thanks, it works! well, now it shows 1s instead of 0s and 2s instead of 1s,
which fine with me...but could you explain me what this function actually
does?
Gabor Grothendieck wrote:
Iff dd is your data frame then:
dd$prev - ave(dd$of, dd$Id, FUN = function(x) c(NA, head(x, -1)))
On
Hello Everyone,
Let's say I have a vector as follows:
vowels = c(aa, aa, ah, ao, eh)
I want to make a factor consisting of two levels: the things that are
aa and those that are not aa. How would I do that?
I tried using factor(vowels, levels=c(aa, ~aa), exclude=NULL), but
this gives me
Hi Thibault
Please give a short example of what the data looks now like and what you
want it to look like.
regards
Christiaan
2009/10/22 Thibault Grava gr...@unbc.ca
Hello,
I'm using R to run a acoustic analysis software called Seewave. I ask the
code to extract a list of variables from
Hi,
I'm trying to build R packages. My package name is TEST.
(/Users/apple/Documents/R/TEST)I type something in Terminal.But it showed an
error...
applede-macbook-pro:~ apple$ R CMD BUILD TESTError: cannot change to directory
'anRpackage1'
I don't know where is wrong...Hope that you could help
Hi R all
Does anyone know how to get predictions and/or put prediction intervals on
model averaged results from dRedging::model.avg for a poisson GLM.
Predict doesn't work in this case like it does dirently on a glm?
glm_aab_null- glm(aabundance~1, family=poisson, data=tenuredata)
.
Hi everybody,
I'm using the method described here to make a linear regression:
http://www.apsnet.org/education/advancedplantpath/topics/Rmodules/Doc1/05_Nonlinear_regression.html
## Input the data that include the variables time, plant ID, and severity
time -
Next time if dd is your data frame display dput(dd) so we know exactly
what you have.
From your output I guessed it was numeric but the behavior of the 0/1
column that you describe is consistent with it being a factor rather
than numeric. Try
dd$of - as.numeric(levels(dd$of)[dd$of])
or
Hi all,
Is there a possibility of getting the interactions feature values for the
variables of a dataset by applying the Random Forest algorithm?
Thanks,
Chrysanthi
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Try this;
factor(vowels == aa, labels = c(aa, ~aa))
On Thu, Oct 22, 2009 at 4:36 AM, nty...@clovermail.net wrote:
Hello Everyone,
Let's say I have a vector as follows:
vowels = c(aa, aa, ah, ao, eh)
I want to make a factor consisting of two levels: the things that are aa
and those that
That has not yet been implemented in the R version of the package.
Best,
Andy
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Chrysanthi A.
Sent: Thursday, October 22, 2009 6:40 AM
To: r-help@r-project.org
Subject: [R]
Hi JIm
Not exactly what I want, as I can't draw the lines corresponding to the
different curves in the plot as I can do in legend, but may be still of
help.
Thkx, Javier
Jim Lemon-2 wrote:
On 10/21/2009 01:30 AM, jcano wrote:
betav-c(0.78,0.94,0.88,0.41,0.59,4.68)
Dear R community,
probably my question is obvious but I did not find any solution yet by
browsind the mailing list results.
I need to perform a simple ols regression in a dataset with cross section
data, where no temporal dimension is inserted. In this data set there are
missing values. I would
I am using twoord.plot and my Y axis units on the left y overlap. I tried
using cex.axis in my par command but that only adjusted the x units, not the
y. cex.axis in twoord.plot did not help. How do I adjust Y units in the
twoord.plot?
example:
twoord.plot (lx=myear, ly=z, rx=myear,
Dear R-Users,
I would like to have the opinion of the list on the following matter. I
have this generic function that creates multiple lattice scatterplots
per page based upon different subsets of the same dataset. The use of
different line/point colors/symbols in each plot is based upon a
Gina, at the terminal, make sure you are in directory
Users/apple/Docments/R
When you type R CMD build TEST
Sounds like you might be in the TEST directory, not the R directory. The
build system looks for a subdirectory with your package name when it starts.
Bryan
*
Bryan Hanson
Aneeta,
You will have to have a seasonal component built into your model,
because the seasonal variation does matter, particularly -where- you are
geographically (San Diego, Chicago, Denver, Miami are very different).
Generally, there is a sinusoidal daily temperature variation, but
frontal
See if this gives you a hint:
size - 50
num - 100
sum(size[1:(num/2)])
-Peter Ehlers
Manny Gomez wrote:
Dear R users,
I'd like to ask u whether you know how to sort out the following:
I'm trying to reproduce a dataset of clusters, and for that I need to build
up a cluster index (inside
On Oct 22, 2009, at 6:56 AM, Francesca Pancotto wrote:
Dear R community,
probably my question is obvious but I did not find any solution yet by
browsind the mailing list results.
I need to perform a simple ols regression in a dataset with cross
section
data, where no temporal dimension is
Hi,
I don't know if it helps, but looking at the output of xyplot you can
extract the legend (a grid.frame) as follows,
library(grid)
library(lattice)
p = xyplot(x~y, group=x,data=data.frame(x=1:10,y=1:10),
auto.key=list(space=right))
legend = with(p$legend$right,
You're too vague about what you want.
Do you want to convert
time freq
1 10
2 20
3 30
to
tf
1 10
2 20
3 30
(use paste())
or to
x
1
2
3
10
20
30
(use c())
or ??
-Peter Ehlers
Thibault Grava wrote:
Hello,
I'm using R to run a acoustic analysis
--- On Tue, 10/20/09, hadley wickham h.wick...@gmail.com wrote:
From: hadley wickham h.wick...@gmail.com
Subject: Re: [R] Putting names on a ggplot
To: John Kane jrkrid...@yahoo.ca
Cc: m...@z107.de, R R-help r-h...@stat.math.ethz.ch
Received: Tuesday, October 20, 2009, 10:59 AM
On Sun,
On 10/22/2009 9:48 AM, rkevinbur...@charter.net wrote:
I am having a hard time interpreting the results of the 'shapiro.test' for
normality. If I do ?shapiro.test I see two examples using rnorm and runif. When
I run the test using rnorm I get a wide variation of results. Most of this may
be
Hi
I have a sunflowerplot, in which the x and y axis cover different ranges
(different orders of magnitude). Is it possible to specify different
digits for the x and y axis?
Rainer
--
Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology,
UCT), Dipl. Phys. (Germany)
Hi,
I have a vector with elements
rs.id=c('rs100','rs101','rs102','rs103')
And a dataframe 'snp.id'
1 SNP_100 rs100
2 SNP_101 rs101
3 SNP_102 rs102
4 SNP_103 rs103
Task
Hi all,
I would like to calculate the area under the ROC curve for my predictive
model. I have managed to plot points giving me the ROC curve. However, I do
not know how to get the value of the area under.
Does anybody know of a function that would give the result I want using an
array of
Hi,
You need to be a lot more specific about what you want. Are you trying to
solve a partial differential equation (PDE) in time and space, in spherical
coordinates? Do you have a closed form solution that provides the values of
the dependent variable (e.g. concentration or temperature) as
Markus Weisner-2 wrote:
I am working on a new package to do fire department analysis. I am
working
with emergency dispatch data from different agencies that all contain the
same information but have slightly different formats. Typically the
variable names and date-time formats are
Hi,
I understand that the glm.fit calls LINPACK fortran routines instead of
LAPACK because it can handle the 'rank deficiency problem'. If my data
matrix is not rank deficient, would a glm.fit function which runs on
LAPACK be faster? Would this be worthwhile to convert glm.fit to use
LAPACK?
olivier.abz wrote:
Hi all,
I would like to calculate the area under the ROC curve for my predictive
model. I have managed to plot points giving me the ROC curve. However, I do
not know how to get the value of the area under.
Does anybody know of a function that would give the result I want
Well, you can use the trapezoidal rule to numerically calculate any area under
the curve. I don't know if a specific exists but you could create one. The
principle is basically to compute the area between two successive points of
your profile with:
AREA=0.5*(Response1 + Response2)/(Time2-Time1)
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of olivier.abz
Sent: Thursday, October 22, 2009 7:23 AM
To: r-help@r-project.org
Subject: [R] How to calculate the area under the curve
Hi all,
I would like to calculate
I assume that you have an ordered pair (x, y) data, where x = sensitivity, and
y = 1 - specificity. Your `x' values may or may not be equally spaced. Here
is how you could solve your problem. I show this with an example where we can
compute the area-under the curve exactly:
# Area under
See package ROCR. Then see ?performance; in the details, it describes a
measure of auc.
Tom Fletcher
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of olivier.abz
Sent: Thursday, October 22, 2009 9:23 AM
To: r-help@r-project.org
Ted,
LAPACK is newer and is supposed to contain better algorithms than LINPACK. It
is not true that LAPACK cannot handle rank-deficient problems. It can.
However, I do not know if using LAPACK in glm.fit instead of LINPACK would be
faster and/or more memory efficient.
Ravi.
Hi,
On Oct 22, 2009, at 10:12 AM, Praveen Surendran wrote:
Hi,
I have a vector with elements
rs.id=c('rs100','rs101','rs102','rs103')
And a dataframe 'snp.id'
1 SNP_100 rs100
2 SNP_101 rs101
3 SNP_102 rs102
4
Try this:
merge(rs.id, snp.id, by.x = 1, by.y = 2)$V2
or:
with(snp.id, V2[match(rs.id, V3)])
On Thu, Oct 22, 2009 at 12:12 PM, Praveen Surendran
praveen.surend...@ucd.ie wrote:
Hi,
I have a vector with elements
rs.id=c('rs100','rs101','rs102','rs103')
And a dataframe 'snp.id'
Dear All,
Is there some way of drawing a boxplot, with R, when one does not have
the original continuous data, but only the data grouped in classes?
The function boxplot() can only deal with original data.
Thanks in advance,
Paul
__
Paul Smith phh...@gmail.com wrote
Is there some way of drawing a boxplot, with R, when one does not have
the original continuous data, but only the data grouped in classes?
The function boxplot() can only deal with original data.
It's not clear how the data are, now. What are the classes? Are
Dear R users,
I'm performing some GLMMs analysis with a negative binomial link.
I already performed such analysis some months ago with the lmer() function but
when I tried it today I encountered this problem:
Erreur dans famType(glmFit$family) : unknown GLM family: 'Negative Binomial'
Does
Hello Brothers/Sister,
I am trying to create a GUI for some formulas
in R using tcltk and then convert them into R package . Can anyone
please tell me how to create a package having GUI capabilities in it.
I would be thankful if somebody can help me.May god bless every one
On Thu, Oct 22, 2009 at 04:36:22PM +0100, Paul Smith wrote:
Is there some way of drawing a boxplot, with R, when one does not have
the original continuous data, but only the data grouped in classes?
The function boxplot() can only deal with original data.
Do you mean a numeric vector grouped
On Thu, Oct 22, 2009 at 5:31 PM, Jakson A. Aquino
jaksonaqu...@gmail.com wrote:
Is there some way of drawing a boxplot, with R, when one does not have
the original continuous data, but only the data grouped in classes?
The function boxplot() can only deal with original data.
Do you mean a
Greetings,
I am recoding a dummy variable (coded 1,0) so that 0 = 2. I am using
the line
sciach$dummyba[sciach$ba==0] - 2
I notice that it creates a new column dummyba, with 0 coded as 2 but
with 1's now coded as NA. Is there a simple way around this in the line
I am using, or do I need
Use ifelse:
sciach$dummyba - ifelse(sciach$ba == 0, 2, 1)
On Thu, Oct 22, 2009 at 2:37 PM, David Kaplan
dkap...@education.wisc.edu wrote:
Greetings,
I am recoding a dummy variable (coded 1,0) so that 0 = 2. I am using the
line
sciach$dummyba[sciach$ba==0] - 2
I notice that it creates a
Function arma is crashing in some (pathological, but crashing is never good)
cases.
For example:
library(tseries)
arma(c(2.01, 2.22, 2.09, 2.17, 2.42), order=c(1,0))
I came to that pathological series while doing test cases; probably there
are crashing cases with longer series.
Alberto
David Kaplan wrote:
Greetings,
I am recoding a dummy variable (coded 1,0) so that 0 = 2. I am using
the line
sciach$dummyba[sciach$ba==0] - 2
I notice that it creates a new column dummyba, with 0 coded as 2 but
with 1's now coded as NA. Is there a simple way around this in the line
I am
On 10/22/2009 12:37 PM, David Kaplan wrote:
Greetings,
I am recoding a dummy variable (coded 1,0) so that 0 = 2. I am using
the line
sciach$dummyba[sciach$ba==0] - 2
I notice that it creates a new column dummyba, with 0 coded as 2 but
with 1's now coded as NA. Is there a simple way
I'm not clear on why we are emphasizing the trapezoidal rule when the
Wilcoxon approach gives you everything plus a standard error.
Frank
Ravi Varadhan wrote:
I assume that you have an ordered pair (x, y) data, where x = sensitivity, and
y = 1 - specificity. Your `x' values may or may not
Another pathological test.
arima does not crash for that series that crashes arma:
arima(c(2.01, 2.22, 2.09, 2.17, 2.42), order=c(1,0,0))
However, arima crashes for this:
arima(c(1.71, 1.78, 1.95, 1.59, 2.13), order=c(1,0,0))
arima seems pretty consistent in its crashing behaviour, since
Well, I wonder which 'Sister' you're thinking of. I can think of
more than a few ... Sarah? Ulrike? Heather? ...
And I'm not so sure God would bless an old sinner like me.
But as to your query: perhaps you could get some ideas from
the Greg Snow's TeachingDemos package.
-Peter Ehlers
SANKET
Hello,
I am generating .png images using the Cairo package in a for loop
(looping on the number of zones, number of zones equals the number of
plots to create based on different zone data). When I run the R
script the .png files are created but they are all black? If I comment
out the
Hi everyone,
I am wondering if there exists a stepwise regression function for the
Bayesian regression model. I tried googling, but I couldn't find anything.
I know step function exists for regular stepwise regression, but nothing
for Bayes.
Thanks
--
View this message in context:
Dear all,
If I have the following data frame:
set.seed(21)
df1 - data.frame(col1=c(rep('a',5), rep('b',5), rep('c',5)),
col4=rnorm(1:15))
col1 col4
1 a 0.793013171
2 a 0.522251264
3 a 1.74641
4 a -1.271336123
5 a 2.197389533
6 b 0.433130777
7
Thanks for the suggestion.
I found some documentation on why accessing a data.gram using the
matrix notation (e.g., [i,j]) is so expensive, which was the cause of
the problem.
regards,
Roberto
On Thu, Oct 22, 2009 at 12:05 AM, Jim Holtman jholt...@gmail.com wrote:
try running Rprof on the
On Thu, Oct 22, 2009 at 12:46 PM, Douglas M. Hultstrand
dmhul...@metstat.com wrote:
Hello,
I am generating .png images using the Cairo package in a for loop (looping
on the number of zones, number of zones equals the number of plots to create
based on different zone data). When I run the R
Hi Folks,
I have been using contour() to produce some contour plots
(of a spatially-smooted density produced by kde2d()), with
very satisfactory results.
I now want access to the coordinates of the points on the
contours, and it would seem that contour() does not return a
value, so there is
On Thu, Oct 22, 2009 at 6:19 PM, Alberto Monteiro
albm...@centroin.com.br wrote:
Another pathological test.
arima does not crash for that series that crashes arma:
arima(c(2.01, 2.22, 2.09, 2.17, 2.42), order=c(1,0,0))
However, arima crashes for this:
arima(c(1.71, 1.78, 1.95, 1.59,
Allan.Y wrote:
Hi everyone,
I am wondering if there exists a stepwise regression function for the
Bayesian regression model. I tried googling, but I couldn't find
anything. I know step function exists for regular stepwise regression,
but nothing for Bayes.
Why? That seems so ...
ROBARDET Emmanuelle wrote:
Dear R users,
I'm performing some GLMMs analysis with a negative binomial link.
I already performed such analysis some months ago with the lmer() function
but when I tried it today I encountered this problem:
Erreur dans famType(glmFit$family) : unknown GLM
On 10/22/09, Peter Ehlers ehl...@ucalgary.ca wrote:
But as to your query: perhaps you could get some ideas from
the Greg Snow's TeachingDemos package.
Also, check the related Rcmdr plug-in [1].
Liviu
[1] http://cran.r-project.org/web/packages/RcmdrPlugin.TeachingDemos/index.html
Hi Ted,
Here's what the last example in ?contour says:
## contourLines produces the same contour lines as contour
So, even without digging into the source, I would guess
that your hunch is correct.
Cheers,
Peter Ehlers
(Ted Harding) wrote:
Hi Folks,
I have been using contour() to produce
A follow-up to my previous query.
Say one of the results returned by contourLines() is
C.W - contourLines()
Then C.W is a list of (in this case 28) lists,
each of which is a list with components $level (a single number),
$x (the x coords of the points on a contour at that level) and
$y
Hi Peter!
Many thanks -- I had overlooked that line.
Ted.
On 22-Oct-09 18:53:25, Peter Ehlers wrote:
Hi Ted,
Here's what the last example in ?contour says:
## contourLines produces the same contour lines as contour
So, even without digging into the source, I would guess
that your
On 22/10/2009 2:57 PM, (Ted Harding) wrote:
A follow-up to my previous query.
Say one of the results returned by contourLines() is
C.W - contourLines()
Then C.W is a list of (in this case 28) lists,
each of which is a list with components $level (a single number),
$x (the x coords of
I wish to save a scatter plot comprising approx. 2 million points
in order to include it in a LaTeX document.
Using 'pdf(...)' produces a file of size about 20 MB, which is
useless.
Using 'cairo_pdf(...)' produces a smaller file, around 3 MB. This
is still too large. Not only that the document
Ben Bolker wrote:
Allan.Y wrote:
Hi everyone,
I am wondering if there exists a stepwise regression function for the
Bayesian regression model. I tried googling, but I couldn't find
anything. I know step function exists for regular stepwise regression,
but nothing for Bayes.
Why? That
On 22-Oct-09 19:03:06, Duncan Murdoch wrote:
On 22/10/2009 2:57 PM, (Ted Harding) wrote:
A follow-up to my previous query.
Say one of the results returned by contourLines() is
C.W - contourLines()
Then C.W is a list of (in this case 28) lists,
each of which is a list with
On Thu, Oct 22, 2009 at 1:39 PM, Ben Bolker bol...@ufl.edu wrote:
ROBARDET Emmanuelle wrote:
Dear R users,
I'm performing some GLMMs analysis with a negative binomial link.
I already performed such analysis some months ago with the lmer() function
but when I tried it today I encountered
carferper wrote:
Hello veryone,
I am interested in the diffusion of particles inside a sphere, and its
release through a small pore on the sphere surface. Unfortunately, I have
not found the way to do this in R. Could you help me?
Thank very much in advance for your help
I have
Dear Lasse,
This won't answer your specific questions and I apologize for that.
AFAIK, pdf() produces uncompressed PDFs only. But you could use tools
like pdftk to compress your PDFs. About the PNGs, you can always set
the 'res' argument to improve resolution, but it won't beat the PDFs.
Frank E Harrell Jr f.harr...@vanderbilt.edu wrote
Ben Bolker wrote:
Allan.Y wrote:
Hi everyone,
I am wondering if there exists a stepwise regression function for the
Bayesian regression model. I tried googling, but I couldn't find
anything. I know step function exists for regular
Ted Harding asked:
I can of course get these individually with, for the 5th one for
instance,
C.W[[5]]$level
C.W[[5]]$x
C.W[[5]]$y
But I can't see how to obtain, in one line and without running
a nasty loop, to get all the levels at once!
In other words, I'm looking for an
Hello,
I need to do an intersection between the list elements (partitionslist) and
the columns and rows of a matrix (mm), so that the result will be the sums
of the rows and columns.
Thanks a lot,
Romildo Martins
Example
1.The Intersection and sum betweeen partitionslist[[1]][[2]] and mm is
The problem with the pdf files is that they are storing the information for
every one of your points, even the ones that are overplotted by other points.
The png file is smaller because it only stores information on which color each
pixel should be, not how many points contributed to a
Sorry... example 2 was error!
2009/10/22 Romildo Martins romildo.mart...@gmail.com
Hello,
I need to do an intersection between the list elements (partitionslist) and
the columns and rows of a matrix (mm), so that the result will be the sums
of the rows and columns.
Thanks a lot,
Joel,
The following should answer most of your questions:
time - c(seq(0,10),seq(0,10),seq(0,10))
plant - c(rep(1,11),rep(2,11),rep(3,11))
severity - c(
42,51,59,64,76,93,106,125,149,171,199,
40,49,58,72,84,103,122,138,162,187,209,
Usage
data(gasoline)
Format
A data frame with 60 observations on the following 2 variables.
octane
a numeric vector. The octane number.
NIR
a matrix with 401 columns. The NIR spectrum
and I see the gasoline data to see below
NIR.1686 nm NIR.1688 nm NIR.1690 nm NIR.1692 nm NIR.1694 nm
On 23/10/2009, at 8:20 AM, Peter Flom wrote:
Frank E Harrell Jr f.harr...@vanderbilt.edu wrote
Ben Bolker wrote:
Allan.Y wrote:
Hi everyone,
I am wondering if there exists a stepwise regression function
for the
Bayesian regression model. I tried googling, but I couldn't find
Hi,
On Oct 22, 2009, at 2:35 PM, bbslover wrote:
Usage
data(gasoline)
Format
A data frame with 60 observations on the following 2 variables.
octane
a numeric vector. The octane number.
NIR
a matrix with 401 columns. The NIR spectrum
and I see the gasoline data to see below
NIR.1686 nm
I notice that the intervals package indicates a dependence on R =
2.9.0. Is there some feature of R 2.9 that intervals depends on, or
might it work with R 2.7.1, which I am running?
Thanks.
Ross Boylan
__
R-help@r-project.org mailing list
Hi all,
I would like to invoke a function that takes multiple arguments (some of
which are specified columns in the data frame, and others that are
independent of the data frame) on split parts of a data frame, how do I do
this?
For example, let's say I have a data frame
fitness_data
name
Aneeta,
If I understand the figure at
http://db.csail.mit.edu/labdata/labdata.html this problem deals
with sensors in a lab that is probably isolated from outdoor
temperature changes.
I assume the predictive model must detect when a rampaging 800
pound gorilla messes with a sensor. Do we
Hi all,
I have re-begging to use R.
I need to using a matrix Y and X with the first row been the name to get a
ols regression and to plot a two elements graph...on the top
the scattergraph and the line representing the regressiónbelow the
resids...at the same time I want to plot with small
On Thu, Oct 22, 2009 at 8:28 PM, Greg Snow greg.s...@imail.org wrote:
The problem with the pdf files is that they are storing the information for
every one of your points, even the ones that are overplotted by other points.
The png file is smaller because it only stores information on which
I just realized my earlier post of my question below was not in
Plain Text mode, hence the repeat post...apologies!
Kavitha
On Thu, Oct 22, 2009 at 4:19 PM, Kavitha Venkatesan
kavitha.venkate...@gmail.com wrote:
Hi all,
I would like to invoke a function that takes multiple arguments (some of
For getting the details in the outer points, here is what I do.
1. use hexbin to create the big central blob (but with additional info).
2. use the chull function to find the outer points and save their indices in
another vector
3. use chull on the rest of the points (excluding those found
Hi,
I'm running into a problem subsetting a data frame that I have never
encountered before:
dim(chkPd)
[1] 32136
df = head(chkPd)
df
PNWB Sire Dam MG SEX
601 1001 715349 61710 61702 67F
969 1001_1 511092 616253 615037 168F
986
Alternatively you could install the NRAIA package and collapse the
construction of the plot to a call to plotfit as shown in the
enclosed.
Note that this is a poor fit of a logistic growth model. There are no
data values beyond the inflection point so the Asym parameter (the
asymptote) is poorly
Hi Sean,
Comment in line below.
On Thu, Oct 22, 2009 at 5:39 PM, Sean MacEachern sean.mace...@gmail.com wrote:
Hi,
I'm running into a problem subsetting a data frame that I have never
encountered before:
dim(chkPd)
[1] 3213 6
df = head(chkPd)
df
PN WB Sire
Hi Ista,
I think I'm suffering long dayitis myself. You are probably right. I
don't use subset that often. I typically use brackets to subset
dataframes. Essentially what I am trying to do is take my original
dataframe (chkPd) and subset it using a smaller dataframe with some
matching PN IDs.
Is this what you want?
df = data.frame('id'=c(1:100),'res'=c(1001:1100))
dfb=df[1:10,]
dfc = df[df$id %in% dfb$id,]
Still not sure, but that's my best guess. Going back to your original
data you can try
dfb = chkPd[chkPd$PN %in% df$PN,]
Hope it helps,
Ista
On Thu, Oct 22, 2009 at 6:10 PM,
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Hi,
I have something strange here...
I want to subset a large sparse matrix, with the subset being still in
sparse form. Easily achievable with mm[i,,drop=F] , right? Well, it
doesn't work on the matrix I'm working on.
This is a very large wikipedia
Works perfectly!
Thanks to all who responded.
Sean
On Thu, Oct 22, 2009 at 6:24 PM, Ista Zahn istaz...@gmail.com wrote:
Is this what you want?
df = data.frame('id'=c(1:100),'res'=c(1001:1100))
dfb=df[1:10,]
dfc = df[df$id %in% dfb$id,]
Still not sure, but that's my best guess. Going back
Hello,
I've been fitting a random effects model using nlme to some data, but I am
discovering that the variation in my random effect is very small. As a result,
I would like to replace it as a fixed effect (i.e. essentially fit the same
model but with no random effect).
As I understand it
Hello
I was wondering if there is any possibility of estimating a linear
regression model with GARCH errors in R? I tried using fGarch library, but I
could only find for instance how to estimate arma-garch models or similar
combinations.
Thank you,
Ioana
--
View this message in context:
I am having trouble with the recode function that is provided in the
CAR package. I trying to create a new factors based on existing factors.
E.g.
x - as.factor(1:20)
y - recode(x, 1:5='A'; 6:10='B'; 11:15='C'; 16:20='D' )
y
[1] A A A A A 6 7 8 9 A A A A A A A A A A A
Levels: 6 7 8 9 A
Hi Nikhil,
The problem is that your initial as.factor(0) causes x to have
values of 1 : 20 instead of 1 : 20. There are two possible
solutions:
1)
library(car)
x - 1:20
y - as.factor(recode(x, 1:5='A'; 6:10='B'; 11:15='C'; 16:20='D' ))
y
[1] A A A A A B B B B B C C C C C D D D D D
Levels:
help(gnls, pack=nlme)
hth,
Kingsford Jones
On Thu, Oct 22, 2009 at 4:36 PM, Michael A. Gilchrist mi...@utk.edu wrote:
Hello,
I've been fitting a random effects model using nlme to some data, but I am
discovering that the variation in my random effect is very small. As a
result, I would
* Message by -Greg Snow- from Thu 2009-10-22:
If you want to go the pdf route, then you need to find some way
to reduce redundant information while still getting the main
points of the plot. With so many point, I would suggest
looking at the hexbin package (bioconductor I think) as one
Hi everyone,
I'm sure this has been asked before, but I could not find the right
search terms to find it.
If I want to see what summarizing a model fit with lm() does, I can
write summary.lm. But what if I want to see what summarizing a model
fit with lmer() (lme4 package) does?
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