On Tue, 3 Aug 2010, j.delashe...@ed.ac.uk wrote:
Quoting Prof Brian Ripley rip...@stats.ox.ac.uk:
See also Omegahat package Rcompression (a copy of which for Windows is
on CRANextras).
Thank you, I will look into that.
But I would do this via unzip, modify, zip
once or even 10 times,
Hi
r-help-boun...@r-project.org napsal dne 03.08.2010 19:02:41:
question again.
here is part of my original code for my research question. originally I
used a
loop with matrix form to do the work and it works (though looks ugly). I
have
tried the methods mentioned in the above posts
Hi all,
I would like to draw a side by side bar plot. How can I adjust the the font
size for the x-axis? Furthermore, I'm not sure what to write for 'at=?'. I
tried cex.axis and cex.lab but still fail. Here is my data and code:
t(all)
0-100 100-150 150-200 200-250 250-300
I made some design matrix X(in linear regression model)
I there any method to normalize X?
--
View this message in context:
http://r.789695.n4.nabble.com/question-tp2312938p2312938.html
Sent from the R help mailing list archive at Nabble.com.
__
Sometimes we try to make things behave the way we think they ought to and find
it surprisingly difficult. Later we discover that our original premise was
flawed and we wasted our time trying to force fit our ideas to work.
Since all of the i-th elements of the columns of a data table are
Hi!
Your example is not really easy to copy paste. Next time, try using
dput(), this would help a lot.
I haven't tried, but I think that cex.axis is correct. The problem is in
your call to barplot(). You have to specify xaxt=n so that the x-axis
won't be plotted. You then plot it with your
Hi
I'm using geoR package to perform linear spatial interpolation(OK).
The function likfit() fails to compute REML.
The error meassage is : Error in solve.default(v$varcov, xmat);
How I can find out that likfit() is failed to process and retrieving the error
message ?
Thank you so much for
Hi
I made some design matrix X(in linear regression model)
I there any method to normalize X?
You can normalize a matrix column-wise like this:
# m is a matrix
apply(m, 2, function(x) x / max(x) )
Or normalize row-wise like this:
t(apply(m, 1, function(x) x / max(x) ))
I'm sure there are
Hi list,
As title, under windows, the method of ~/.inputrc won't work, is it
possible to let Rterm be vi style on windows?
P.S. I've sent it serveral days ago, get no response, so I send it again
to check if I've failed to send it.
--
Regards,
Yue Wu
Key Laboratory of Modern Chinese
So you *can* do it. There is a question
remaining of whether you *should* do it.
More R-ish is to use a list.
On 04/08/2010 01:39, jim holtman wrote:
?paste
?assign
On Tue, Aug 3, 2010 at 8:29 PM, Hey Skyheyskywal...@yahoo.com wrote:
hey, Rers
in Stata therer is a loop command ` ' which
Dear list,
here are my two data frames:
av -
structure(list(DESCRIPTION = c(COFFEE C Sep/10, COPPER Sep/10,
CORN Dec/10, CRUDE OIL miNY Sep/10, GOLD Aug/10, HENRY HUB
NATURAL GAS Sep/10,
PALLADIUM Sep/10, SILVER Sep/10, SOYBEANS Nov/10, SPCL HIGH
GRADE ZINC USD,
SUGAR NO.11 Oct/10, WHEAT
Hi,
I have a dataframe with a rather complicated descriptive column (V9):
test3[(1:3), ]
V1 V4 V5
10 1 4559 7173
17 1 58954 59871
19 1 357522 358458
V9
10 ID=NM_182905.1;Name=NM_182905;Alias=FLJ00038;Note=hypothetical protein
LOC375690
17
Hi
you tried OK
result - merge(zz, av, by=DESCRIPTION, all=TRUE)
and as you did not specify what to do when one value is NA here is one
possible solution
rowSums(cbind(result$PL.x, result$PL.y), na.rm=T)
Regards
Petr
r-help-boun...@r-project.org napsal dne 04.08.2010 11:52:00:
Dear list,
On Wed, Aug 4, 2010 at 12:29 AM, Suphajak Ngamlak
supha...@phatrasecurities.com wrote:
Dear all,
I would like to do sample statistics, e.g. mean, median from very large
dataset. This is part of commands I use routinely with several dataset
so I would like to make it into function. The
I'm dealing with count data that's nested and has spatial dependence.
I ran a glmm in lmer with a random factor for nestedness. Spatial dependence
seems to have been accommodated by model. However I can't add a variance
strcuture to this model (to accommodate heterogeneity).
Is there a model that
On Wed, Aug 4, 2010 at 6:00 AM, LogLord nils.sch...@web.de wrote:
Hi,
I have a dataframe with a rather complicated descriptive column (V9):
test3[(1:3), ]
V1 V4 V5
10 1 4559 7173
17 1 58954 59871
19 1 357522 358458
V9
10
Godfrey van der Linden wrote:
G'day, All.
I've recently started writing up my dissertation and need to consider how to
store the research results. I'd like to use the binary compressed save() format
as it is considerably smaller than the raw data. However, will a future R be
able to read it
Questions about geoR should be directed to R-SIG-GEO.
Anyway, you should provide more info about your problem.
Read the Posting Guide.
Have you tried changing the model? Sometimes falling back from Matern to
exponential or Gaussian allows successful convergence.
HTH
Rubén
Hi
Is there any way to find out at what points two timeseries cross over
perhaps with some information on how they cross?
Alternatively, I could write a loop that solves pairs of adjacent points
using linear algebra but looking for a simpler way.
Thanks in advance,
Blair
[[alternative
Hi,
a - c(ID=NM_182905.1;Name=NM_182905;Alias=FLJ00038;Note=hypothetical
protein
+ LOC375690
+ ,ID=NM_001005484;Alias=OR4F5;Note=olfactory receptor%2C family 4%2C
+ subfamily F
+ ,ID=NM_001005224.1;Name=NM_001005224;Alias=OR4F3;Note=olfactory
+ receptor%2C family 4%2C subfamily F
+ )
Yes, ?jpeg
--- On Tue, 8/3/10, linda.s samrobertsm...@gmail.com wrote:
From: linda.s samrobertsm...@gmail.com
Subject: Re: [R] save plot
To: gavin.simp...@ucl.ac.uk
Cc: r-help@r-project.org
Received: Tuesday, August 3, 2010, 5:36 PM
[I presume you addressed this to
Duncan Murdoch for a
Hi,
I am trying to plot several time series plots with R, but I can't seem to get
the x-axis properly formatted. What I am doing at the moment is:
dates - seq(as.Date(2007/06/10, %Y/%m/%d),
as.Date(2010/03/28, %Y/%m/%d), 7)
par(mfrow=c(5,2))
plot(DateJonEnd1, End1Jon, main=Weekly Training at
L.S.
I am trying to get data from an excel sheet using the RODBC library, quite
like the example on page 34 of the 'Data manipulation with R' book written
by Phil Spector. It seems very straightforward, yet I get the same error
everytime I try it, with different excel sheets and trying to tweek
Hi, I want to split a text to seperate numerical and non-numerical portions of
that. For example suppose I have a text abc 3456 and I want to split in 2
parts like abc 3456.
Is there any function to do that?
Thanks,
__
R-help@r-project.org mailing
On 04/08/2010 5:38 AM, Sander wrote:
L.S.
I am trying to get data from an excel sheet using the RODBC library, quite
like the example on page 34 of the 'Data manipulation with R' book written
by Phil Spector. It seems very straightforward, yet I get the same error
everytime I try it, with
TY Petr, it works. I will then replace NA by 0.
2010/8/4 Petr PIKAL petr.pi...@precheza.cz:
Hi
you tried OK
result - merge(zz, av, by=DESCRIPTION, all=TRUE)
and as you did not specify what to do when one value is NA here is one
possible solution
rowSums(cbind(result$PL.x, result$PL.y),
On Wed, Aug 4, 2010 at 7:04 AM, Megh Dal megh700...@yahoo.com wrote:
Hi, I want to split a text to seperate numerical and non-numerical portions
of that. For example suppose I have a text abc 3456 and I want to split in
2 parts like abc 3456.
Is there any function to do that?
If the
On Aug 4, 2010, at 7:04 AM, Megh Dal wrote:
Hi, I want to split a text to seperate numerical and non-numerical
portions of that. For example suppose I have a text abc 3456 and I
want to split in 2 parts like abc 3456.
Is there any function to do that?
?strsplit
?regex
--
David
On Aug 3, 2010, at 7:59 PM, Johann Hibschman wrote:
Duncan Murdoch murdoch.dun...@gmail.com writes:
On 29/07/2010 6:18 PM, chipmaney wrote:
-Why does R recognize '[' as a function?
Because it is a function.
More explicitly, '[' is a string. sapply then calls match.fun to look
up
Dear all
Is there an easier way to retrieve the name of an object? For example,
tmp - 1:10
as.character(quote(tmp))
[1] tmp
as.character(quote(mtcars$cyl))
[1] $ mtcars cyl
as.character(quote(mtcars$cyl))[3]
[1] cyl
The last call more than anything seems a hack. Is there a better
Dear List,
(self-contained example + version info at the bottom)
I'm having trouble producing a barplot using the functions in ggplot2. When I
use the position=dodge option, the bars are plotted but also a number of
spurious markers. More specifically, a number of black dots are plotted in
I am not sure what you mean by a hack? Can you elaborate further,
give details on the problem you are trying to solve.
Does this work as a lesser hack:
tail(as.character(quote(mtcars$cyl)),1)
[1] cyl
tail(as.character(quote(cyl)),1)
[1] cyl
On Wed, Aug 4, 2010 at 7:47 AM, Liviu Andronic
Tryu this:
deparse(substitute(mtcars))
On Wed, Aug 4, 2010 at 8:47 AM, Liviu Andronic landronim...@gmail.comwrote:
Dear all
Is there an easier way to retrieve the name of an object? For example,
tmp - 1:10
as.character(quote(tmp))
[1] tmp
as.character(quote(mtcars$cyl))
[1] $
Try this:
library(RDCOMClient)
## Criando a conexao com o SAP
oLogonControl - COMCreate(SAP.Logoncontrol.1)
## Conectando com o usuario ao SAP
g_connection - oLogonControl$NewConnection()
g_connection[[System]]- PD4
g_connection[[ApplicationServer]] - server_ip
On Aug 4, 2010, at 7:52 AM, Dieter Vanderelst wrote:
Dear List,
(self-contained example + version info at the bottom)
I'm having trouble producing a barplot using the functions in
ggplot2. When I use the position=dodge option, the bars are
plotted but also a number of spurious markers.
Can I make a group of jpeg instead of pdfs?
Thanks.
Linda
On Wed, Aug 4, 2010 at 6:47 AM, John Kane jrkrid...@yahoo.ca wrote:
Yes, ?jpeg
--- On Tue, 8/3/10, linda.s samrobertsm...@gmail.com wrote:
From: linda.s samrobertsm...@gmail.com
Subject: Re: [R] save plot
To:
On 08/04/2010 05:38 PM, Roslina Zakaria wrote:
Hi all,
I would like to draw a side by side bar plot. How can I adjust the the font
size for the x-axis? Furthermore, I'm not sure what to write for 'at=?'. I
tried cex.axis and cex.lab but still fail. Here is my data and code:
t(all)
?jpeg
On 04-Aug-10 14:28, linda.s wrote:
Can I make a group of jpeg instead of pdfs?
Thanks.
Linda
On Wed, Aug 4, 2010 at 6:47 AM, John Kanejrkrid...@yahoo.ca wrote:
Yes, ?jpeg
--- On Tue, 8/3/10, linda.ssamrobertsm...@gmail.com wrote:
From: linda.ssamrobertsm...@gmail.com
Subject: Re:
On 04/08/2010 7:47 AM, Liviu Andronic wrote:
Dear all
Is there an easier way to retrieve the name of an object? For example,
tmp - 1:10
as.character(quote(tmp))
[1] tmp
as.character(quote(mtcars$cyl))
[1] $ mtcars cyl
as.character(quote(mtcars$cyl))[3]
[1] cyl
The last call more
Hi guys,
I have a large text file with a bunch of Time in HH:MM format, what would
be the best way to process it into a Time Object so that I can use
comparisons like (1:001:15) or (13:002:00) to both return true.
Right now if I do a comparison like (3:00 1:59) I get a true, but if I
do (3:00
On Wed, Aug 4, 2010 at 8:43 AM, allany all...@cmu.edu wrote:
Hi guys,
I have a large text file with a bunch of Time in HH:MM format, what would
be the best way to process it into a Time Object so that I can use
comparisons like (1:001:15) or (13:002:00) to both return true.
Right now if I
Olga
There was a presentation at the London R user group about doing what you
ask using the bigmemory package
(http://www.londonr.org/LondonR-20090331/realtimeR.pdf). I believe you
can do the same thing with ff. This wouldn't share the whole workspace
but just the objects you choose.
Regards
You might also find match and pmatch of use here.
On Aug 3, 2010, at 12:05 AM, Daniel Malter wrote:
Hi, I have a good grasp of grep() and gsub() for finding and
extracting
character strings. However, I cannot figure out how to use a search
term
that is stored in a
Hi Arnaud,
It is slightly confusing what you are asking but if you just want a
dataframe with the two zz$PL and av$PL columns concatenated then
merge is not what you are after.
Try something like this.
t - rbind(data.frame(PL=zz$PL), data.frame(PL=av$PL))
t
PL
1 3075.00
Try this,
source(http://gridextra.googlecode.com/svn/trunk/inst/test/expressions.r;)
library(grid)
e = expression(p[Wilcoxon], *2.2%*%10^{-16})
grid.expr(e)
HTH,
baptiste
On Aug 4, 2010, at 9:56 AM, Johannes Graumann wrote:
Hi Baptiste,
This is, I fear a bit beyond my level of competency
Dear Steven,
Thanks for your post. It explains a lot.
The reason I wanted to use multiple sessions simultaneously is that
because when I run something it usually takes quite a while (calculating
distance matrices), meaning that I have to wait before R can handle the
next task. Also RAM is
Another way is
t1- c(3:00,1:59,3:00,2:00)
t2 - strptime(t1, format=%H:%M)
t2[-4]t2[-1]
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail.com
On Aug 4, 2010, at 8:47 AM, Gabor Grothendieck wrote:
On Wed, Aug 4, 2010 at 8:43 AM, allany
haenl...@gmail.com wrote:
I'm sorry -- I think I chose a bad example. Let me start over again:
I want to estimate a moderated regression model of the following form:
y = a*x1 + b*x2 + c*x1*x2 + e
Based on my understanding, including an interaction term (x1*x2) into the
regression in addition
The reason you see the exra markers is that the first part of the command
qplot(DT$N,DT$D,fill=factor(DT$C)) already plots the individual points.
You didn't see it with geom_bar(stat = identity) simply because the
stacked bars made the previous layer invisible. To see this you can use the
ggplot
Hi,
I'm trying to use grid.polygon() to plot several polygons at once, with a
view to putting coloured polygons beneath a curve. I'm struggling just to
get the grid.polygon to plot anything
# PLOT SOME POINTS
x - 1:100
y - 1:100*0.5 + 3
plot(x, y, pch = .)
# PLOT 2
Seems to me it may be worth stating what may be elementary to some on this list:
- If all relevant variables are included in the model and the true model is
indeed linear, then all least squares estimated coefficients are unbiased. [
David Ruppert once said about the three kinds of lies:
On Aug 4, 2010, at 9:32 AM, Ally wrote:
Hi,
I'm trying to use grid.polygon() to plot several polygons at once,
with a
view to putting coloured polygons beneath a curve. I'm struggling
just to
get the grid.polygon to plot anything
# PLOT SOME POINTS
x - 1:100
y -
Thanks David,
I looked briefly through the bigmemory package and it looks very
promising. As I mentioned in my previous post to Steven I am calculating
distance matrices, and if it is my bottleneck (I am nearly sure it is) I
believe that the bigmemory will be particularly useful as it is designed
I have the following array:
head(stocks)
DATE TICKER PERMNO EXCHCD TSYMBOL TRDSTAT SHROUTPRC RET
1 19950131 EWST 10001 3EWST A 2224 -7.75000
-0.031250
2 19950228 EWST 10001 3EWST A 2224 7.54688
-0.026210
3 19950331 EWST 10001 3
Thanks, this indeed solved the problem.
Regards,
Dieter
On 4/08/2010 15:21, Shentu wrote:
The reason you see the exra markers is that the first part of the command
qplot(DT$N,DT$D,fill=factor(DT$C)) already plots the individual points.
You didn't see it with geom_bar(stat = identity) simply
On Aug 4, 2010, at 9:46 AM, Leigh E. Lommen wrote:
I have the following array:
head(stocks)
DATE TICKER PERMNO EXCHCD TSYMBOL TRDSTAT SHROUTPRC
RET
1 19950131 EWST 10001 3EWST A 2224 -7.75000
-0.031250
2 19950228 EWST 10001 3EWST A
Is this what you mean?
x=c(1,2,2,3,4,5,6,3,2,1)
y=c(2,3,4,2,1,2,3,4,5,6)
matplot(cbind(x,y),type=l)
which(diff(sign(x-y))!=0)+1
[1] 4 8
--
View this message in context:
http://r.789695.n4.nabble.com/Finding-points-where-two-timeseries-cross-over-tp2313257p2313510.html
Sent from the R help
Dear Duncan
On Wed, 04 Aug 2010 08:33:49 -0400
Duncan Murdoch murdoch.dun...@gmail.com wrote:
As other have pointed out, in a function you can use substitute(arg)
to retrieve the expression passed as arg, and
deparse(substitute(arg)) to turn it into a string that's suitable for
using as a
Hello Wolfgang.
I'd appreciate if you could help me check whether I am doing the proper
thing to do an arm-level meta-analysis with metafor and what differences
there might be in trying to do the same with lme and lm.
I am following the arm based model described in section 3.2 of the
Salanti's
Dear list;
I just created a utility function that replicates what I have done in
the past with Excel or OO.org by putting a formula of the form
=sum($A1:A$1) in an upper-corner of a section and then doing a fill
procedure by dragging the lower-rt corner down and to the right. When
I added a parse argument to grid.table so that when switched to TRUE (default
FALSE) all the text strings are interpreted as expressions (inspired by
ggplot2::geom_text),
d - data.frame(alpha, beta)
grid.table(d, parse=T)
you'll need revision 258 of gridExtra for this to work (googlecode now,
Hi Leigh,
Several aspects of your email make it challenging to offer advice. We
are missing part of your data (e.g., 'T' in 1:T), and it is not
terribly clear what 't' in (t-5) to t(19950630) to (t+7) is supposed
to mean (thought I would hazard the guess time, and that it is
represented by your
Great! I will give it a try ASAP!
Thanks!
Joh
On Wednesday 04 August 2010 16:47:12 baptiste Auguié wrote:
I added a parse argument to grid.table so that when switched to TRUE
(default FALSE) all the text strings are interpreted as expressions
(inspired by ggplot2::geom_text),
d -
On Wed, 4 Aug 2010, David Winsemius wrote:
Dear list;
I just created a utility function that replicates what I have done in the
past with Excel or OO.org by putting a formula of the form =sum($A1:A$1) in
an upper-corner of a section and then doing a fill procedure by dragging
the lower-rt
Hi,
I'm trying to use RWeka and followed the following example from the
RWeka manual.
## Use some example data.
w - read.arff(system.file(arff,weather.nominal.arff,
package = RWeka))
## Identify a decision tree.
m - J48(play~., data = w)
m
## Use 10 fold cross-validation.
e -
Just a little difference.
ecdf.tbl - function (.dat) {
.dat - as.matrix(.dat)
na.m - is.na(.dat)
.dat[na.m]-0 # Assign NA a value 0
res - apply(t(apply(.dat,1,cumsum)),2,cumsum)
res[na.m] - NA # Assign NA back
return(res)
}
-
A R
Hi R,
Seems like the maximum seasonal 'q' parameter for the ?arima is 350. Any
way, where we can increase this? Since I am working on 3 year (q=252*3)
and 5 year(q=252*5) returns, I may require this option. Thanks.
fit=arima(r,c(3,0,0),seasonal = list(order = c(0, 0, 500), period =
Hi,
Try help($).
dwt - function(ld, filter='d8', n.levels=lev, boundary=reflaction)
{
list(W=ld, V=n.levels,filter=filter)
}
dec - dwt(1:10, filter='d8', n.levels=10, boundary=reflaction)
dec$W
dec$V
dec$filter
dec$W - sqrt(1:10)
dec
-
A R learner.
--
View this message in
On Aug 4, 2010, at 11:06 AM, Charles C. Berry wrote:
On Wed, 4 Aug 2010, David Winsemius wrote:
Dear list;
I just created a utility function that replicates what I have done
in the past with Excel or OO.org by putting a formula of the form
=sum($A1:A$1) in an upper-corner of a section
Hello,
I have a data set which is similar to the following data
mice - rep(letters[1:4],10)
outcome - sample(c(0,1),length(mice),replace=T)
group - c(rep(A,length(mice)/2),rep(B,length(mice)/2))
my.data - data.frame(mice,outcome,group)
my.sort.data - my.data[order(my.data[,1]),]
I would like
In general, the lapply(split(...)) construction should never be used.
Use tapply() or by() instead, along the lines of
by(dataframe,with(yourdata,list(your columns)), function(...),...)
If you find this complexity annoying, then look into Hadley Wickham's
plyr package for simpler constructions,
Hello!
I have a code for converting monthly values into weekly values:
monthly-data.frame(month=c(20100301,20100401,20100501,20100601,20100301,20100401,20100501,20100601),monthly.value=c(100,NA,200,300,10,NA,20,30),market=c(Market
A,Market A,Market A,Market A,Market B,Market B,Market
B,Market B))
Sorry, I found the solution:
library(zoo)
z - read.zoo(monthly, split = market)
all.dates - seq(start(z), as.Date(as.yearmon(end(z)), frac = 1), by = day) ##
mondays - all.dates[weekdays(all.dates) == Monday] ##
weeks - na.locf(z, xout = mondays, na.rm = FALSE, maxgap=0)
do.call(rbind, by(weeks,
I'm trying to get help to print the help pages in html format to the
terminal. This is in order to be able to see the html help files remotely.
If I do
printURL = function(file) {a=readLines(url(file));cat(a,sep=\n)}
options(browser=printURL)
options(help_type=html)
then invoking help with
?scale
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of leepama
Sent: Tuesday, August 03, 2010 10:55 PM
To:
In general, the lapply(split(...)) construction should never be used.
Why? What makes it so bad to use?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Hello,
This will put the results of the Fisher test in a list, with each
element of the list being the results for mouse type a, b, c, and d.
mice - rep(letters[1:4],10)
outcome - sample(c(0,1),length(mice),replace=T)
group - c(rep(A,length(mice)/2),rep(B,length(mice)/2))
my.data -
Hi,
I am running a Cox Mixed Effects Hazard model using the library coxme. I
am trying to model time to onset (age_sym1) of thought problems (e.g.
hearing voices) (sym1). As I have siblings in my dataset, I have
decided to account for this by including a random effect for family
(famid). My
Do ?'for' to learn about for loops. Look at FAQ 7.21 for the other part of the
question (the most important part of the answer is the end where it tells you
to use lists instead like Patrick suggests).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
On Wed, Aug 4, 2010 at 1:28 PM, Greg Snow greg.s...@imail.org wrote:
Do ?'for' to learn about for loops.
and put that in quotes since its a reserved word:
?for
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
On Wed, Aug 4, 2010 at 10:17 AM, Joshua Wiley jwiley.ps...@gmail.com wrote:
Hello,
This will put the results of the Fisher test in a list, with each
element of the list being the results for mouse type a, b, c, and d.
mice - rep(letters[1:4],10)
outcome -
I am sorry, I'd like to split my column (names) such that all the
beginning of a string (X..) is gone and only the rest of the text is
left.
x-data.frame(names=c(X..aba,X..abb,X..abc,X..abd))
x$names-as.character(x$names)
(x)
str(x)
Can't figure out how to apply strsplit in this situation -
Dimitri Liakhovitski wrote:
I am sorry, I'd like to split my column (names) such that all the
beginning of a string (X..) is gone and only the rest of the text is
left.
x-data.frame(names=c(X..aba,X..abb,X..abc,X..abd))
x$names-as.character(x$names)
(x)
str(x)
Can't figure out how to apply
Have a look at the
stringr
package
It simplifies such things...
Contact
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Try this:
gsub(X\\.\\., , x$names)
On Wed, Aug 4, 2010 at 2:42 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
I am sorry, I'd like to split my column (names) such that all the
beginning of a string (X..) is gone and only the rest of the text is
left.
On Wed, Aug 4, 2010 at 1:35 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Wed, Aug 4, 2010 at 1:28 PM, Greg Snow greg.s...@imail.org wrote:
Do ?'for' to learn about for loops.
and put that in quotes since its a reserved word:
?for
Sorry, I missed the fact that Greg actually did
Hi,
You already have great solutions. I just wanted to point out that
A) strsplit() works on the entire column automatically so you would
not need a loop
B) with the argument stringsAsFactors = FALSE, your character data
will not be converted to factor, so you would not need to convert it
back.
On 04/08/2010 12:44 PM, Michael Lachmann wrote:
I'm trying to get help to print the help pages in html format to the
terminal. This is in order to be able to see the html help files remotely.
If I do
printURL = function(file) {a=readLines(url(file));cat(a,sep=\n)}
options(browser=printURL)
On Aug 4, 2010, at 1:42 PM, Dimitri Liakhovitski wrote:
I am sorry, I'd like to split my column (names) such that all the
beginning of a string (X..) is gone and only the rest of the text is
left.
I could not tell whether it was the string X.. or the pattern X..
that was your goal for
Thank you very much, everyone!
Dimitri
On Wed, Aug 4, 2010 at 2:10 PM, David Winsemius dwinsem...@comcast.net wrote:
On Aug 4, 2010, at 1:42 PM, Dimitri Liakhovitski wrote:
I am sorry, I'd like to split my column (names) such that all the
beginning of a string (X..) is gone and only the rest
I just tried it under OSX and linux, and both get stuck. Maybe R on windows
forks the R help server?
But what you suggested works perfectly. Thanks!
Do you by chance know if there was something equivalent in R 2.10.x and/or
2.9.x
Duncan Murdoch-2 wrote:
On 04/08/2010 12:44 PM, Michael
Hi R-users,
Since R.2.11 aggregate can now deal with non-scalar functions, which is
very useful to me.
However, I have a question about how best to process the output.
test - data.frame(a = rep(c(g1, g2), each = 50), b = runif(100))
res - aggregate(test$b, list(group = test$a), function(x)
Dear colleagues,
I have a problem which has bitten me occasionally. I often need to
prepare graphs for many variables in a data set, but seldom for all.
or for any large number of sequential or sequentially named variables.
Often I need several graphs for different subsets of the dataset
for a
Hi,
I am sorry for that I canât determine which R-sig list to post
questions about package ff.
Now I have made a little progress with this:
read.dbres.ffdf - function(
res){
data1 - fetch(res, 0)
if (nrow(data1) == 0){
Hi!I'm doing a little data importing from .cel files,
setwd(/home/mandova/celfiles)
mydata-ReadAffy()
Error in sub(^/?([^/]*/)*, , filenames, extended = TRUE) :
unused argument(s) (extended = TRUE)
Then I tried
filenames-paste(GSM,c(seq(138597,138617,1)),.cel,sep=)
Hello
dec=dwt(ld, filter='d8', n.levels=lev, boundary=reflaction);
dwt is an function which returns W, V, filter
dec is an object which consist of W, V, filter and others
How to change values (of W, V, filter...) when i have dec
thank you!
--
View this message in context:
Hello
I do wavelet transform by using this code:
dec=dwt(ld, filter='d8', n.levels=lev, boundary=reflaction);
dec consists of the decomposition coefficients and other
How can I change the coefficients the decomposition manualy?
--
View this message in context:
Hi R-users,
Since R.2.11 aggregate can now deal with non-scalar functions, which is very
useful to me.
However, I have a question about how best to process the output.
test - data.frame(a = rep(c(g1, g2), each = 50), b = runif(100))
res - aggregate(test$b, list(group = test$a), function(x)
I am sorry, someone said that strsplit automatically works on a
column. How exactly does it work?
For example, if I want to grab just the first (or the second) part of
the string in nam1 that should be split based on ..
x-data.frame(nam1=c(bbb..aba,ccc..abb,ddd..abc,eee..abd),
Hi R-users,
I'm using R 2.11.1, mgcv 1.6-2 to fit a generalized additive mixed model.
I'm new to this package...and just got more and more problems...
1. Can I include correlation and/or random effect into gam( ) also? or only
gamm( ) could be used?
2. I want to estimate the smoothing function
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