> x<-ts(sales,frequency=4,start=c(2010,1))
> x
Qtr1 Qtr2 Qtr3 Qtr4
2010 1.8 8.0 6.0 3.0
2011 2.0 11.0 7.0 3.5
2012 2.5 14.0 8.0 4.2
2013 3.0 15.2 9.5 5.0
> plot.ts(x)__
R-help@r-project.org mailing list
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Hi,
try
?axis.Date
works similar to the axis function.
I hope that helps!
daniel
Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ;
meghatalmazó: 水静流深 [1248283...@qq.com]
Küldve: 2013. december 15. 9:17
To: r-help
Tárgy: [R] how can i
seq(as.Date("2001/1/1"),as.Date("2010/1/1"),"years")
seq(as.Date("2001/1/1"),as.Date("2010/1/1"),"weeks")
seq(as.Date("2001/1/1"),as.Date("2010/1/1"),"days")
why there is no
seq(as.Date("2001/1/1"),as.Date("2010/1/1"),"quarters") ?
[[alternative HTML version deleted]]
__
> x=read.table(text="
+ Qtr1 Qtr2 Qtr3 Qtr4
+ 2010 1.8 8.0 6.0 3.0
+ 2011 2.0 11.0 7.0 3.5
+ 2012 2.5 14.0 8.0 4.2
+ 2013 3.0 15.2 9.5 5.0",sep="",header=TRUE)
> x
Qtr1 Qtr2 Qtr3 Qtr4
2010 1.8 8.0 6.0 3.0
2011 2.0 11.0 7.0 3.5
2012 2.5 14.0 8.0 4.2
2013 3.0 15.2 9.5
On 13-12-15 6:43 AM, 水静流深 wrote:
seq(as.Date("2001/1/1"),as.Date("2010/1/1"),"years")
seq(as.Date("2001/1/1"),as.Date("2010/1/1"),"weeks")
seq(as.Date("2001/1/1"),as.Date("2010/1/1"),"days")
why there is no
seq(as.Date("2001/1/1"),as.Date("2010/1/1"),"quarters") ?
There's no need for it. Jus
On 13-12-15 6:48 AM, 水静流深 wrote:
x=read.table(text="
+ Qtr1 Qtr2 Qtr3 Qtr4
+ 2010 1.8 8.0 6.0 3.0
+ 2011 2.0 11.0 7.0 3.5
+ 2012 2.5 14.0 8.0 4.2
+ 2013 3.0 15.2 9.5 5.0",sep="",header=TRUE)
x
Qtr1 Qtr2 Qtr3 Qtr4
2010 1.8 8.0 6.0 3.0
2011 2.0 11.0 7.0 3.5
2012 2.5
t<--4:4
y<-c(5,7,10,13,15,16,14,12,11)
plot(t,y,type="l")
how can i add a curve y=0.83*t-0.44*t^2 in the graph?
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
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PLEASE do re
t<- -4:4
y<-c(5,7,10,13,15,16,14,12,11)
plot(t,y,type="l", ylim=c(-20,20))
curve(0.83*x-0.44*x^2, add=TRUE)
I added ylim to plot so that your curve is on the graph. Maybe you wanted to
use different parameters.
Try going through a basic manual of R!
daniel
_
kindly help find out why the following code is given me error
time<-c(80,80,90,90,85,85,85,85,85,85,92,77,85,85)
> tem<-c(170,180,170,180,175,175,175,175,175,175,175,175,182.07,167.93)
> yield<-c(80.5,81.5,82,83.5,83.9,84.3,84,79.7,79.8,79.5,78.4,75.6,78.6,77)
> Block<-c("B1","B1","B1","B1","B1"
There is no variable called x1 in the chem dataframe.
Did you mean
gg1=rsm(yield~ FO(x1,x2),data=ggg,subset = (Block == "B1"))
hth.
daniel
Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ;
meghatalmazó: IZHAK shabsogh [ishaqb...@yahoo
I would like to do this:
f <- function(formula, data=NULL) {
gg <- sqrt
model.frame(formula, data=data)
}
x <- y <- 1:10
f(y ~ gg(x))
Error in eval(expr, envir, enclos) : could not find function "gg"
Is there a simple way to get access to gg from within the model.frame
invocation inside f?
The following works because model.frame looks for things in environment(formula)
and ancestral environments thereof. It puts the new things in a child
environment
of the original environment(formula) so it does not alter the original
environment.
f2 <- function (formula, data = NULL)
{
env
The response below asks what I actually did.
I defined a function (details omitted; it computes the data frame
LRtest.out); arguments include "path\\filename.csv" to which I want to write
a data frame using write.csv( ). Repeated executions of the function
(without the file( ) and close( )
On 15.12.2013 20:13, John Karon wrote:
The response below asks what I actually did.
I defined a function (details omitted; it computes the data frame
LRtest.out); arguments include "path\\filename.csv" to which I want to
write a data frame using write.csv( ). Repeated executions of the
funct
According to the documentation (?as.vector), "All attributes are removed from
the result if it is of an atomic mode, but not in general for a list result."
data.frames are lists so
> is.vector(as.vector(x))
[1] FALSE
However if you convert using unlist() or as.matrix(), you will get a vector:
Thank you Bill, that worked perfectly.
Frank
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, r
Hello,
My x-y scatterplot produces a very ragged best-fit line when imported into
Word.
* >plot (data.file$x, data.file$y, type = "p", las=1, pch=20,ylab =
expression("Cover of Species y" ~ (m^{2}~ha^{-1} )),xlab =
expression("Cover of Species x" ~ (m^{2}~ha^{-1})) )>lines
See ?png and argument 'pointsize'. You can increase that as you
increase the dimensions of the output image.
/Henrik
On Sun, Dec 15, 2013 at 3:00 PM, david hamer wrote:
> Hello,
>
> My x-y scatterplot produces a very ragged best-fit line when imported into
> Word.
>
>
>
> * >plot (data.file$x,
And possibly better, argument 'res', e.g.
png("R.graph.png", width=1200, height = 700, res=144)
plot(...)
dev.off()
Default corresponds to res=72.
/Henrik
On Sun, Dec 15, 2013 at 3:13 PM, Henrik Bengtsson wrote:
> See ?png and argument 'pointsize'. You can increase that as you
> increase the
On 13-12-15 6:00 PM, david hamer wrote:
Hello,
My x-y scatterplot produces a very ragged best-fit line when imported into
Word.
Don't use a bitmap format (png).
Don't produce your graph in one format (screen display), then convert to
another (png). Open the device in the format you want for
On 16/12/13 12:23, Duncan Murdoch wrote:
[After a number of other "Don'ts"]
Don't trust the preview to tell you the quality of the graph, try
printing the document.
Word isn't quite as bad as it appears.
Don't use Word.
Fortune?
cheers,
Rolf Turner
_
Hello all,
I have two copies of the same data; I want to compare trend matrices for
degrees 3 and 4, so I name the copies "triplo" and "quatro".
> triploSurface<-surf.ls(3,triplo)
> triploMat<-trmat(triploSurface,-150.,150.,-150.,150., 300)
[This produces the expected triploSurface
Hi,
On itself one way of doing this is:
lines(seq(-4,4,0.1),sapply(seq(-4,4,0.1),function(t)0.83*t-0.44*t^2))
The curve does not fit in the scale, however. Are you sure the formula is
correct? Then you have to adapt the scale:
plot.new()
plot.window(xlim=c(-4,4), ylim=c(-10,16))
axis(1)
axis(2)
t
Hi all,
I'm kinda new in R programming and I need some help preparing a database to
run logistic regression.
I have data in a tuple form:
*id cat val*
1 A 2
1 C 4
3 B 1
5 A 2
6 A 3
6 B 5
6 C 2
8 B 5
8 D 2
9 D 3
and would like to have it like:
*id catA
Thanks enormously, Bill! I'll run with this for a while, and let you know
how it works for me.
Yours, andrewH
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_
Thanks, Enrico! I did this, and if I did it right, there are no nonstandard
characters. So now I am suspecting a size limit internal to the filehash
package, and trying to chase that down. Your help is much appreciated.
On Mon, Dec 9, 2013 at 11:11 PM, Enrico Schumann wrote:
> On Mon, 09 Dec 201
Thanks, Earl. Your utility ran like a charm, and confirmed that my effort
to adapt Enrico's code to this purpose had not gone astray, which is to
say, I found no funky characters. Your help is greatly appreciated.
Sincerely, andrewH
On Tue, Dec 10, 2013 at 7:35 AM, Earl F Glynn [via R] <
ml-node+
Hi R mailing listers:
Assume that there are two sets of data (denoted as A and B) with the same
size, say 100X1. And I try to select a new set of data (100X1) that has a
better normal feature from these two sets. Better normal feature means
that the histogram shape of the constructed set of dat
input <- "
ty
1 5.3
2 7.2
3 9.6
4 12.9
5 17.1
6 23.2"
dat<-read.table(textConnection(input),header=TRUE,sep="")
t<-dat[,1]
y<-dat[,2]
`y=3.975*(1.341^t)` is the resule of fit,how can i use `nls`
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