Hi,
Im just doing some Beta-Regressions with the betareg package. My question is
now, is there a possibility
to calculate the marginal effects with the betareg package or is there
another package which can handle marginal effects on regression output for
the beta class?
I try to calculate the
This was garbled en route, but
(a) Your R is far too old: please update as the posting guide asked of
you. (You seem to be using a pre-release of 2.11.0.)
(b) You need to set DISPLAY, as the message says. It usually is set
at a Mac OS X console, so perhaps you need to ask on R-sig-mac
On Mon, 22 Aug 2011, thmsfuller...@gmail.com wrote:
Hi,
'R CMD INSTALL -l' will stop if some error is encountered. I don't
find in the manual an option to keep the command running. Is there
such an option?
No. Why would you want to install a package with errors?
If you mean that
R CMD
Hi
as far as I understand your question it seems to me that
round(5*(1.4^(0:10)))
gives you your sequence
and
cumsum(round(5*(1.4^(0:10
gives you summary sequence.
Regards
Petr
I definitely used too much lines s of code because I still don't know
how to
do some staff.
but it
Dear all
I'm familiarising myself with Ridge Regressions in R and the following
is bugging me: How does one get p-values for the coefficients obtained
from MASS::lm.ridge() output (for a given lambda)? Consider the
example below (adapted from PRA [1]):
require(MASS)
data(longley)
gr -
Without a reproducilbe example, it is hard to give you good advise...
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens ashz
Verzonden: maandag 22 augustus 2011 23:53
Aan: r-help@r-project.org
Onderwerp: [R] Time series and
Hi,
When using the glmnet() function of the package glmnet, A series of
coefficients is returned for a list of descending lambda values.
I am unable to locate anything in the documentation that explains HOW this
choice of lambda series is made. (There is documentation about how to choose
my
At 02:15 23/08/2011, Nathan Miller wrote:
Hi all,
See comment in-line
I have a data set that looks a bit like this.
feed1
RFU Site Vial Time lnRFU
1 811 10 10.702075
2 4752111 20 10.768927
3 4290511 30 10.66674
4 4686711 40
I searched archives for how to do quantile regressions with complex survey
data, and there was nothing that could be helpful to a first time user. I am
looking for equivalent of the functions in quantreg package. A vignette
and/or examples will be very helpful. Is there someone on this list who
Dear colleagues,
I would like to run a cluster analysis on a number of variables. They are
Likert Scale (0 to 10), but they also have a Don't know' option at the end of
the scale. Apparently, with the 'Don't Know' option in place, they cannot be
considered to be linear or ordinal. How can
On 08/23/2011 08:39 PM, Iasonas Lamprianou wrote:
Dear colleagues,
I would like to run a cluster analysis on a number of variables. They are Likert
Scale (0 to 10), but they also have a Don't know' option at the end of the
scale. Apparently, with the 'Don't Know' option in place, they
Checked the authors' paper on page 8 it explains how the \lambda sequence is
chosen. I guess this is what you were looking for.
http://www.stanford.edu/~hastie/Papers/glmnet.pdf
Eugen Pircalabelu
(0032)471 842 140
(0040)727 839 293
From: Noah Silverman
Hi Roger,
Maybe I'm missing a clue. Here's an example:
x-rnorm(50)
x1-rnorm(50)
y-x+x1+rnorm(50)
out-rq(y~x+x1, tau=1:9/10)
plot(summary(out))
plot.out-plot(summary(out))
#I change the variable names
dimnames(plot.out)-list(c(intercept, sex, inc))
#Look at the plot
#it's ONE simple
Thank you for the hint. But this means that I will still have missing responses
in my data and this defeats my purpose to analyse the data using cluster
analysis.
Dr. Iasonas Lamprianou
Department of Social and Political Sciences
University of Cyprus
From:
Dear all,
I am using Fisher r to z transformation for pooling partial correlation
estimates over multiple imputed data (number of imputations = 200). The number
of observations in my data is 190. Unfortunately, when i calculate p values for
the pooled estimates, some of them are p 1 (ranging
Hi All!
I am interested in testing whether the means for the data I am investigating
are equal to a specific value - let's say 0.01. I have already run a
one-way ANOVA and know that the differences in the means are not
significant, so now I want to know what values the means take on. otestme
is
On 08/23/2011 09:07 PM, Iasonas Lamprianou wrote:
Thank you for the hint. But this means that I will still have missing
responses in my data and this defeats my purpose to analyse the data
using cluster analysis.
If it is really necessary to avoid missing values, and there is an
indifference
sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)
locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1]
Hi,
I am using this script to read a xlsx file to a data frame:
library(xlsx)
File - file.path(d:, car , car95-99.xlsx)
B_car - read.xlsx(File, raw_data)
Car2x - data.frame(month = B_car$Date,Ch = B_car$Ch.des,
lat=B_car$Latitude)
The last row in the data.frame is always NA, how can I remove
Dear R list, I have one very elementary question regrading correlation between
two variables.
x = c(44,46,46,47,45,43,45,44)
y = c(44,43,41,41,46,48,44,43)
cov(x, y)
[1] -2.428571
However, if I try to calculate the covariance using the formula as
covariance = sum((x-mean(x))*(y-mean(y)))/8
well, you don't have the correct denominator, i.e., n-1, with n denoting
the sample size. Have a look at the *Details* section of the online help
file for cov(), and try also
sum((x-mean(x))*(y-mean(y)))/7
cov(x, y)
I hope it helps.
Best,
Dimitris
On 8/23/2011 1:18 PM, Vincy Pyne wrote:
In addition, something has gone wrong, Vincy, with your data x,y
between evaluating cov(x,y) and evaluating your explicit formula.
If I repeat your commands:
x = c(44,46,46,47,45,43,45,44)
y = c(44,43,41,41,46,48,44,43)
cov(x, y)
# [1] -2.428571
sum((x-mean(x))*(y-mean(y)))/8
# [1]
Hello!
with the R system command I would like to call a perl script which needs an
input directory and an output directory in form of a path. When I put in the
path directly it works. The script line looks as follows:
system(perl '../path1' '../path2' '../path3')
If I store the path in
Dear list members,
I want to apply AR(1)-GARCH(1,1) model in order to conduct a test of
structural shifts in conditional correlations which I previously estimated.
To be more exact, first, I estimate the conditional correlations using the
DCC-GARCH model. Now I want to check whether these
See ?paste
or use system2(perl, shQuote(c(p1, p2, p3)))
On Tue, 23 Aug 2011, syrvn wrote:
Hello!
with the R system command I would like to call a perl script which needs an
input directory and an output directory in form of a path. When I put in the
path directly it works. The script line
There is probably a more elegant way to do this, but this worked for me:
# a function to identify the first occurrence of a run of values
first - function(x) {
l - length(x)
c(1, 1-(x[-1]==x[-l]))
}
# identify the first occurrence of a run of Rev values
df$frst - first(df$Rev)
#
ashz wrote on 08/23/2011 03:25:57 AM:
Hi,
I am using this script to read a xlsx file to a data frame:
library(xlsx)
File - file.path(d:, car , car95-99.xlsx)
B_car - read.xlsx(File, raw_data)
Car2x - data.frame(month = B_car$Date,Ch = B_car$Ch.des,
lat=B_car$Latitude)
The last row
Hi,
it works great with the paste command! thanks a lot!
Best,
syrvn
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So, you are looking for confidence intervals for each factor level?
You can use the predict() function to do that.
fit - aov(values ~ ind, data=otestme)
newdat - data.frame(ind=factor(levels(otestme$ind)))
cbind(newdat, predict(fit, newdata=newdat, interval=confidence))
Jean
Anna Dunietz wrote
Dear Mr. Dimitris and Mr Harding, thanks a lot for your guidance. It will be
interesting to find out how the Excel deals with this formula. I will try it.
Thanks again.
Regards
Ashok
--- On Tue, 8/23/11, ted.hard...@wlandres.net ted.hard...@wlandres.net wrote:
From: ted.hard...@wlandres.net
Dear Mr Dimitris and Mr Harding, by mistake I have typed my colleagues name
(i.e. Ashok) while thanking you. Please excuse me for that.
Regards
Vincy
--- On Tue, 8/23/11, ted.hard...@wlandres.net ted.hard...@wlandres.net wrote:
From: ted.hard...@wlandres.net ted.hard...@wlandres.net
Subject:
Hmmm, maybe I will try this as an excercise, thank you Jim
Dr. Iasonas Lamprianou
Department of Social and Political Sciences
University of Cyprus
From: Jim Lemon j...@bitwrit.com.au
To: Iasonas Lamprianou lampria...@yahoo.com
Cc: r-help@r-project.org
On Aug 23, 2011, at 8:21 AM, Jean V Adams wrote:
ashz wrote on 08/23/2011 03:25:57 AM:
Hi,
I am using this script to read a xlsx file to a data frame:
library(xlsx)
File - file.path(d:, car , car95-99.xlsx)
B_car - read.xlsx(File, raw_data)
Car2x - data.frame(month = B_car$Date,Ch =
syrvn ment...@gmx.net writes:
If I store the path in a variable/object and call the perl script again it
does not run and I don't know how to overcome that issue.
p1 - ../path1
p2 - ../path2
p3 - ../path3
system(perl p1 p2 p3)
You want something like:
system(paste(perl, p1, p2, p3))
Can you please make an example available that reproduces the error, e.g.
some code that produces a similar csv file like yours?
Uwe Ligges
On 23.08.2011 06:29, Mitu De wrote:
Hello,
I am using Windows 7 and R version 2.13.1. I was trying to read a csv file
which has about 1.5 million
On Tue, Aug 23, 2011 at 7:54 AM, JC Matthews j.c.matth...@bristol.ac.uk wrote:
Thankyou for your replies, you've answered my question and given me more to
think on. I guess it is unwise to draw any conclusions from the
standardised results for these reasons.
No, by all means try to draw
2011/8/22 Uwe Ligges lig...@statistik.tu-dortmund.de
On 22.08.2011 18:43, Rainer M Krug wrote:
Hi
I want to create a dependence diagram of a subset of the packages on CRAN
and would therefore like to read the DEACRIPTION files into a list. The
list
should be as follow for each package:
On 8/23/2011 3:35 AM, Liviu Andronic wrote:
[snip]
But how does one obtain the customary 'lm' summary information for the
model above? I tried supplying the chosen lambda to Design::ols()
using its 'penalty' argument, but strangely the results differ. See
below.
require(Design)
Can someone show me how to modify one (R.oo) class's object inside another
(R.oo) class's method? Is that possible with the R.oo package? A quick
example or reference to an example would be outstanding...
Thanks,
Ben
[[alternative HTML version deleted]]
Thanks Dennis! Worked perfectly. I keep forgetting that plyr can split data
based on multiple subsetting variables.
Thanks so much,
Nate
On Mon, Aug 22, 2011 at 10:12 PM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
You're kind of on the right track, but there is no conditioning
formula in
Divide by 8 leads biased estimator of covariance.
R cov function calculates unbiased estimator(divide by (sample size)-1).
Regards,
Kohta
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Hi,
I've got an R script that I'm trying to turn into a ArcGis script tool so
that I can run it from ModelBuilder in Arc. Arc isn't giving me any errors
when I run the model holding the current tool, but the run time for the R
script is 0 seconds. I don't know if the parameters aren't being
I you had posted your code which gave the results, we would have seen that
you switched your variables.
Bart
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Hi,
I used simple linear regression with the R software and EXCEL on the same
data. Although , I find the same R2=0.84, I find different estimated values
(intercept and slope). For the R software (slope =0.0009, Intercept =
-0.1478), for EXCEL (slope =927.7, Intercept = 154,41).
When I use the
Thankyou for your replies, you've answered my question and given me more to
think on. I guess it is unwise to draw any conclusions from the
standardised results for these reasons.
James.
--On 22 August 2011 17:30 +0100 ted.hard...@wlandres.net wrote:
On 22-Aug-11 15:37:40, JC Matthews
Hi,
i´ve read all that i could about the post-hoc test for kruskall-Wallis, now
i know that i need to run Duncan test, but i´m not able to find in any
package or script. I s anybody knows where can i find it?
Thanks
-
Mario Garrido Escudero
PhD student
Dpto. de Biología Animal, Ecología,
Dear R-users,
I need to produce a histogram where for every breaks there are the mean of the
data.
I tried tu use the function hist(x, break=20 ... ) but this return the
numerosity for every breaks, not the mean.
Any hint?
Thanks in advance,
francesco
Dear All,
I am quite a newbie to R. Trying to learn it these days. Sorry for asking
such a basic question, but could you kindly tell me how to map unique string
values to integers in a data frame? I have a graph which has, most of its,
vertices' attributes in a string format. I would like to
Hi ,
I know this question has been asked twice in the past but to my knowldege,
it still hasn't been solved.
I am doing a zero inflated binomial model using the VGAM package, I need to
obtain p values for my Tvalues in the vglm output. code is as follows
Thanks, I see the problem. R did the opposite of Excel. I invert the
position of variables in my code and the result is now correct.
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Sent from the R help
Hi all,
I am trying to do spectral analysis of a time series data. But I am not sure
how to do it. Can anyone tell me if there are any package I can use to do
the analysis using fast fourier transform?
Thanks in advance.
Cassie
[[alternative HTML version deleted]]
Hi
Dear All,
I am quite a newbie to R. Trying to learn it these days. Sorry for
asking
such a basic question, but could you kindly tell me how to map unique
string
values to integers in a data frame? I have a graph which has, most of
its,
vertices' attributes in a string format. I would
Hi
Hi all,
I am trying to do spectral analysis of a time series data. But I am not
sure
how to do it. Can anyone tell me if there are any package I can use to
do
the analysis using fast fourier transform?
Maybe ?fft or ?spectrum
Regards
Petr
Thanks in advance.
Cassie
I do not believe your code (minimal as it is) would work: the correct
argument is breaks. More generally, do you really mean to say that hist(x,
breaks = 20) immediately returns the bin counts? It doesn't on my machine
and if you knew how to get the counts, you should be able to get the
midpoints
Dear R users,
I’d like to pose aquestion about pMCMC and HDP.
I have performed a mixed logistic regression by MCMCglmm (a very good package)
obtaining the following results:
Iterations = 250001:799901
Thinning interval = 100
Sample size = 5500
DIC: 10.17416
G-structure: ~ID_an
post.mean
Hi everybody,
I'm fairly new to survival analysis with R and have some questions how to
apply and interpret the coxph and related functions:
I have time-dependent covariates with several measurements per subject with
constant delta t. The covariates change in each time step.
I fitted the
On Aug 23, 2011, at 7:48 AM, Komine wrote:
Hi,
I used simple linear regression with the R software and EXCEL on the
same
data. Although , I find the same R2=0.84, I find different estimated
values
(intercept and slope). For the R software (slope =0.0009, Intercept =
-0.1478), for EXCEL
Dear list,
I have a table with entries for 20 animals - x and y coordinates to analyze
movement with the package adehabitat.
The package does all the necessary analyses, but I need to create an object
of class ltraj for each animal first. For this kind of object I need to
define xy
Hi all,
I have an object that looks (roughly) like the following:
l - list(a = matrix(rnorm(9), 3), b = matrix(rnorm(9), 3), c =
matrix(rnorm(9), 3))
l$a[3,] - sample(c(Message 1, Message 2, Message 3))
l$b[3,] - sample(c(Message 1, Message 2, Message 3))
l$c[3,] - sample(c(Message 1, Message
Is there anything in R similar to spectogram command in matlab?
On Tue, Aug 23, 2011 at 10:20 AM, Petr PIKAL petr.pi...@precheza.cz wrote:
Hi
Hi all,
I am trying to do spectral analysis of a time series data. But I am not
sure
how to do it. Can anyone tell me if there are any package
Try this:
subset(as.data.frame(do.call(rbind, lapply(l, [, , 1))), row3 == Message 1)
On Tue, Aug 23, 2011 at 1:28 PM, Lara Poplarski larapoplar...@gmail.com wrote:
Hi all,
I have an object that looks (roughly) like the following:
l - list(a = matrix(rnorm(9), 3), b = matrix(rnorm(9), 3), c
His is better, but you can also use a for loop...
out-data.frame(rows=1:3)
for(i in 1:3){
if(l[[i]][3]=='Message 1') {
out$V1[i]-l[[i]][1]
} else {
out$V1[i]-NA
}
}
but shouldn't if your list is very long
On Tue, Aug 23, 2011 at 9:35 AM, Henrique Dallazuanna www...@gmail.comwrote:
Daniel Malter daniel at umd.edu writes:
id-rep(c(1:100),each=2)
obs-rep(c(0:1),100)
d-rep(sample(c(-1,1),100,replace=T),each=2)
base.happy-rep(rnorm(100),each=2)
happy-base.happy+1.5*d*obs+rnorm(200)
data-data.frame(id,obs,d,happy)
I am statistically confused tonight. When the
Hi everyone,
I have the following problem. I have some small p-values but when I use
qnorm(1-4e-30)
I get an error.
Is there anyway to get around this?
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
How about
qnorm(4e-30, lower.tail = FALSE) ?
You cannot subtract 4e-30 from 1 and expect to get something else than
1 (exactly).
Peter
On Tue, Aug 23, 2011 at 10:21 AM, Jim Silverton jim.silver...@gmail.com wrote:
Hi everyone,
I have the following problem. I have some small p-values but when
On Aug 23, 2011, at 12:21 PM, Jim Silverton wrote:
Hi everyone,
I have the following problem. I have some small p-values but when I use
qnorm(1-4e-30)
I get an error.
Is there anyway to get around this?
Here is a hint:
qnorm(1 - 4e-30)
[1] Inf
qnorm(1)
[1] Inf
# See ?all.equal
Hi,
Looking at a large data set with many factors.
I would like to expand each factor variable into multiple new variables for
each level. (0,1) coding.
My first though was just to code a big nasty loop, to take each level and cbind
a column onto my data set. But, that seems painful. There
On Aug 23, 2011, at 12:34 PM, Noah Silverman wrote:
Hi,
Looking at a large data set with many factors.
I would like to expand each factor variable into multiple new variables for
each level. (0,1) coding.
My first though was just to code a big nasty loop, to take each level and
On Tue, Aug 23, 2011 at 10:34 AM, Noah Silverman noahsilver...@ucla.edu wrote:
Hi,
Looking at a large data set with many factors.
I would like to expand each factor variable into multiple new variables for
each level. (0,1) coding.
My first though was just to code a big nasty loop, to
Thank you very much. I have managed to do the mappings. And sorry for not
mentioning the data type. It was a data frame.
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fantastic Michael! The code V = sort(unique(ave(x,cut(x,HistOut$breaks
works good. I got what I need! that is 20 mean values. I calculate the first
value in excel, it's the same (79.61429).
HistOut= hist(data$ DMP_m3.jaso..10, freq=F, breaks =20,
xlim=c(0,3000),col=grey26,
The Predict.Plot function in the TeachingDemos package can help you visualize
interactions. It will work best if Month is treated as a continuous variable.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original
Hello everyone,
I am trying to figure out a way of replacing missing observations in one of
the variables of a data frame by values of another variable. For example,
assume my data is X
X -as.data.frame(matrix(c(9, 6, 1, 3, 9, NA, NA,NA,NA,NA,
6, 4, 3,NA, NA, NA, 5, 4, 1, 3),
Hi,
On Tue, Aug 23, 2011 at 12:29 PM, StellathePug ritacarre...@hotmail.com wrote:
Hello everyone,
I am trying to figure out a way of replacing missing observations in one of
the variables of a data frame by values of another variable. For example,
assume my data is X
X
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Ista Zahn
Sent: Tuesday, August 23, 2011 11:06 AM
To: StellathePug
Cc: r-help@r-project.org
Subject: Re: [R] Replacing NAs in one variable with values of another
variable
R 2.13
Vista
Colleagues,
I have encountered a problem with Windows environment variables that I don't
understand. My R code is designed to execute another program that uses Intel
Fortran. Intel Fortran apparently has some complicated issues regarding
environment variables. As a result,
m.fenati at libero.it m.fenati at libero.it writes:
Dear R users,
I’d like to pose aquestion about pMCMC and HDP.
I have performed a mixed logistic regression by
MCMCglmm (a very good package)
obtaining the following results:
[snip]
post.mean l-95% CI u-95% CIeff.samp
ID_an
@Josh Thanks! That's exactly what I need.
@ Marc. I want to do this manually because I want to do many things with the
data beyond R's built in functions.
--
Noah Silverman
UCLA Department of Statistics
8117 Math Sciences Building
Los Angeles, CA 90095
On Aug 23, 2011, at 10:43 AM, Joshua
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Dennis Fisher
Sent: Tuesday, August 23, 2011 11:40 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] Setting Windows environment variables via a batch file
R 2.13
Vista
On Tue, 23-Aug-2011 at 12:22AM -0400, Jim Holtman wrote:
| if you are running on Windows, the install AutoHotKey and this will allow
you to setup some macros to accomplish this. this is what notepad++ does and
it is easybto extend.
|
| Sent from my iPad
|
| On Aug 22, 2011, at 9:07, Robert
cassie jones cassiejones26 at gmail.com writes:
Is there anything in R similar to spectogram command in matlab?
Don't know what it does, perhaps you could tell us, but:
have you *looked* at example(spectrum) ... ?
Ben Bolker
__
Greetings all,
I'm porting an algorithm from MATLAB to R, and noticed some minor
discrepancies in small decimal values using rowSums and colSums which
are exacerbated after heavy iteration and log space transformation.
This was rather perplexing as both programs claimed and appeared to use
Hi Roger,
Maybe I'm missing a clue. Here's an example:
x-rnorm(50)
x1-rnorm(50)
y-x+x1+rnorm(50)
out-rq(y~x+x1, tau=1:9/10)
plot(summary(out))
plot.out-plot(summary(out))
#I change the variable names
dimnames(plot.out)-list(c(intercept, sex, inc))
Why not just change the names of the
Hi,
I have a very large XTS object. It is about 600,000 entries over 1 year time.
I would like to subset a specific piece, by number of days not a specific
date.
The way I do it now is awkward. Would love to find a way to do this easier to
generate a new object just containing the days I
Not directly related to what you said below, but might I suggest that for
numerical work all.equal() might be a little more robust in a
computationally heavy implementation.
x = c(0.812672,0.916541,0.797810) #dont' call variables c -- just a bad idea
y = x[1]+x[2]+x[3]
sum(x) ==y
[1] FALSE
?all.equal
--
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
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Simple question but searching rseek did not yield the results I wanted.
Question: Is there a way to open a help manual for a package from within R.
For instance I would like to type a function in r for the tm package and R
would open that PDF as seen here:
After loading the package, does help.start() do what you want?
-- Bert
On Tue, Aug 23, 2011 at 2:32 PM, Tyler Rinker tyler_rin...@hotmail.com wrote:
Simple question but searching rseek did not yield the results I wanted.
Question: Is there a way to open a help manual for a package from
Dear R-users,
I am trying to get the plyr syntax right, without much success.
Given:
d- data.frame(cbind(x=1,y=seq(20100801,20100830,1)))
names(d)-c(first, daterep)
d2-d
# I can convert the daterep column in place the classic way:
d$daterep-as.Date(strptime(d$daterep, format=%Y%m%d))
# How to
I don't think help.start is what I'm looking for but I may be doing it wrong.
I tried:
library(tm)
help.start(tm)
This may be inappropriate as it returns:
library(tm)
help.start(tm)
Error in if (update) make.packages.html(temp = TRUE) :
argument is not interpretable as logical
Just
On Tue, Aug 23, 2011 at 8:17 PM, Daniel Lai dan...@bccrc.ca wrote:
Greetings all,
I'm porting an algorithm from MATLAB to R, and noticed some minor
discrepancies in small decimal values using rowSums and colSums which are
exacerbated after heavy iteration and log space transformation. This
Hi Jean,
On Tue, Aug 23, 2011 at 6:16 PM, jjap jean.plamon...@fpinnovations.ca wrote:
Dear R-users,
I am trying to get the plyr syntax right, without much success.
Given:
d- data.frame(cbind(x=1,y=seq(20100801,20100830,1)))
names(d)-c(first, daterep)
d2-d
# I can convert the daterep
Jean,
Ista is right, but:
In your function you are asking as.Date to convert the whole data.frame df
rather than just your daterep column.
out-ddply(d2, .(daterep), function(df)
as.Date(strptime(df$daterep,format='%Y%m%d')))
str(out)
'data.frame':30 obs. of 2 variables:
$ daterep: num
On Aug 23, 2011, at 6:16 PM, jjap wrote:
Dear R-users,
I am trying to get the plyr syntax right, without much success.
Given:
d- data.frame(cbind(x=1,y=seq(20100801,20100830,1)))
names(d)-c(first, daterep)
d2-d
# I can convert the daterep column in place the classic way:
On Aug 23, 2011, at 6:38 PM, David Winsemius wrote:
On Aug 23, 2011, at 6:16 PM, jjap wrote:
Dear R-users,
I am trying to get the plyr syntax right, without much success.
Given:
d- data.frame(cbind(x=1,y=seq(20100801,20100830,1)))
names(d)-c(first, daterep)
d2-d
# I can convert the
David,
For me, on a windows machine, help(package=) results in a summary window
opening in R rather than the detailed help manual that is available through
CRAN. Others suggested help.start() which takes me to a CRAN library site but
requires me to still click with the mouse to locate the
Try:
help(package=tm)
(You do not need library(). )
--
David.
On Aug 23, 2011, at 6:17 PM, Tyler Rinker wrote:
I don't think help.start is what I'm looking for but I may be doing
it wrong. I tried:
library(tm)
help.start(tm)
This may be inappropriate as it returns:
library(tm)
Hi Barry,
Shock and horror indeed, addition is not _associative_, at least for
floating point arithmetic [1].
Flipping the order of the operands seems to be the explanation of the
discrepancies between R and MATLAB as you suggest:
In R:
a = 0.812672
b = 0.916541
c = 0.797810
On Tue, Aug 23, 2011 at 6:56 PM, Daniel Lai dan...@bccrc.ca wrote:
Hi Barry,
Shock and horror indeed, addition is not _associative_, at least for
floating point arithmetic [1].
Flipping the order of the operands seems to be the explanation of the
discrepancies between R and MATLAB as you
Hi
Are you looking for the axTicks() function (or possibly the pretty()
function) ?
Paul
On 19/08/2011 3:51 a.m., David A. Johnston wrote:
Hi R list,
I like the default ticks that are set up using grid.xaxis() or grid.yaxis()
with no arguments. Finding good values for the 'at' argument is
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