 2013 at 10:11 PM, æ°´éæµæ·± <1248283...@qq.com> wrote:
>> i have write a function to convert decimal number into binary number in R.
>>
>> dectobin<-function(x){
>> as.numeric(intToBits(x))->x1
>> paste(x1,collapse="")->x2
>> as.numeric(gsub("0+$","",x2))->x3
>> return
x<-c(1,4,9,20,3,7)
i want to get a serie c(5,13,29,23,10).
y <- c()
for (i in 2:length(x)){
y[i-1] <- x[i-1]+x[i]}
is there more simple way to get?
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