Dear All:
I create a file named :P_Value with only one simple function:
P_Value - function( Table ) {
S = fisher.test(Table, alternative = two.sided);
return(S$p.value);
}
However, it seems that it's impossible to use this function
how about:
d=data.frame(ht=rnorm(20,60,20),sex=sample(0:1,20,rep=T)); d
with(d,by(ht,sex,mean))
with(d,by(ht,sex,function(x)mean(x=60)))
hth, david freedman
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Thanks.
Estimation of the same model with the same dataset gives different results
when tsls (from package sem) is used as opposed to ivreg() (from package
AER); both parameter estimates and standard errors are different. This is
intriguing. Can anyone throw some light on this? Is there any
Dear R users,
I have a univariate time series and I want to examine the presence of
seasonality in the series without using graphical methods like
autocorrelation(acf) as I need to automate it in a function. Is there any
other way by which I could find the presence of seasonality and if
Dear Lowie,
To use a function in R that you have saved in a file, you need to
source it in first (see ?source). Try this,
source(file=pathtoyourfile/P_Value.R) #or whatever your file is named exactly
Once you do that, you should be able to use your function. If you
always want it to load at
-Mensaje original-
De: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] En nombre de Ozan Bakis
Enviado el: domingo, 02 de mayo de 2010 21:25
Para: r-help@r-project.org
Asunto: [R] adding year information to a scatter plot
Hi R users,
I would like to add
Lowie,
You can save the function in a file called
'P_value.txt' or something like it
(I prefer 'P_value.r').
Then you load the function into your R session by
Using:
Source(P_value.txt,sep=) ).
It is better if you keep all your R functions into
The same directory, say directory my_rfunc,
Then
Try using
source(P_Value)
before anything else.
Also, P_Value can be written as a one-liner:
P_Value - function(Table) fisher.test(Table)$p.value
so you don't really need a separate function at all.
Bill Venables
CSIRO/CMIS Cleveland Laboratories
-Original Message-
From:
I could use some help generating a time series for the Mackey-Glass equation:
dx/dt = 0.2 x(t-tau)/(1 + x(t-tau)^10) - 0.1 x(t) for the case there tau = 17.
I tried the ddesolve package but the dde(...) function seems to hang, not
producing anything. Can someone show me the R script how to do
On 03.05.2010 04:26, Claudia Penaloza wrote:
I ran this code (several times) from the Quick-R web page (
http://www.statmethods.net/advstats/cart.html) but my cross-validation
errors increase instead of decrease (same thing happens with an unrelated
data set).
Why does this happen?
Since
Move everything in ./inst/doc that is required for the vignette.
Best,
Uwe Ligges
On 03.05.2010 05:00, Sébastien Bihorel wrote:
Dear R-users,
I am going through the last steps of package prep before submission to CRAN
and I have the following problem. My package contains a single vignette
On 02.05.2010 21:55, galen kaufman wrote:
Dear R Community,
I am trying to run a command line in R that will open an external program, have
it import a specific input file, run the program, then close the program. The
command line that I got from the developer of the model to do this looks
Thanks David Henrique,
I've been using R for over two years and always used cbind or rbind, that
was what I was taught by several folk, and on training courses, you learn
something new every day!
Cheers,
Ross
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Hello,
I have used layout() to produce to 2 plots on a page, leaving a plotting space
above them. I would like Legend, which will actually be a real legend, to be
centered above the two graphs. Right now I am only able to position Legend
above the second graph that I create... obviously I am
Dear R Users,
what is the best way to trace the origin of warning messages?
If an error occurs I can use traceback() to see where it comes from. I
would like to do similar investigation, where a warning message
originates from.
Is there an option to turn warnings to errors?
I tried the
Dear R-users,
I am confused about defining S3 methods when objects or methods involve
several dots. Here is my problem.
Package coda defines the following two generic methods:
as.mcmc(x,...)
as.mcmc.list(x,...)
I want to add an S3 method for the second method for my object of class
twDEMC
Hi
r-help-boun...@r-project.org napsal dne 01.05.2010 00:13:06:
Dear R list,
Our statisticians usually give us results back in a PDF format. I would
like
to be able to copy and past tables from R output directly into a
Microsoft
Word table since this will save us tons of time, be more
Hi
r-help-boun...@r-project.org napsal dne 30.04.2010 23:11:54:
Hello David,
On Apr 30, 2010, at 11:00 PM, David Winsemius wrote:
Note: Loops may be just as fast or faster than apply calls.
How come!? is this true also for other similar functions: lapply, tapply
and sapply?
Then
Use options(warn=2)
Uwe Ligges
On 03.05.2010 08:55, Thomas Wutzler wrote:
Dear R Users,
what is the best way to trace the origin of warning messages?
If an error occurs I can use traceback() to see where it comes from. I
would like to do similar investigation, where a warning message
On Mon, 3 May 2010, Thomas Wutzler wrote:
Dear R Users,
what is the best way to trace the origin of warning messages?
options(warn=2)
at least for the first warning.
If an error occurs I can use traceback() to see where it comes from. I
would like to do similar investigation, where a
To continue Petr suggestion, a simple variation for this would be to use
sink()
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
On 05/02/2010 10:00 PM, Duncan Murdoch wrote:
...
I've seen this description a couple of times lately, and I think it's
worth pointing out that it's misleading. The deparse(substitute(x))
trick returns the *expression* that was passed to the argument x.
Sometimes that's the name of a variable,
Jim Lemon wrote:
On 05/02/2010 10:00 PM, Duncan Murdoch wrote:
...
I've seen this description a couple of times lately, and I think it's
worth pointing out that it's misleading. The deparse(substitute(x))
trick returns the *expression* that was passed to the argument x.
Sometimes that's the
Hello,
I can't find how to get de column name from a data.frame dollar reference.
To make it simple, I'd like to obtain Bar from a foo$Bar notation.
I've tried col.names(foo$Bar), names(foo$Bar) and so on without sucess.
Regards
Blaise
[[alternative HTML version deleted]]
Well, it's may not be the best way to do so, but you may use this:
colnames(foo[n])
where n is the column number, or the column name inside '' ''.
The $ can't be used because it turn the column content into a vector,
loosing its properties (e.g.: column and row names). But when you using
[],
On 03-May-10 11:09:43, adrien Penac wrote:
Hello,
I can't find how to get de column name from a data.frame dollar
reference.
To make it simple, I'd like to obtain Bar from a foo$Bar notation.
I've tried col.names(foo$Bar), names(foo$Bar) and so on without sucess.
Regards
Blaise
You
as.character(quote(foo$Bar)[[3]])
[1] Bar
Hint: this is nothing to do with data frames ($ applies to lists). $
is an operator, so foo$Bar is a call. quote() stops it being
evaluated, [[3]] selects the third of the elements (which are $, foo,
Bar) and as.character turns the name into a
Good afternoon,
I am looking for a way to do a scatterplot of 4 values summing to 1
inside a 3D symplex, i.e. an equilateral pyramid. With the function
triax.plot I can do that with 3 values summing to 1, but I can't find
an equivalent with an extra dimension.
Thanks to whoever can help me!
Thank a lot for these answers.
Some of you wondered why I needed
to do that!
In fact, I have not so big data.frame whith many
columns (98) and many of them are similar (many binary answers, some
factor data and a few quantitative datas).
As I am a lazy guy, I
wanted to do a sort of function
See the describe functions in these packages:
Hmisc
pysch
prettyR
On Mon, May 3, 2010 at 8:12 AM, adrien Penac farfel...@yahoo.fr wrote:
Thank a lot for these answers.
Some of you wondered why I needed
to do that!
In fact, I have not so big data.frame whith many
columns (98) and many of
On 05/03/2010 05:15 PM, John Poulsen wrote:
Hello,
I have used layout() to produce to 2 plots on a page, leaving a plotting space above them. I would
like Legend, which will actually be a real legend, to be centered above the two graphs.
Right now I am only able to position Legend above the
On 05/03/2010 09:43 PM, Gabriele Esposito wrote:
Good afternoon,
I am looking for a way to do a scatterplot of 4 values summing to 1
inside a 3D symplex, i.e. an equilateral pyramid. With the function
triax.plot I can do that with 3 values summing to 1, but I can't find
an equivalent with an
To whoever it may correspond,
My name is Oscar Saenz and I am working on my thesis in Spain.
I am using GAMs in R and, now that I have estimated my models, I need to plot
the predicted smooth functions against the dependent variable (just as in
Carlslaw et al. 2007, fig. 1*).
Otherwise,
Hi,
The model specification formula language introduced in Chambers and
Hastie potentially handles rather complex models. Typically the user
specifies the model and in a separate argument specifies the method. For
example, one specifies a general linear model with glm(formula,family).
But
Dear R users,
I think i have a simple question which i want to explain by an example;
i have several 2-digit industry codes that i want to use for conducting
by-industry analysis but i think there is a problem with the degrees of freedom!
for example, when i do my analysis without any
Thanks Tal/Petr, I regularly use the clipboard option but for most of
what I do people would prefer to see things in proportional fonts and
I'd love to be able to reformat tabulations easily to get nice layouts.
Of course, I've got very adept at global search and replace of
successive spaces with
Try, unique(dataset[,1:a]), where a is the number of columns that you have.
1:a would apply the unique to all columns.
-
Lanna Jin
lanna...@gmail.com
510-898-8525
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R 2.10.0
windows XP
I am trying to get the coefficients returned from lmer. Normally I would use
the names function to get the objects returned by the function. This does not
work in lmer, but it does for lm:
- lmer(Reaction ~ Days + (1|Subject) + (0+Days|Subject), sleepstudy)
names(fm2)
Hello,
Hierarchical factors are a very common data structure. For instance, one
might have municipalities within states within countries within
continents. Other examples include occupational codes, biological
species, software types (R within statistical software within analytical
software),
Thanks Tal Thomas, I am now experimenting with both SWord and R2wd and
both are certainly a huge step forward for me, tied as I am to Word and
the Windoze/M$ world for now.
Chris
Tal Galili sent the following at 01/05/2010 09:44:
Hi all,
I forwarded this question to the r-com mailing
as a r noob i am having another problem:
i have a big dataframe where each row corresponds to one entry and each
column is a field...
for instance, i have the column ID and time and many more...
Id like to get a dataframe where all IDs are just included once (some users
with that ID might have
Hi Marshell,
What exactly do you mean by handles this kind of data structure?
What do you want R to do?
Best,
Ista
On Mon, May 3, 2010 at 9:44 AM, Marshall Feldman ma...@uri.edu wrote:
Hello,
Hierarchical factors are a very common data structure. For instance, one
might have municipalities
Did you try: if x is the data frame, unique(x)?
-
Lanna Jin
lanna...@gmail.com
510-898-8525
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if that doesn't work, maybe also try:
if x is your data frame with length a columns, unique(x[,1:a]).
-
Lanna Jin
lanna...@gmail.com
510-898-8525
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Sent from the R help
thanks for your effort.
to be more precise:
ID , OS, time and many more are the columns.
each entry is a row.
when I do:
x - unique(dataset$ID)
It just gives me a list of all IDs (levels).
I want to get a dataframe where just one entry (row) for each ID is
included...
like:
userA , Win,
hello,
i did median based linear regression and computed conf.intervals for my
coefficients.
but something must have went wrong as the estimates are out of estimates
confidence bounds.
does someone know what is the problem here.
i get warning messages about wilcox.test is not able to do
On May 3, 2010, at 9:41 AM, John Sorkin wrote:
R 2.10.0
windows XP
I am trying to get the coefficients returned from lmer. Normally I
would use the names function to get the objects returned by the
function. This does not work in lmer, but it does for lm:
When you read the help page for
Hi John,
David's advice is probably sufficient, but I thought I would just add
the following example, showing how to find out what extractor
functions have been defined for a particular s4 class:
class(fm2) #find out the class of fm2 (mer)
showMethods(classes=mer) # find out what methods are
Hi Serdal,
There is a lot of confusion here (how much is yours and how much is
mine remains to be seen). See specific comments in line.
On Mon, May 3, 2010 at 9:19 AM, serdal ozusaglam
saint-fi...@hotmail.com wrote:
Dear R users,
I think i have a simple question which i want to explain by an
On May 3, 2010, at 10:38 AM, Ista Zahn wrote:
Hi Serdal,
There is a lot of confusion here (how much is yours and how much is
mine remains to be seen). See specific comments in line.
Also inline comments.
On Mon, May 3, 2010 at 9:19 AM, serdal ozusaglam
saint-fi...@hotmail.com wrote:
Does anyone know how to deal with DATES in MySQL database connections?
dbWriteTable converts dates in a data frame to a character.
Thanks!
--
Dr. Brett Ginsburg
Assistant Professor
Department of Psychiatry
The University of Texas Health Science Center at San Antonio
San Antonio, TX 78229
plus, it took long time to complete this analysis for only 24 records!
Although I get the error msg, the result seems to be correct.
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I am conducting a very simple t test for two genes using lapply (i try to
avoid loop since i will have thousands of genes later on). however, I got
strange error msg like the followings. It looks that R is complaining my
factor has only one level, which is not the case (I check many times).
See
example(gam)
particularly
plot(gam.object,se=TRUE).
On Mon, May 3, 2010 at 5:20 AM, Oscar Saenz de Miera saenz...@yahoo.comwrote:
To whoever it may correspond,
My name is Oscar Saenz and I am working on my thesis in Spain.
I am using GAMs in R and, now that I have estimated my
I dont want to apply the unique for all columns but just the ID column.
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names()
-
Lanna Jin
lanna...@gmail.com
510-898-8525
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__
could you please elaborate a little more on that?
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Thanks for getting back so quickly Ista,
I was actually casting about for any examples of R software that deals
with this kind of structure. But your question is a good one. Here are a
few things I'd like to be able to do:
* Store data in R at the finest level of detail but easily refer to
Thank you for your advice, ill try to be more explicit now, i wasnt in the
first mail because i thought it is a simple question to answer,
so; i have a panel data which contains 48858 observations during 3 year
therefore, there are 146574 observations in total,
i have 22 different industries
Thank you for your advice, ill try to be more explicit now, i wasnt in
the first mail because i thought it is a simple question to answer,
so; i have a panel data which contains 48858 observations during 3 year
therefore, there are 146574 observations in total,
i have 22 different
On 05/03/2010 08:47 AM, Chris Evans wrote:
Thanks Tal Thomas, I am now experimenting with both SWord and R2wd and
both are certainly a huge step forward for me, tied as I am to Word and
the Windoze/M$ world for now.
Chris
Note that many of the general solutions offered produce documents
Hi Serdal,
There is a lot of confusion here (how much is yours and how much is
mine remains to be seen). See specific comments in line.
Also inline comments.
On Mon, May 3, 2010 at 9:19 AM, serdal ozusaglam
saint-fi...@hotmail.com wrote:
Dear R users,
I think i have a simple
Hi:
Here are three solutions; since this question comes up fairly often, you can
find
other solutions in the R-help archives.
(1) Use functions from base R: split the data frame by ID, extract the first
record from each split and slurp them together with rbind():
do.call(rbind,
How about
yourdata[ !duplicated(yourdata$ID), ]
?
See ?duplicated for more information.
HTH,
Jorge
On Mon, May 3, 2010 at 9:04 AM, someone wrote:
as a r noob i am having another problem:
i have a big dataframe where each row corresponds to one entry and each
column is a field...
for
x0=rnorm(100)
y0=rpois(100,3)+1
ind=as.data.frame(table(y0))
ind1=ind[,1]
ind2=ind[,2]
phi=NULL
for (i in 1:length(ind2)){
phi[i]=sum(x0[y0==ind1[i]])/ind2[i]
}
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I am trying to run na.approx on a zoo object in which some of the columns
contain nothing by NA values. When I do this I get the following error:
Error in approx(x[!na], y[!na], xout, ...) :
need at least two non-NA values to interpolate
Is there a way I can use na.approx with my dataset so
On May 3, 2010, at 11:43 AM, serdal ozusaglam wrote:
Hi Serdal,
There is a lot of confusion here (how much is yours and how much is
mine remains to be seen). See specific comments in line.
Also inline comments.
On Mon, May 3, 2010 at 9:19 AM, serdal ozusaglam
saint-fi...@hotmail.com
Hi,
I have plotted a cdf using the ecdf function with plot(). I want to
highlight/identify points on the same plot. Also, because this is a cdf I
am not sure of the y coords for the point, otherwise I thought of using
highlight in the NCStats package.
Thanks
--
View this message in
On May 3, 2010, at 9:59 AM, someone wrote:
I dont want to apply the unique for all columns but just the ID
column.
dataset[ !duplicated(dataset$ID), ]
--
David Winsemius, MD
West Hartford, CT
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Hi Marshall,
I'm not aware of any packages that implement these features as you
described them. But most of the tasks are already fairly easy in R --
see below.
On Mon, May 3, 2010 at 11:18 AM, Marshall Feldman ma...@uri.edu wrote:
Thanks for getting back so quickly Ista,
I was actually
Here is a workaround:
library(zoo)
# test data
z - zoo(cbind(1:5, NA, c(1:3, NA, 5), NA))
ix - colSums(!is.na(z)) 0
z[, ix] - na.approx(z[, ix])
On Mon, May 3, 2010 at 12:41 PM, Abiel X Reinhart
abiel.x.reinh...@jpmchase.com wrote:
I am trying to run na.approx on a zoo object in which some
as.vector(do.call(c, lapply(split(x0, y0), mean)))
Test with data generated according to your code:
phi=NULL
for (i in 1:length(ind2)){
+ phi[i]=sum(x0[y0==ind1[i]])/ind2[i]
+ }
phi
[1] -0.18922774 0.36333115 -0.04295032 -0.13892563 -0.03968301 0.33326034
[7] 0.28649576 -0.03786830
On May 3, 2010, at 12:28 PM, SamT wrote:
Hi,
I have plotted a cdf using the ecdf function with plot().
Code would make this concrete.
I want to
highlight/identify points on the same plot. Also, because this is a
cdf I
am not sure of the y coords for the point, otherwise I thought of
Try this:
rowsum(x0, y0)[,1]/table(y0)
On Mon, May 3, 2010 at 2:11 PM, song song rprojecth...@gmail.com wrote:
x0=rnorm(100)
y0=rpois(100,3)+1
ind=as.data.frame(table(y0))
ind1=ind[,1]
ind2=ind[,2]
phi=NULL
for (i in 1:length(ind2)){
phi[i]=sum(x0[y0==ind1[i]])/ind2[i]
}
Or better:
tapply(x0, y0, mean)
On Mon, May 3, 2010 at 2:35 PM, Henrique Dallazuanna www...@gmail.comwrote:
Try this:
rowsum(x0, y0)[,1]/table(y0)
On Mon, May 3, 2010 at 2:11 PM, song song rprojecth...@gmail.com wrote:
x0=rnorm(100)
y0=rpois(100,3)+1
ind=as.data.frame(table(y0))
I believe you are trying to find the mean of x0 for each level (group) of y0.
try this:
by(x0, y0, mean)
or if you want a vector (e.g., to merge into a matrix)
c(by(x0, y0, mean))
Best regards,
Josh
On Mon, May 3, 2010 at 10:11 AM, song song rprojecth...@gmail.com wrote:
x0=rnorm(100)
Hello all,
I believe this can be done using bootstrap, but I am wondering if there is
some other way that might be used to tackle this.
#Let's say I have two pairs of samples:
set.seed(100)
s1 - rnorm(100)
s2 - s1 + rnorm(100)
x1 - s1[1:99]
y1 - s2[1:99]
x2 - x1
y2 - s2[2:100]
#And both yield
I will sur that problem is coming from the first part of the test.
data[pa,k] is a vector because pa is a vector.
Coud you help me to solve this error.
Best Regards
2010/4/30 Duncan Murdoch murdoch.dun...@gmail.com
On 30/04/2010 4:19 AM, anderson nuel wrote:
Dear r-help,
Could you
hell all:
I have a vector as follows:
head(res)
1007_s_at.value 1053_at.value117_at.value121_at.value 1255_g_at.value
0.225801033 0.009747404 0.709517954 0.008825740 0.496859178
1294_at.value
0.005091231
after I convert the res into a data frame I got
Can you reverse the color scheme order in rcolorbrewer. If so, how?
[[alternative HTML version deleted]]
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
?rev
On Mon, May 3, 2010 at 4:06 PM, Steve Hempell sthemp...@gmail.com wrote:
Can you reverse the color scheme order in rcolorbrewer. If so, how?
[[alternative HTML version deleted]]
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On Mon, May 3, 2010 at 7:21 PM, wei x1 weix1_2...@hotmail.com wrote:
hell all:
I have a vector as follows:
head(res)
1007_s_at.value 1053_at.value 117_at.value 121_at.value
1255_g_at.value
0.225801033 0.009747404 0.709517954 0.008825740 0.496859178
On 03/05/2010 2:43 PM, anderson nuel wrote:
I will sur that problem is coming from the first part of the test.
data[pa,k] is a vector because pa is a vector.
Coud you help me to solve this error.
If you want to test that all elements match, you should use
all(data[pa, k] == df[, j])
For simplicity we'll assume my code looks as follows
plot(ecdf(1:1000))
which produces a diagonal line.
Ideally, I'd like to highlight or label in a different color certain X
values (say 50,55,60,65)
Is there a function which will allow me to do this?
--
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On May 3, 2010, at 3:16 PM, SamT wrote:
For simplicity we'll assume my code looks as follows
plot(ecdf(1:1000))
which produces a diagonal line.
Ideally, I'd like to highlight or label in a different color certain X
values (say 50,55,60,65)
Is there a function which will allow me to do
I tried with this:
sessionInfo()
R version 2.10.1 (2009-12-14)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
Thanks David, the text() worked for me. I wasnt able to correctly use plot
with points and the ecdf function.
David Winsemius wrote:
Yes, ecdf() which returns a function
points(c(50,55,60,65), ecdf(1:1000)( c(50,55,60,65) ), col=c(red,
green, yellow, blue) )
Draws colored
Many thanks to both David and Ista. I think that my question regarding
extracting items returned from lmer points out a weakness of R, viz. less than
perfect documentation. I know that R is written by volunteers, all of whom I
cherish, yet having to ask the listserve for a question as basic as
After having read through some literature on R, I am happy with collecting
it, and visualizing numbers. Unfortunately I do find it hard to compare and
visualize string data. The data is from a LimeSurvey with lots of YES/NO or
multi-choice questions (check-boxes)
Among the check-boxes there a
Dear all,
I'm looking for an implementation of the generalized extended Kalman filter
for survival data, presented in this article Fahrmeir (1994) - 'dynamic
modelling for discrete time survival data'. The same author also publish a
Bayesian version of the algorithm 'dynamic discrete-time
The situation arises where I open a file to write a data.frame to it. with
write.table.
multiple lines are written to the file and the file is kept in Append=TRUE
mode.
If one sets the col.names to the names of the variables being written, you
have output
that looks like this...
name1 name2
Dear John,
The R documentation is of course not perfect, but it is (in my
estimation) quite good. See more specific responses below, keeping in
mind that I'm actually pretty ignorant of this things myslef.
On Mon, May 3, 2010 at 4:44 PM, John Sorkin jsor...@grecc.umaryland.edu wrote:
Many thanks
Hi there,
This will sound very stupid because I just started using R but I see you had
similar problems.
I just loaded a very large dataset (2950*6602) from csv into R. The format
is ticker=row, date=column.
Every time I want to compute basic operations, R returns In Ops.factor: not
meaningful
On 03-May-10 21:19:34, steven mosher wrote:
The situation arises where I open a file to write a data.frame
to it. with write.table.
multiple lines are written to the file and the file is kept in
Append=TRUE
mode.
If one sets the col.names to the names of the variables being
written,
Hi Steve,
I think you just need to set col.names = FALSE (instead of col.names
=NULL) on subsequent writes.
-Ista
On Mon, May 3, 2010 at 5:19 PM, steven mosher mosherste...@gmail.com wrote:
The situation arises where I open a file to write a data.frame to it. with
write.table.
multiple lines
On May 3, 2010, at 6:22 PM, vincent.deluard wrote:
Hi there,
This will sound very stupid because I just started using R but I see
you had
similar problems.
I just loaded a very large dataset (2950*6602) from csv into R. The
format
is ticker=row, date=column.
Not a particularly
In the ESL(2) (by F-H-T), they have labels on their Lasso graphs
(specifically for the South African Heart Disease data).
So my question is, how do I label each variable in a similar way on my
graph.
The following code should produce a plot with 9 enumerated variables:
library(glmnet)
Hello,
I am new to R and need some help with the legend. How can I add a legend for
two variables (in two columns) each having multiple values to be explained
in the legend. For example:
Var 1 Var 2
symbol - Higher symbol - Higher
symbol - Avgsymbol
All-
Thank you in advance for any help you might be able to lend. Here is
my issue. I am trying to open a fairly large .dat file. The file
originally was downloaded as a GZ file but I unzipped it (with 7-zip) into
it's current 1.86 gig .dat format. I know that the data is just a plain
Dear R gurus,
A re-post due to suggestions from moderators.
Thanks to tips from Gabor and Felix, I was able to make some progress.. How
do I control the position of panel.text? I would like the text appear at a
specific position (say, top right corner) inside a panel.
Below is the sample
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