Another approach:
foo <- function(t){
bm <- ceiling(t/15)
s <- cut(t,breaks=15*(0:bm),labels=1:bm)
s <- as.numeric(levels(s)[s])
t^(s+1)
}
This idea can be generalised .
cheers,
Rolf Turner
On 27/03/12 15:31, R. Michael Weylandt wrot
Hello helpful R folks,
I am simply trying to graph a quarter circle centered at the origin in the
first quadrant. When I set the xlim of the plot to the radius of the
circle, the plot appears correct. However, I'd like to see a slight
extension of the axes beyond the domain of the function itself
Hi
I have records like like this
X1 X2 State
34 72 state1
9 63 state1
49 31 state1
60 34 state1
80 73 state1
60 20 state2
59 87 state2
88 20 state2
71 66 state2
65 56 state2
59 16
AJ: This is something to learn a lesson from. A question, starved of
preparation, can't help anybody to help you.
James
On Mar 25, 2012 9:00 PM, "Rolf Turner" wrote:
> On 26/03/12 00:18, Anjana Thampi wrote:
>
>> How do you decompose inequality in R, say by gender?
>>
>>
> This has to be one of
On 2012-03-26 05:09, John D. Muccigrosso wrote:
I cannot for the life of me figure this out:
What's the parameter to fill in with color circles made with circles()? col changes the
line color, but all I see in the help is a reference to "additional graphic
parameters", and no examples via goog
Hi,
On Mon, Mar 26, 2012 at 6:46 PM, nserdar wrote:
> Hi
>
> I should stochastic receding control process for portfolio
>
> optimization, so that I need a convex optimisation package
>
> in R .
>
> Please let me know if any package is available for convex optimisation
> packages.
This is probab
One way is to simply nest your ifelse()s:
y <- ifelse(t < 15, t^2, ifelse(t < 30, t^3, t^4))
Michael
On Mon, Mar 26, 2012 at 7:48 PM, Aimee Jones
wrote:
> Dear all,
>
> I'm aware if y has two separate functions (depending on the conditions
> of x) you can use the ifelse function to separate y i
library(reshape2)
x <- structure(list(ExpName = structure(c(1L, 1L, 1L, 1L, 2L, 2L,
2L, 3L, 3L), .Label = c("MM1-x1", "MM1-X2", "MMX-X3"), class = "factor"),
Identifier = structure(c(1L, 2L, 3L, 4L, 1L, 2L, 4L, 1L,
2L), .Label = c("GSK", "GSK2", "GSK3", "GSK4"), class = "factor"),
valu
Hi
I am trying to estimate bottom temperatures over a particular depth range,
based on one dataset containing synthetic temperature profiles (for set depths)
and another that contains information on bathymetry. I need to do this for
multiple regions and thus would ideally like an automated solut
Hi
I should stochastic receding control process for portfolio
optimization, so that I need a convex optimisation package
in R .
Please let me know if any package is available for convex optimisation
packages.
Regards,
Serdar
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Dear all,
I'm aware if y has two separate functions (depending on the conditions
of x) you can use the ifelse function to separate y into two separate
functions depending on input. How do you do this if there a multiple
different conditions for x?
for example,
y fits the following between t>0 &
Thank you for your insights, to give a little more background on what i am
looking for:
-I am trying to handle large sets of data (between 20 000 to 50 000)
entries, but the number of distinct values is very low (maybe 20 - 30, 50 in
extreme cases). Hence handling of data should be as automated a
Hi:
sorry for asking simple question (not simple for me though).
I have a data frame something like this:
ExpName Identifier value
MM1-x1 GSK 0.02
MM1-x1 GSK2 0.001
MM1-x1 GSK3 0.04
MM1-x1 GSK4 0.1
MM1-X2 GSK 0.4
MM1-X2 GSK2 0.1
MM1-X2 GSK4 0.002
MMX-X3 GSK 0.4
MMX-X3 GSK2 0.23
I am
Benilton,
Try this:
read.table(textConnection(gsub('","', "','", gsub('^\"|\"$', "'",
readLines('../teste.csv', sep = ',', quote = "'", header = TRUE)
On Mon, Mar 26, 2012 at 8:09 PM, Benilton Carvalho
wrote:
> I need to read in csv files, created by 3rd party, with fields
> containing sing
While not having a perfect solution, I have made enough progress to be able to declare
"good enough", thanks in particular to Duncan Murdoch and Yihui Xie.
First, the font can be made smaller so output fits on a line and does not overflow the
margins. This is accomplished by putting the command
I need to read in csv files, created by 3rd party, with fields
containing single quotes (as shown below).
"header1","header2","header3","header4"
"field1r1","field2r1","field3r1","field4r1"
"field1r2","field2r2","field3r2PartA), field3r2PartB Very" Long","field4r2"
"field1r3","field2r3","field3r3"
Dear R wizards---
I have a wrapper on mclapply() that makes it a little easier for me to
do multiprocessing. (Posting this may make life easier for other
googlers.) I pass a data frame, a vector that tells me what rows
should be recomputed, and the function; and I get back a vector or
matrix of
On Mar 26, 2012, at 1:59 PM, Strassburger, Daniel wrote:
I haven't been successful in converting my colleagues to the world
of R yet they wish to share collected data so that they may analyze
it in SPSS.
I know how to write to an SPSS file and it opens fine, but my
problem is that it onl
Hi Himanshu,
the use of optim is described on its help page.
In addition to this package FME provides additional functionality for
fitting ODE models. See FME help files, package vignettes and the
example below for details.
Hope it helps
Thomas Petzoldt
knitr gives you full power to do whatever you want with the output, so
you can use all sorts of LaTeX environments or packages to customize
your output. For example, in this case, I think listings may be
helpful; it has an option breaklines=true which allows you to break
lines.
Here is a minimal e
I followed this case while it was ongoing.
It was a very interesting example of basic mistakes but also (for me) of
journal politicking.
Keith Baggerly and Kevin Coombes wrote a great paper - "DERIVING
CHEMOSENSITIVITY FROM CELL LINES: FORENSIC BIOINFORMATICS AND REPRODUCIBLE
RESEARCH IN H
Sorry- I was so excited it was working that I didn't see your reply:
"What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it."
Answer is that I was trying to figure out the most efficient way (mainly in
time moreso than physical space) to save my o
Testing it on the full 1000 simulations:
-used 54GB RAM.
-after saving then removing - using rm() - the memory usage did decrease
significantly.
-I guess my test run was too small to show a noticeable difference.
-thanks again.
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'sink' would probably not help since it is just capturing output to
the console and you are still doing the binary to character
conversion. 'save' helps to avoid that. You might want to see if
there is any difference with using compression (which I think is the
default for 'save') as opposed to w
And to answer your question, Jim,
" Are you just going to read it back into R for processing? if so 'save'
will probably be faster."
Yes, that is my intent. This is a great solution - infinitely better than
the writing I was trying to do.
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On 2012-03-26 12:57, Sam Steingold wrote:
when subsetting a matrix results in a single row, it is converted to a
vector, not a matrix.
how do I avoid this?
Check ?"[" and note the 'drop=' argument.
Peter Ehlers
1. __GOOD__
edges<- get.edges(g,E(g))
edges
[,1] [,2]
[1,]02
On 26-03-2012, at 21:57, Sam Steingold wrote:
> when subsetting a matrix results in a single row, it is converted to a
> vector, not a matrix.
> how do I avoid this?
>
> 1. __GOOD__
>
>> edges <- get.edges(g,E(g))
>> edges
> [,1] [,2]
> [1,]02
> [2,]03
> [3,]04
> [4
On 2012-03-26 09:09, Sebastián Daza wrote:
Thank you Peter. The problem with you solution is that it doesn't
represent the actual values of disruption. Look this example:
person<- rep(1:2, each=4)
income<- c(100, 120, 150, 200, 90, 100,120, 150)
disruption<- c(0,0,0,1,0,1,1,0)
time<- rep(c(1:4),
Thank you, Jim! I was just trying this as you wrote. Testing it on a small
sample - it seems to work!
I am curious - I removed the data that was stored in memory (using rm()),
and checked to see that it was gone using ls() - and it was; but I didn't
see a concurrent reduction in the Memory Usage o
when subsetting a matrix results in a single row, it is converted to a
vector, not a matrix.
how do I avoid this?
1. __GOOD__
> edges <- get.edges(g,E(g))
> edges
[,1] [,2]
[1,]02
[2,]03
[3,]04
[4,]05
[5,]11
[6,]04
[7,]06
[8,]
Have you tried 'save' to save the output. 'write' is probably
spending a lot of time converting to character. Are you just going to
read it back into R for processing; if so 'save' will probably be
faster.
On Mon, Mar 26, 2012 at 12:53 PM, Diann Prosser wrote:
> Dear all,
> I am working with la
Running findFn('linear programming') from the sos package brings up
several possibilities that look promising.
On Sun, Mar 25, 2012 at 5:48 AM, agent dunham wrote:
> Dear Community,
>
> I've a Work -Shift Scheduling Problem I'd like to solve via constraint
> linear programming.
>
> Maybe somethin
I haven't been successful in converting my colleagues to the world of R yet
they wish to share collected data so that they may analyze it in SPSS.
I know how to write to an SPSS file and it opens fine, but my problem is that
it only includes the existing data - none of the variables I created wi
Dear Researchers,
Sorry for this email but I am not a statistician, and for this I have this
problem to understand. Thanks in Advance for help and suggestions.
Gianni
I have 21 classes (00, 01, 02, 04, ,020) with different length. I did a
kruskal wall test in R with the following code
krusk
Great :)
From: Sarah Goslee
Cc: R help
Sent: Monday, March 26, 2012 6:51 PM
Subject: Re: [R] Predefined set of gray scale colors
You might start with
?gray
Sarah
> Dear all,
> I am printing 7 different data sets to the same plots, if I use some colors
On 26/03/2012 1:34 PM, John Benning wrote:
Hi,
*Warning: R newb here. *
I've got a 3D scatterplot (
http://www.fleetwoodswoodworks.com/scatterPlot.jpg) which I want to drape
in color, with the draped plane corresponding to the mean z values on the
plot (similar to: http://www.fleetwoodswoodwork
Warning: This has little directly to do with R, although R and related
tools (e.g. sweave and other reproducible research tools) have a
natural role to play.
The IOM report:
http://www.iom.edu/Reports/2012/Evolution-of-Translational-Omics.aspx
that arose out of the Duke Univ. genomics testing sc
Then there's something missing in what you tell us.
# this is an awkward and inefficient way of constructing a data frame
city<-data.frame(city="Barcelona",cod=1)
city<-rbind(city,data.frame(city="Madrid",cod=2))
city<-rbind(city,data.frame(city="Lisbon",cod=3))
city<-rbind(city,data.frame(city="
Fortunes candidate?!
-- Bert
On Mon, Mar 26, 2012 at 10:24 AM, Sarah Goslee wrote:
< The OP wrote>
"The problem is that it gives the result that I want."
: That's a new sort of problem.
--
Bert Gunter
Genentech Nonclinical Biostatistics
Internal Contact Info:
Phone: 467-7374
Website:
ht
On Sun, Mar 25, 2012 at 3:50 PM, stivi wrote:
> Hello,
>
> my question is if anyone has any good ideas how to create a Markov Chain
> from ordered data. So, I have some sort of time series, and if value1
> happens as time1 and value2 happens at time2 I record this as an update to
> the probability
'elem' is an attribute (the name) of the 'data' variable (bad name for
a variable), not something that is independently accessible. By very
(exceptionally!) loose analogy, think of it as being something like a
field/property of an object -- you can't get to it directly (and
that's a good thing) but
Perhaps something like this,
x <- seq(0, 10, length = 1000)
y <- x %%1
Michael
On Mon, Mar 26, 2012 at 10:18 AM, mail me wrote:
> Hi:
>
> I am trying to create a sawtooth waveform. I used the following
>
> x <- runif(500, min = -2, max = 2)
> y <- (1 -abs(x3))* ((x3) <= 1)
> combined <- data.fr
"The problem is that it gives the result that I want."
That's a new sort of problem.
You show two different merge() commands. What do you expect to happen,
vs what does happen?
Sarah
On Mon, Mar 26, 2012 at 10:12 AM, MSousa wrote:
> Thanks for the reply.
>
> I am using the function you gave
Hello,
I'm wondering what was the year (or year range) of collection for the
data included in the 'diamonds' dataset in ggplot2.
This information would be very helpful in interpreting the 'price' variable.
Thank you!
Marina Doucerain
__
R-help@r-pr
Hi,
*Warning: R newb here. *
I've got a 3D scatterplot (
http://www.fleetwoodswoodworks.com/scatterPlot.jpg) which I want to drape
in color, with the draped plane corresponding to the mean z values on the
plot (similar to: http://www.fleetwoodswoodworks.com/draped3Dscatterplot.jpg).
Does anyone k
Yes! Thanks. It was just the "NA" value instead of the "as.null" that does
the trick. Correct code for the original piecewise I stated (for those who
might be looking later) is:
f <- function(x){
ifelse((-1 < x & x < 1),x^2,ifelse((2http://r.789695.n4.nabble.com/Simple-questio
HI, thanks Weidong, Henrique,
this works for the example (may be a bad example, but it should be as
simple as possible),
This is however only a workaround to name the columns of a dataframe,
but is not addressing the problem itself, that is, how can I substitute
a string (used as argument in
thanks for the reply,
But does not the results I need. What is confusing is that when making
the first merge it gives only two cities when it should take three
travel<-data.frame(pos=1,Source=1,Destine=2)
travel<-rbind(travel,data.frame(pos=1,*Source=1*,Destine=3))
travel<-rbind(travel,data.f
Hi,
Does anyone have any suggestions for plotting confidence intervals onto
fitted non-linear models?
Thank you for your help!
Kate
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Sent from the R help mailing list
Hello,
Can someone please help me out with the optim() function.
# parameters
pars<- c(a1= 0.9, a2= 0.7, a3= 0.06, a4=0.02)
y<- c(Y=0.2, Z=0.1)
Ymax<- c(0.8)
fucntion deriv
derivs <- function(time, y, pars) {
with (as.list(c(y, pars)), {
dy = a1*Y*(1-Y/Ymax) - a2*(Y+0.001)
dz = a3*Y- a4*
"The problem is that it gives the result that I want"
I'm not sure that's what one would usually identify as a
problemcan you say a little more specifically what you are looking
to do?
Michael
On Mon, Mar 26, 2012 at 10:12 AM, MSousa wrote:
> Thanks for the reply.
>
> I am using the func
Hi All,
I am trying to develop a GLM model in which the dependent data is a
proportion data with 4 categories. I want to develop this model as an
alternative method to Multinomial Logit Model for comparison of
results. Here is the sample data:
CategorySharesY x1 x2
1 0.79705215
Hello,
>
> The problem is that it gives the result that I want
>
Oh dear, it shouldn't?
>
> The idea is based on the column of source and intended Identify the
> cities and put a new data structure
>
> The idea is something like this.
> pos Source city Destine city_destine
> 1
Me again,
what I really dont understand is the behaivour of R here.
notice, elem is not written in "brackets", why it is interpreted as
string and not as a variable?
I would expect an error "elem not found" because it is not defined as
variable, but in contrary it is accepted as column name, see
Hi:
I am trying to create a sawtooth waveform. I used the following
x <- runif(500, min = -2, max = 2)
y <- (1 -abs(x3))* ((x3) <= 1)
combined <- data.frame(x = x3, y = y3)
plot(combined)
and I get a triangular waveform, not sawtooth. Can someone give a
solution to create a sawtooth waveform?
T
thanks,
is working
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__
R-help@r-project.org mailing list
https://stat.eth
It worked perfectly!
Thank you
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-p
You really need to read one of many beginners guides to R; but I will say
this much:
YOUR_DATA <- read.table("nigeria slr misc/Nigeria India /DBHHISTOGRAM.txt",
header=TRUE)
then you should be able to plot it like I said previously.
If that still doesn't work, you need to read the posting guide
If there is a header in your .txt-file you should have header=TRUE, if there
is no header you should have header=FALSE.
You have to save your dataset in a variable, for example:
myData<-read.table("data.txt",header=TRUE)
Then you can make a histogram by:
hist(myData)
--
View this message in co
this is what I do, IT READS THE TABLE, But then I am stuck, I removed the
header
read.table("nigeria slr misc/Nigeria India /DBHHISTOGRAM.txt", header=FALSE)
hist("nigeria slr misc/Nigeria India /DBHHISTOGRAM.txt"$V1)
Error in "nigeria slr misc/Nigeria India /DBHHISTOGRAM.txt"$V1 :
$ operator i
Thanks for the reply.
I am using the function you gave me.
complete.travel<-merge(travel, city, by.x = "Source", by.y = "cod", all =
TRUE)
complete.travel<-merge(travel, city, by.x = "Destine", by.y = "cod", all =
TRUE
The problem is that it gives the result that I wa
Dear all,
I am working with large matrices (19.6 million elements * 1000 simulations)
and am trying to get around memory problems and vector length issues. I’ve
split the inputs so that the output vector length will not exceed 2^31.
Working on a 64bit machine with 80GB RAM, I still get close to th
You might start with
?gray
Sarah
On Mon, Mar 26, 2012 at 12:49 PM, Alaios wrote:
> Dear all,
> I am printing 7 different data sets to the same plots, if I use some colors
> like (red,blue,yelow...) the different datasets are easily recognized. I
> would like to use some gray scale colors for
Hi All,
I need to normalize/scale string variable which represents interests of
customers (e.g., 'cycling, rollerblading, swimming' etc).
Does anybody know how to do this, I want then use it along with other numeric
variables for SVM classification.
Appreciate for any advice.
-Alex
[
Dear all,
I am printing 7 different data sets to the same plots, if I use some colors
like (red,blue,yelow...) the different datasets are easily recognized. I would
like to use some gray scale colors for distinguising the datasets. Is there any
ready set with grayscale colors (usually the prede
Thanks a lot :)
That helped !
Alex
From: R. Michael Weylandt
Cc: R help
Sent: Monday, March 26, 2012 4:48 PM
Subject: Re: [R] many curves to the same plot
You can use lines() with ecdf objects or use the
not-at-all-surprisingly-named add = TRUE parameter of
On 25.03.2012 22:10, jiefan wrote:
I have met into this problem when I tried to run panel regression by plm.
My code:
library(plm)
indus<- read.csv(file="full.csv",header=TRUE)
industry<-as.data.frame(indus)
reg<-lm(LnTSO2 ~ LnPGDP + LnPGDP2 + LnSOES + LnCOES + LnLIMD +
LnSHOLD + LnPRIV +
Thank you Peter. The problem with you solution is that it doesn't
represent the actual values of disruption. Look this example:
person <- rep(1:2, each=4)
income <- c(100, 120, 150, 200, 90, 100,120, 150)
disruption <- c(0,0,0,1,0,1,1,0)
time <- rep(c(1:4),2)
dat <- as.data.frame(cbind(person
Hello all
This is my first posting for some years. I am back
using R again and must say I do like the language
(regarding scripting, I also use matlab, perl, and bash).
My question involves plotting a Pareto frontier in
three dimensions. This is strictly a exercise in
visualization, I make no
casperyc hotmail.co.uk> writes:
>
I don't know what is wrong with your Maple calculations, but I think
you should check them carefully, because:
(1) As Petr explained, the value of the integral will be < 0.5
(2) The approach of Peter still works and returns : 0.4999777
(3) And the same result c
On Mar 25, 2012, at 6:17 PM, MBoersma wrote:
Hi guys,
I'm quite new to R but quite enthousiastic. I'm trying to rewrite a
bit of
code in order to make it faster, so instead of nesting for loops I'm
trying
some apply functions.
It is an urban myth that using 'apply' functions will delive
Hi Till,
you need a plug-in for SPSS to run R, which is bound to a fixed R
version. SPSS 19 uses R2.10 and SPSS 20 uses R2.12 (both not up-to-date
R versions). So you have to download this plugin (after registering on
IBM Support) for your spss version and the according R version - or
start with a
On 26/03/2012 9:30 AM, John C Nash wrote:
Yihui is correct that an error is generated. I should have mentioned that this
is
something I wanted to show in a vignette, and I got 103 characters as the
length of the
maximal line from a sink() file. Yihui's use of nchar is more elegant.
Thus the qu
Hi
If you went through help page you probably could find it yourself in
shorter time then you spend writing email and waited for others to write
back.
see
?plot.stepfun
> Dear all,
> I would like to plot under the same plot
> many curves like
>
> mycdf<-ecdf(runif(100))
> plot(mycdf)
> curve(
On Mar 26, 2012, at 16:31 , Duncan Murdoch wrote:
> I think you will need to ask this question in an SPSS help forum:
>
> - Version 2.10.x of R is quite old, and is no longer supported.
> - The error message is coming from SPSS, not R
and actually, it doesn't look like an error message at all,
You can use lines() with ecdf objects or use the
not-at-all-surprisingly-named add = TRUE parameter of curve().
Michael
On Mon, Mar 26, 2012 at 10:44 AM, Alaios wrote:
> Dear all,
> I would like to plot under the same plot
> many curves like
>
> mycdf<-ecdf(runif(100))
> plot(mycdf)
> curve(mycd
Dear all,
I would like to plot under the same plot
many curves like
mycdf<-ecdf(runif(100))
plot(mycdf)
curve(mycdf,from=-1,to=1)
mycdf<-ecdf(runif(200)+3)
plot(mycdf)
curve(mycdf,from=-1,to=1)
as you can see each new call to curve redraws (erases) the window. How I can
have an effect like
plo
On 26/03/2012 7:32 AM, Till Below wrote:
Dear all,
I am trying to conduct an enhanced version of factor analysis with a
SPSS interface that allows to use R. This approach has been suggested in
the recent article:
Basto, M. and J.M. Pereira An SPSS R-Menu for Ordinal Factor Analysis.
Journal of
On 26/03/2012 6:24 AM, Henrik Alsing Friberg wrote:
Hi R-help users..
I mistakenly posted a now rejected bug report
(https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=14855), but the
subject still bugs me..!
If I have a package called MyPackage with C code inside, then library
or require(MyP
>
> On similar lines, how do I assign the position of the an occurence of a
> value say 7 in a data vector to another data vector?
>
> > nFields
> [1] 6 6 6 6 6 6 7 7 6 6 6 7 6 6 6 6
> I need the output data vector showing occurence of 7 as:
> [1] 7 8 12
which(nFields==7)
Regards
Petr
>
Please provide your code as is requested in the posting guidelines.
At a guess, you have read in your data into a data.frame called dhb with a
variable called dhb.
Try hist(dhb$dhb) and see what happens.
To see a bit more of what you've got try
str(dhb) to see what the sdtructure of the data is
It isn't quite clear to me that cod in data frame city and pos in
travel are are actually the same index, but if so then you can easily
use merge() for this task.
Sarah
On Mon, Mar 26, 2012 at 5:41 AM, MSousa wrote:
>
> Hello,
>
> I have two different dataset, and wanted to join the two.
> Fo
Yihui is correct that an error is generated. I should have mentioned that this
is
something I wanted to show in a vignette, and I got 103 characters as the
length of the
maximal line from a sink() file. Yihui's use of nchar is more elegant.
Thus the question remains of how to avoid margin over-r
Hi R-help users..
I mistakenly posted a now rejected bug report
(https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=14855), but the
subject still bugs me..!
If I have a package called MyPackage with C code inside, then library
or require(MyPackage) will call loadNamespace() which will look for
Hi,
I'm using the SEM package to estimate a model with a binary variabele as
dependent variable.
In the literature I have to use then the correlation matrix, made by
function hetcor(). Literature also says that the standard errors are not
correct then.
My question is if somebody knows what the imp
Hello,
I have two different dataset, and wanted to join the two.
For example I have a table of codes of cities, and other with with the
codes of travels, [source for the destine].
what he wanted was to have a new data.frame with all the information
city<-data.frame(city="Barcelona",cod=1)
Hi everyone,
does anyone know, which test of uniformity is run under the command
range.circular() in the circular statistics package? It didn't say anything in
the description of this command.
Your help would be appreciated a lot.
Benedikt
[[alternative HTML version deleted]]
_
Dear all,
I am trying to conduct an enhanced version of factor analysis with a
SPSS interface that allows to use R. This approach has been suggested in
the recent article:
Basto, M. and J.M. Pereira An SPSS R-Menu for Ordinal Factor Analysis.
Journal of Statistical Software 46, pp. 1-29.
Hi All,
I need to scale variable of tokens (several words per observation) to be able
to use it for svm().
For example, i have varible x4=(how,grow,tree)
Any ideas how to scale to use in svm()?
Thank you,
-Alex
[[alternative HTML version deleted]]
Hello, I didn't quite understand what you need, but maybe you can have a look
here:
www.slideshare.net/whitish/textmining-with-r
R code fragments are in appendixes of the presentation.
Hope this will help,
-Alex
From: r-help-boun...@r-project.org [r-help-
On 26/03/2012 8:09 AM, John D. Muccigrosso wrote:
I cannot for the life of me figure this out:
What's the parameter to fill in with color circles made with circles()? col changes the
line color, but all I see in the help is a reference to "additional graphic
parameters", and no examples via go
Dear all,
I have a data-set of 91 variables.
It is related with a qualitative analysis I made to analyse a few
phenomena. For this reason the questions are grouped in 10 clusters, since
they express different phenomena of the analysis. I made the subset to
collect the questions and represent the p
Gabor,
That does it! I can't thank you enough
Many thanks,
Tom
On Mon, Mar 26, 2012 at 7:22 AM, Gabor Grothendieck wrote:
> On Sun, Mar 25, 2012 at 10:20 PM, Thomas Adams
> wrote:
> > Gabor,
> >
> > Thank you for your help -- it did help me a lot. However, with my data:
> >
> >lead_time
Your data set probably has character variables. You can transform them
into factors before running na.roughfix.
Weidong Gu
On Mon, Mar 26, 2012 at 8:17 AM, silje skår wrote:
> I have a question regarding NA in randomForest (in R). I have a dataset
> which include both numerical and non-numerical
I have a question regarding NA in randomForest (in R). I have a dataset
which include both numerical and non-numerical variables, and the data
includes some NA. I tried to use na.roughfix but then i get an error
message "na.roughfix only works for numeric or factor". I also tried
rfImpute but this
I cannot for the life of me figure this out:
What's the parameter to fill in with color circles made with circles()? col
changes the line color, but all I see in the help is a reference to "additional
graphic parameters", and no examples via google.
Thanks!
John Muccigrosso
__
Should of seen that! It works now.
Thank you Duncan,
Aurelie
On 2012-03-26, at 8:39 AM, Duncan Murdoch wrote:
> On 12-03-26 7:17 AM, Aurelie Cosandey Godin wrote:
>> Dear list,
>>
>> I am getting an error message with "pin" and I am not quiet sure how to fix
>> it.
>>
>> library(maps)
>> op<-
On 12-03-26 7:17 AM, Aurelie Cosandey Godin wrote:
Dear list,
I am getting an error message with "pin" and I am not quiet sure how to fix it.
library(maps)
op<-par(mar=c(1,1,1,1))
map(xlim=c(-40,-45),ylim=c(60,65), boundary=T, fill=T, col='gray95')
Error in par(pin = p) :
invalid value sp
On Sun, Mar 25, 2012 at 10:20 PM, Thomas Adams wrote:
> Gabor,
>
> Thank you for your help -- it did help me a lot. However, with my data:
>
> lead_time cycle r_squared fcst_date
> 1 6 0 5.405095e-02 07/31/2010
> 2 12 0 5.521620e-06 07/31/2010
> 3 18 0 1
Dear list,
I am getting an error message with "pin" and I am not quiet sure how to fix it.
library(maps)
op<-par(mar=c(1,1,1,1))
map(xlim=c(-40,-45),ylim=c(60,65), boundary=T, fill=T, col='gray95')
>Error in par(pin = p) :
invalid value specified for graphical parameter "pin"
par("pin")
> pa
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