On Thu, 23 Sep 2010, Peter Ehlers wrote:
On 2010-09-21 5:51, Nikhil Kaza wrote:
example(factor)
iris1$Species<- factor(iris1$Species, drop=T)
will get you what you need.
Hmm, doesn't work for me. ?factor does not list a 'drop='
argument.
I suspect
iris1$Species <- [iris1$Species, drop=TR
Hello,
Not sure about packages to suggest but some of this can be done with
base stats...
# Generate densities using common from, to and n args
male <- rnorm(100, 80, 10)
female <- rnorm(100, 60, 10)
male.d <- density(male, from=0, to=150, n=1024)
female.d <- density(female, from=0, to=150, n=102
Hi,
I want to find a value of n1. I used the following code but I am getting the
error -
Error in as.vector(x, mode) :
cannot coerce type 'closure' to vector of type 'any'
n=10
a_g<-(1/(n*(n-1)))*((pi/3)*(n+1)+(2*sqrt(3)*(n-2))-4*n+6)
a_s<-function(n1)
{
t1=(n1-1)/2;
(t1*(gamma(t1)/g
darckeen hotmail.com> writes:
> This is an example of the type of problem, and how i'm currently using
> optim() to solve it.
I would classify this as an integer programming (IP) problem, it is not
even non-linear, as there are no continuous variables. Your approach with
optim() will not work
On 2010-09-23 17:57, array chip wrote:
Yes, it does what I want. Thank you Peter! Just wondering what else grid.pars
controls? not just the symbol in legend, right?
John
You can have a look at ?gpar (after loading package 'grid').
For example, your original request could be handled with
setti
Yes, it does what I want. Thank you Peter! Just wondering what else grid.pars
controls? not just the symbol in legend, right?
John
- Original Message
From: Peter Ehlers
To: array chip
Cc: "r-help@r-project.org"
Sent: Thu, September 23, 2010 4:34:44 PM
Subject: Re: [R] how to make p
On 2010-09-23 17:15, array chip wrote:
Now I got point character thicker using panel.points(lwd=2), But I can't make it
thicker in the legend of the plot. Here is an example:
xyplot(1:10~1:10,groups=rep(1:5,2),col=1:2,pch=0:1,type='p',cex=2,panel=panel.points,lwd=3,
key=list(x=0.7,y=0.2,corner=
Now I got point character thicker using panel.points(lwd=2), But I can't make
it
thicker in the legend of the plot. Here is an example:
xyplot(1:10~1:10,groups=rep(1:5,2),col=1:2,pch=0:1,type='p',cex=2,panel=panel.points,lwd=3,
key=list(x=0.7,y=0.2,corner=c(0,0),type='p',
points=list(col=1:2,pc
Thank you Greg. I also got it work by using panel.points (lwd=2) instead of
using panel.xyplot()
- Original Message
From: Greg Snow
To: array chip ; "r-help@r-project.org"
Sent: Thu, September 23, 2010 2:48:06 PM
Subject: RE: [R] how to make point character thicker in xyplot?
Th
On Thu, Sep 23, 2010 at 5:53 PM, Struve, Juliane wrote:
> Hello,
>
> thank you very much for replying. I have tried this, but I get error
> message
>
> "Error in .subset(x, j) : invalid subscript type 'list'" after
>
> z1 <- read.zoo(textConnection(Lines1), skip = 1, index = list(1, 2), FUN =
> dt
There is probably a simpler way, but if you want full customization, look at
panel.my.symbols in the TeachingDemos package.
-Original Message-
From: array chip
Sent: Thursday, September 23, 2010 2:52 PM
To: r-help@r-project.org
Subject: [R] how to make point character thicker in xyplot?
Is there anyway to make plotting point character being thicker in xyplot? I
mean
not larger which can achieved by "cex=2", but thicker. I tried lwd=2, but it
didn't work. I know "lwd" works in regular plot() not only for lines, but also
for points. For example
plot(1:10, lwd=2)
Thanks
John
Include individual as a factor in your dataset, and use ggplot2:
library(ggplot2)
ggplot(aes(x=Date, y=Distance, color=Individual), data=data) + geom_line()
ought to do it.
Jonathan
On Thu, Sep 23, 2010 at 9:31 AM, Struve, Juliane wrote:
> Sorry for posting this questions twice, but my previo
On 2010-09-21 5:51, Nikhil Kaza wrote:
example(factor)
iris1$Species<- factor(iris1$Species, drop=T)
will get you what you need.
Hmm, doesn't work for me. ?factor does not list a 'drop='
argument.
-Peter Ehlers
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North
There was a typo error in my code below. See the inserted correction.
On 23-Sep-10 17:05:45, Ted Harding wrote:
> On 23-Sep-10 16:52:09, Duncan Murdoch wrote:
>> On 23/09/2010 11:42 AM, wangguojie2006 wrote:
>>> b<-runif(1000,0,1)
>>> f<-density(b)
>>
>> f is a list of things, including x value
Patrick--
One other option in addition to Thierry's suggestion, within R you might
also consider the ordinal package, which handles random-intercept models.
That said, if you are used to SPSS, I suspect this will be a titanic
pain trying to move to R (part. if just for this one analysis...).
On 23/09/2010 2:10 PM, rtist wrote:
I was wondering if anyone has a way out of the limitation that you must use
equal length
x,y, and z sequence lengths.
For instance,
x<-seq(1,100)
y<-seq(1,100)
z<-rnorm(100)
scatterplot3d(z,x,y)
works fine.
However, if I get some results that has a different
You guys are really good.
Thanks.
--
View this message in context:
http://r.789695.n4.nabble.com/help-in-density-estimation-tp2552264p2552484.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://st
I was wondering if anyone has a way out of the limitation that you must use
equal length
x,y, and z sequence lengths.
For instance,
x<-seq(1,100)
y<-seq(1,100)
z<-rnorm(100)
scatterplot3d(z,x,y)
works fine.
However, if I get some results that has a different y subset length, such as
x<-seq(1,10
Hi All
I have been trying to plot irregular XYZ data on world / country map, where X
and Y are llongitude and latitude coordinates and Z is a variable. My problem
is I get contour lines over water bodies. The method I follow is:>
library(akima)> library(maps)> library(mapdata)> library(gpclib)>
You could do:
b <- runif(1, 0, 1)
tmp <- density(b, from=0.5, to=0.5, n=1)
tmp$y
As one direct approach.
You could also look at the logspline package for an alternative for density
estimation that provides you with a density function (and also allows for
finite domains).
--
Gregory (Gre
On Thu, Sep 23, 2010 at 9:50 AM, Struve, Juliane
wrote:
> Dear list,
>
> I would like to create a time series plot in which the paths of several
> individuals are stacked above each other, with the x-axis being the total
> observation period of three years ( 1.1.2004 to 31.12.2007) and the y-axi
On Thu, Sep 23, 2010 at 12:18 PM, rasanpreet kaur suri
wrote:
> hi guys..thx for help
> i have to perform a calculation
> P-B/T-B
> where P is the values in pdf and B is the values in bdf and T in tdf
You have 69 rows in your pdf data.frame, and something like 20 rows in
bdf and tdf, so my origin
On 23-Sep-10 16:52:09, Duncan Murdoch wrote:
> On 23/09/2010 11:42 AM, wangguojie2006 wrote:
>> b<-runif(1000,0,1)
>> f<-density(b)
>
> f is a list of things, including x values where the density is
> computed,
> and y values for the density there. So you could do it by linear
> interpolation
Hi Gennadi,
I think this is fixed in the latest version of the ncdf package, which you
can access by going to CRAN, then 'packages', then 'ncdf', and clicking on
'URL'. Or here is a direct link:
http://cirrus.ucsd.edu/~pierce/ncdf
Give that a try and let me know if you still have a problem,
--
I've found using the arm package to be very useful.
require(arm)
then use ranef(Full_model) and fixef(Full_model)
On Wed, Sep 22, 2010 at 05:39:41PM +1000, Lorenzo Cattarino wrote:
> Hi again,
>
>
>
> Sorry, probably I was not clear enough.
>
>
>
> I was wondering if there is a way in R
Am 23.Sep.2010 um 18:27 schrieb Ralf B:
> I wonder what the best way is to access those values. I am using the
> following code:
>
> x1 <- c(1,2,1,3,5,6,6,7,7,8)
> x2 <- c(1,2,1,3,5,6,5,3,8,7)
> d1 <- density(x1, na.rm = TRUE)
> d2 <- density(x2, na.rm = TRUE)
> plot(d1, lwd=3, main="bla")
> lin
Hi Sayan,
This is exactly what I was looking for - it worked perfectly.
Many thanks!!
Also, thanks to everyone else for their suggestions.
Pele
--
View this message in context:
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Sent from the R help mailing list ar
Please consider this matrix:
x <- structure(c(5, 4, 3, 2, 1, 6, 3, 2, 1, 0, 3, 2, 1, 0, 0, 2, 1,
1, 0, 0, 2, 0, 0, 0, 0), .Dim = c(5L, 5L))
For each pair of columns, I want to calculate the proportion of entries
different than 0 in column j (i > j) that have lower values than the entries
in the s
On 23/09/2010 11:42 AM, wangguojie2006 wrote:
b<-runif(1000,0,1)
f<-density(b)
f is a list of things, including x values where the density is computed,
and y values for the density there. So you could do it by linear
interpolation using approx or approxfun. For example
> b <- runif(1000,
By range on the y-axis, do you mean distance? It would have to be if time is
on your x? Or am I misreading this?
You could just plot() with the data for your first individual, and then add
additional individuals after that using lines(), specifying a different
colour and/or line type for each indi
Hi, guys,
I'm using kernel "density" estimation. But how can I return to a density
estimation for a fixed point?
For example,
b<-runif(1000,0,1)
f<-density(b)
How can I get the value of density(b) at b=0.5?
Your help is extremely appreciated. Thanks.
Jay
--
View this message in context:
ht
I wonder what the best way is to access those values. I am using the
following code:
x1 <- c(1,2,1,3,5,6,6,7,7,8)
x2 <- c(1,2,1,3,5,6,5,3,8,7)
d1 <- density(x1, na.rm = TRUE)
d2 <- density(x2, na.rm = TRUE)
plot(d1, lwd=3, main="bla")
lines(d2, lty=2, lwd=3)
d1[1]
d1[2]
The last two lines allow m
hi guys..thx for help
i have to perform a calculation
P-B/T-B
where P is the values in pdf and B is the values in bdf and T in tdf
On Thu, Sep 23, 2010 at 7:49 PM, Mike Rennie wrote:
> First, you might want to start by generating a new column to identify your
> 'pdf" and "bdf" or whatever once
Sorry for posting this questions twice, but my previous question was
accidentally sent unfinished.
Dear list,
I would like to create a time series plot in which the paths of several
individuals are stacked above each other, with the x-axis being the total
observation period of three years ( 1
Dear list,
I would like to create a time series plot in which the paths of several
individuals are stacked above each other, with the x-axis being the total
observation period of three years ( 1.1.2004 to 31.12.2007) and the y-axis
being some defined range[min,max].
My data consist of Date/
Hello, I am using survey data with two stage sampling for two sub-populations.
I am looking for a package (or packages) that can do the following for a
measure of weight. The sub-populations are M (male) and F (female).
(1) empirical df and cdf for weight, and compare that across two sub-
popula
??? Use lm's data argument to do this, not the with() construction:
> x <- 1:10
> y <- 2*x+5 +rnorm(10)
> form <- formula(y~x)
> class(form)
[1] "formula"
> lm(form)
Call:
lm(formula = form)
Coefficients:
(Intercept)x
5.7471.921
> dat <- data.frame(x=x,y=y)
> rm(x,y)
>
On 23-Sep-10 15:08:06, Jay Vaughan wrote:
> How do I construct a figure showing predicted value plots for the
> dependent variable as a function of each explanatory variable
> (separately) using the results of a logistic regression? It would also
> be helpful to know how to show uncertainty in the
Are you aware that you can pass the function, itself, as an argument
(R is mostly a functional programming language where language objects
are first class objects).
e.g
> g <- function(x,fun)fun(x)
> g(2,function(x)x^2)
[1] 4
On Wed, Sep 22, 2010 at 4:06 PM, Jonathan Greenberg
wrote:
> R-hel
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Joshua Wiley
> Sent: Thursday, September 23, 2010 8:23 AM
> To: Ralf B
> Cc: r-help Mailing List
> Subject: Re: [R] Length of vector without NA's
>
> Hi Ralf,
>
> The usual way
Hi:
Try ?termplot.
HTH,
Dennis
On Thu, Sep 23, 2010 at 8:08 AM, Jay Vaughan wrote:
> How do I construct a figure showing predicted value plots for the dependent
> variable as a function of each explanatory variable (separately) using the
> results of a logistic regression? It would also be hel
Hi Ralf,
The usual way (as others have shown you), takes advantage of the fact
that the logical values TRUE and FALSE are counted as 1 and 0,
respectively. is.na() returns TRUE if the value is NA, so to find how
many are not NA, the result is reversed using ' ! '. Similar logic
can be used to fi
> this following code:
>
> x<-c(1,2,NA)
> length(x)
>
> returns 3, correctly counting numbers as well as NA's. How can I
> exclude NA's from this count?
>
sum(!is.na(x))
Peter
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r
try this
sum(!is.na(x))
I hope it helps.
Best,
Dimitris
On 9/23/2010 5:08 PM, Ralf B wrote:
Hi,
this following code:
x<-c(1,2,NA)
length(x)
returns 3, correctly counting numbers as well as NA's. How can I
exclude NA's from this count?
Ralf
__
Hi Pele,
I think this should work
file1$state.sum <- rowSums(file2[file1$state,6:10],0)
On Thu, Sep 23, 2010 at 7:46 PM, Steve Lianoglou <
mailinglist.honey...@gmail.com> wrote:
> Hi Pele,
>
> On Wed, Sep 22, 2010 at 12:40 PM, Pele wrote:
> >
> > Hi David - thanks for your suggestion, but I a
Hi,
this following code:
x<-c(1,2,NA)
length(x)
returns 3, correctly counting numbers as well as NA's. How can I
exclude NA's from this count?
Ralf
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the p
How do I construct a figure showing predicted value plots for the dependent
variable as a function of each explanatory variable (separately) using the
results of a logistic regression? It would also be helpful to know how to show
uncertainty in the prediction (95% CI or SE).
Thanks-
This emai
Hi Katrin,
(don't forget to reply-all to R-help messages -- by default they are
only sent to the recipient, and not back to the list).
On Thu, Sep 23, 2010 at 10:03 AM, Katrin Fleischer
wrote:
> The error message is
>
> Error in hdf5load("Jan2006.HDF") : unable to open HDF file: Jan2006.HDF
Dum
First, you might want to start by generating a new column to identify your
'pdf" and "bdf" or whatever once it's merged.
For the merging, see
?merge
But as someone's already pointed out, it's not clear what you are trying to
merge by.
Also, as your example calculations show, you don't need to m
Hi Pele,
On Wed, Sep 22, 2010 at 12:40 PM, Pele wrote:
>
> Hi David - thanks for your suggestion, but I am trying to avoid doing any
> merging and sorting for this step because the real file I will be working
> with has about 20 million records. If I can get this loop or something
> similar to
Hi,
On Thu, Sep 23, 2010 at 7:13 AM, Katrin Fleischer
wrote:
> Dear All,
>
> I have data in HDF file format and would like to read it into R.
> I have tried the package hdf5 without success.
What type of errors are you getting?
> Any ideas and suggestions??
Sure, from:
http://cran.r-project.or
Hello
I would like to pass a model formula as an argument to the with.mids
function from the mice package. The with.mids functon fits models to
multiply imputed data sets.
Here's a simple example
library(mice)
#Create multiple imputations on the nhanes data contained in the mice
package.
Hi,
On Thu, Sep 23, 2010 at 9:04 AM, rasanpreet wrote:
>
> hi guys
> i have multiple data frames which i want to merge.
> there are four of them..eg
Can you provide a (correct) example of what you want your merged
data.frame to look like?
What column do you want to use in your data.frame to mer
This is an example of the type of problem, and how i'm currently using
optim() to solve it.
mydata <- runif(500,-1,1)
myfunc <- function(p,d)
{
print(p <- floor(p))
ws <- function(i,n,x) sum(x[i-n+1]:x[i])
ws1 <- c(rep(NA,p[1]-1),sapply(p[1]:NROW(d),ws,p[1],d))
ws
In your call to polygon(), include lty="dashed" or "dotted" or whatever you
want your line type to look like.
take a good look at all the options in
?par
For everything you can customize in plots. Alternatively, Paul Murrel's R
Graphics book is the best reference I know for this sort of stuff.
You need to take into account the bin width as well (hence the extra multiple
you asked about), but it is simpler to just include prob=TRUE in the hist call,
then you do not need to do any adjustment on the y-values of the reference
distribution.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Dat
Hi:
Evidently I misunderstood your intention, so let's try again. I tried to
find a lattice solution but came up empty. I think this works instead using
base graphics. The idea is to build up vectors of start points (x0, y0),
endpoints (x1, y1) and the color of the segment between each pair of poi
On Sep 22, 2010, at 8:15 PM, Krambrink, Amy M wrote:
> Hello,
>
>
>
> I'm using plot.survfit to plot cumulative incidence of an event.
> Essentially, my code boils down to:
>
> cox <-coxph(Surv(EVINF,STATUS) ~ strata(TREAT) + covariates, data=dat)
>
> surv <- survfit(cox)
>
> plot(surv,mark.time=F
Hi
r-help-boun...@r-project.org napsal dne 23.09.2010 15:10:04:
> Hi group,
>
> I am currently plotting two densities using the following code:
>
> x1 <- c(1,2,1,3,5,6,6,7,7,8)
> x2 <- c(1,2,1,3,5,6,5,7)
> plot(density(x1, na.rm = TRUE))
> polygon(density(x2, na.rm = TRUE), border="blue")
>
>
On 23/09/2010 8:54 AM, Maas James Dr (MED) wrote:
This code worked fine for me, then did some cleaning up of formatting using ESS
(Emacs) and now I get this error, no idea what is causing it, all the
brackets/parentheses seem to be balanced. What have I done wrong?
It would help a lot if yo
On Sep 23, 2010, at 14:54 , Maas James Dr (MED) wrote:
> This code worked fine for me, then did some cleaning up of formatting using
> ESS (Emacs) and now I get this error, no idea what is causing it, all the
> brackets/parentheses seem to be balanced. What have I done wrong?
Look for ") ("
Hi group,
I am currently plotting two densities using the following code:
x1 <- c(1,2,1,3,5,6,6,7,7,8)
x2 <- c(1,2,1,3,5,6,5,7)
plot(density(x1, na.rm = TRUE))
polygon(density(x2, na.rm = TRUE), border="blue")
However, I would like to avoid bordering the second density as it adds
a nasty bottom
On Thu, 23 Sep 2010, ivo welch wrote:
thank you, achim. I will try chol2inv.
sandwich is a very nice package, but let me make some short
suggestions. I am not a good econometrician, so I do not know what
prewhitening is,
In general it means, that you do not look at a (typically multivariate
hi guys
i have multiple data frames which i want to merge.
there are four of them..eg
pdf
SampleID UVDose_J RepairHours Day_0 Day_45 Day_90
1SDM001 1.0 3 485.612 465.142 490.873
2SDM001 1.0 3 503.658 457.863 487.783
3SDM001 1.0 2 533.1
Peter Dalgaard gmail.com> writes:
> > Then I tried:
> >
> >> pdf()
> > Error in pdf() : unable to start device pdf
> > In addition: Warning message:
> > In pdf() : cannot open 'pdf' file argument 'Rplots.pdf'
>
> And your working directory is writable for you? Otherwise, "Change dir"
> from the
Dear All,
I want to upload data in HDF-file format into R. I have tried the
package 'hdf5' without success. I presume that my files are in the hdf4
format and therefore are not readable with this package. I understand
that its possible to convert hdf4 into hdf5 format using C, but I was
wonde
This code worked fine for me, then did some cleaning up of formatting using ESS
(Emacs) and now I get this error, no idea what is causing it, all the
brackets/parentheses seem to be balanced. What have I done wrong?
Thanks
Jim
p0.trial01 <- 0.25
TruOR01 <- 0.80
num.patients.01 <- 50
num.tri
thank you, achim. I will try chol2inv.
sandwich is a very nice package, but let me make some short
suggestions. I am not a good econometrician, so I do not know what
prewhitening is, and the vignette did not explain it. "?coeftest" did
not work after I loaded the library. automatic bandwidth s
I've run into a problem with a fitting procedure I would like R to solve for
me. Basically I have to fit some data to an equation which includes a sum
within the formula e.g.
Y(x;a,b,c) = f_i(x;a,b,c_i) + m*x
where a,b are unknown and f_i(x) is the sum of another function over a known
interval i
I am trying to create a NetCDF file with bathymetry (a matrix) and then to
add a grid type (an integer) as variables.
This is the relevant part of the code:
library(ncdf)
wfile="my_file.nc"
msv=-10
grdtp=2
# Dimensions
d1=dim.def.ncdf("lon","degrees",as.double(lon))
d2=dim.def.ncdf("lat","degrees
On 09/23/2010 09:38 PM, Richard DeVenezia wrote:
New to R. I am trying to create a simple xy plot wherein the line
segment color is determined by a categorical column
The following does not change colors for me, probably because I don't
quite have a handle on either functions or value mapping s
Hi:
time <- c(1, 2, 3, 7,10,11,14,16,20)
pressure <- c(0,10,20,20,50,18,60,65,90)
status <- c(0, 0, 1, 1, 1, 0, 3, 3, 3)
# You want to combine the vectors into a data frame instead
# of concatenating them into one long vector.
df <- data.frame(time, pressure, status)
df
time pressure stat
Hi there,
I ran into a weird problem using the np-package doing some local linear
kernel regression. Whenever I run the function "npregbw(...)" with the
option "regtype=ll" (local linear modelling), my optimal bandwidth is
supposed to be 1278946. This is kind of funny, because my regressor data
(1
Something like this ?
time <- c(1, 2, 3, 7,10,11,14,16,20)
pressure <- c(0,10,20,20,50,18,60,65,90)
status <- c(0, 0, 1, 1, 1, 0, 3, 3, 3)
statusColor <- c("green", "orange", "pink", "red")
plot(time, pressure)
for (i in 2:length(time)) lines(time[(i-1):i], pressure[(i-1):i],
col=statusColo
Hi
?merge
see all.x, all.y parameters
Regards
Petr
r-help-boun...@r-project.org napsal dne 23.09.2010 13:59:18:
> Greetings
> I need to merge two data sets. The first data set contains
information
> on individual organisms roosting in clusters,e.g., cluster one has 5
> individuals: 3 fe
Hi
I am not sure but you probably need to use segments.
?segments
AFAIK for line there is only one colour for whole line possible.
and in your function statusColor you shall not use if but ifelse or better
statusColor <- function (x) {
c("green","orange","pink","red")[x+1]
}
based on fact
Greetings
I need to merge two data sets. The first data set contains information
on individual organisms roosting in clusters,e.g., cluster one has 5
individuals: 3 females, two males, and one juvenile and so on for hundreds
of clusters. The second data set contains temperature data on each o
Hi
the other option to get last item is
tail(pheno.dt$year,1)
Regards
Petr
r-help-boun...@r-project.org napsal dne 23.09.2010 12:02:35:
> Yes, thanks! :)
>
> 2010-09-23 11:55, Ivan Calandra skrev:
> >Hi,
> > xlim and ylim should be given the extremes only:
> >
> > plot(x,y, xlim=c(pheno.d
New to R. I am trying to create a simple xy plot wherein the line
segment color is determined by a categorical column
The following does not change colors for me, probably because I don't
quite have a handle on either functions or value mapping syntax.
--
time <- c(1, 2, 3, 7,10,11,14
Dear All,
I have data in HDF file format and would like to read it into R.
I have tried the package hdf5 without success.
Any ideas and suggestions??
Kind regards,
Katrin
--
Katrin Fleischer
Vrije Universiteit Amsterdam
Faculty of Earth and Life Sciences
Subdepartment Hydrolgy and Geo-Environm
That's golden Henrique, thanks a lot! Worked like a charm even with large
datasets.
On Tue, Sep 21, 2010 at 2:56 PM, Henrique Dallazuanna wrote:
> Try this:
>
> d <- data.frame(A = letters[1:10], B = sample(letters[11:20]), C =
> sample(10))
> xtabs(C ~ A + B, d)
>
> On Tue, Sep 21, 2010 at 8:39
Siterer "Yihui Xie" :
lists are easier to post messages, but I really believe they have too
many disadvantages, e.g. (relatively) difficult to search, dull
interface, HTML not welcome (I don't like HTML in emails, though), no
lively images, no code highlighting, attachments often get chopped
off
darckeen hotmail.com> writes:
>
> Are there any packages that do non-linear integer otimization? Looked at
> lpSolve but i'm pretty sure it only works with linear programming and not
> non-linear, tried "L-BFGS-B" optim after flooring all my params, works
> somewhat but seems really inefficient
Dear Patrick,
You can fit such a model with the MCMCglmm package. Have a look at the vignette
of that package.
install.packages("MCMCglmm")
vignette("CourseNotes", package = "MCMCglmm")
But I'm affraid that this will require some rockclimbing upon the learning
curve if you are a R novice.
HTH
On Tue, 21-Sep-2010 at 12:55PM -0500, Vojtěch Zeisek wrote:
[...]
|> Well, if You think the niche is filled, never mind, but I think R
|> should have an official web forum. It can be stackoverflow (then
|> I'd expect links from main R pages to it) or I (or someone else)
|> can create it. Still, I
depmixS4 has reached some form of maturity and therefore we have bumped its
version number to 1.0-0 which is now on CRAN:
http://cran.r-project.org/web/packages/depmixS4/index.html
depmixS4 fits hidden (latent) Markov models of multivariate, mixed
categorical and continuous data, otherwise known a
Hi:
Look at the examples in ?Reduce
HTH,
Dennis
On Thu, Sep 23, 2010 at 12:31 AM, statquant2 wrote:
>
> Thank you very much, this solve the problem, but more generally is there a
> function that allow to apply a function to a list of object, applying
> recursively the function to each answer..
Yes, thanks! :)
2010-09-23 11:55, Ivan Calandra skrev:
>Hi,
> xlim and ylim should be given the extremes only:
>
> plot(x,y, xlim=c(pheno.dt$year[1],pheno.dt$year[nrow(pheno.dt)]),
> ylim=c(50,150), xlab="Year", ylab="Julian Day")
>^^
Hi,
xlim and ylim should be given the extremes only:
plot(x,y, xlim=c(pheno.dt$year[1],pheno.dt$year[nrow(pheno.dt)]),
ylim=c(50,150), xlab="Year", ylab="Julian Day")
^^ ^^
Does it work now?
Ivan
Le 9/23/2010 11:44, fuge
On Sep 23, 2010, at 11:44 , fugelpitch wrote:
>
> Sorry, but now I understand what you're doing, and it works as a stand-alone
> in the console! :)
>
> But why doesn't it work in the xlim?
You have a colon where a comma would work.
BTW: Wouldn't xlim=range(pheno.dt$year) be more expedient?
>
OK, got it to work now, just messed up the ","' and ":". Thank you!
--
View this message in context:
http://r.789695.n4.nabble.com/referencing-last-row-in-a-column-tp2551645p2551721.html
Sent from the R help mailing list archive at Nabble.com.
__
R-he
Sorry, but now I understand what you're doing, and it works as a stand-alone
in the console! :)
But why doesn't it work in the xlim?
__
> pheno.dt$year[1]
[1] 1877
> pheno.dt$year[nrow(pheno.dt)]
[1] 1916
> plot(x,y, x
Another option for doing opertions by grouping variables is to the the
plyr package.
d <- data.frame(x=1:10,
g1=LETTERS[rep(11:12,each=5)],
g2=letters[rep(21:23,c(3,3,4))]
)
library(plyr)
ddply(d, c("g1", "g2"), function(z){
z$x <- z$x / max(z$x)
z
On Thu, Sep 23, 2010 at 8:05 AM, KAYIS Seyit Ali wrote:
> I need to create eps file which is the required figure format
> of the journal that I want to submit a paper. I am able to
> create files in pdf or wmf format but not in eps format. Is
> there a way to convert pdf or wmf to eps? or alterna
Dear List,
I fitted a bivariate extreme value distribution to my matrix (data). Now
I would like to add ‘quantile curves’ to the plot, similar to
those calculated with
qcbvnonpar()
But I need the
probability of the upper right corner instead of the lower left corner.
(So that the curve
Thanks, but I have multiple entries for each year so the column reads:
year
1877
1877
1877
1877
1878
1878
1878
1878
1879
1879
1879
1879
So I need to pick out the year from the last row rather than just counting
rows.
--
View this message in context:
http://r.789695.n4.nabble.com/referencin
You could do data$year[ nrow(data) ]
On 23 September 2010 18:56, fugelpitch wrote:
>
> I am trying to set u a limit for my plot window according to the range of a
> year column I have.
> The first year in the range is just simply the first row, referred to as
> data$year[1]. How can I find the la
I am trying to set u a limit for my plot window according to the range of a
year column I have.
The first year in the range is just simply the first row, referred to as
data$year[1]. How can I find the last row in a similar way to set the last
year in the range?
What I want to accomplish is the f
Thank you very much, this solve the problem, but more generally is there a
function that allow to apply a function to a list of object, applying
recursively the function to each answer...
mathematically for a 2 argument function f(u,v) you would like a function g
doing
g(u1,u2,u3,u4,u5) =f(f(f(f
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