Hi,
I've solved the problem. This command works
axis(1,at=axTicks(1),label = axTicks(1)/100);
Thanks for your help
~Rajesh
On Mon, May 17, 2010 at 9:55 AM, rajesh j wrote:
> Hi,
> I tried using this in my plot and I get an error saying
> 'at' and 'labels' lengths differ, 8 != 5
>
> ~Rajesh
>
>
Hi,
I tried using this in my plot and I get an error saying
'at' and 'labels' lengths differ, 8 != 5
~Rajesh
On Sat, May 15, 2010 at 10:47 PM, Henrique Dallazuanna wrote:
> Try this:
>
> x <- 1:100
> plot(x, xaxt = 'n')
> axis(1, axTicks(1), pretty(x) / 10)
>
> On Sat, May 15, 2010 at 2:10 PM, r
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of sam.e
> Sent: Sunday, May 16, 2010 10:13 AM
> To: r-help@r-project.org
> Subject: Re: [R] Splines under tension
>
>
> Thank you for the helpful direction to the smoothing spline
Try this:
1 + (1 / log(length(lambda_cor))) * sum((l <- lambda_cor /
length(lambda_cor)) * log(l))
On Sun, May 16, 2010 at 10:43 PM, Roslina Zakaria wrote:
> Hi r-users,
>
> I have this code here, but I just wonder how do I use 'sapply' to make it
> more efficient
>
> lamda_cor <- eigen(winter_
Hi r-users,
I have this code here, but I just wonder how do I use 'sapply' to make it more
efficient
lamda_cor <- eigen(winter_cor)$values
> lamda_cor
[1] 1.3459066 1.0368399 0.8958128 0.7214407
lamda_cxn <- function(dt)
{ n <- length(dt)
term <- vector(length=n, mode="numeric")
Or even:
with(x, a / coredata(a[1]) )
On Sun, May 16, 2010 at 7:48 PM, Gabor Grothendieck
wrote:
> Normally that would be written like this using the coredata extraction
> function which extracts the data portion of a zoo object:
>
> x$a / coredata( x$a[1] )
>
> On Sun, May 16, 2010 at 7:32 PM,
Normally that would be written like this using the coredata extraction
function which extracts the data portion of a zoo object:
x$a / coredata( x$a[1] )
On Sun, May 16, 2010 at 7:32 PM, Sean Carmody wrote:
> Thanks David,
>
> You comment made me realise that whereas when x is a data frame, x$a
Thanks David,
You comment made me realise that whereas when x is a data frame, x$a is a
numeric vector,
when x is of class zoo, x$a is also of class zoo, so the following does what
I was expecting:
x$a/as.numeric(x$a[1])
Sean.
On Sun, May 16, 2010 at 9:25 PM, David Winsemius wrote:
>
> On May
Never mind. Stupid misplaced 's'.
-Andrew
On Sun, May 16, 2010 at 5:39 PM, Andrew Redd wrote:
> Dear R help,
>
> What am I doing wrong here? when I don't specify the priors it works
> just fine but when I specify the priors it breaks. Does anyone know
> why and how I can fix it?
>
>> N=2000
Dear R help,
What am I doing wrong here? when I don't specify the priors it works
just fine but when I specify the priors it breaks. Does anyone know
why and how I can fix it?
> N=2
> ncontrol=ncases=50
> X <- as.matrix(rnorm(N,0,1))
> eta <- -5.3 + X * 1.7
> p <- exp(eta)/(1+exp(eta))
>
And if you do have many variables in one dataframe, you might
wish to construct the formula first using paste():
nm <- c("0", names(d)[-c(1,2)])
fo <- as.formula(paste("~", paste(nm, collapse= "+")))
d <- cbind(d, model.matrix(fo, data=d)
-Peter Ehlers
On 2010-05-16 15:30, Thomas Stewart wr
Maybe this will lead you to an acceptable solution. Note that changed how
the data set is created. (In your example, the numeric variables were being
converted to factor variables. It seems to me that you want something
different.) The key difference between my code and yours is that I use the
Greetings
I suppose a simple matter for R experts but for me...
Am using GSA and have the GSA.obj (class GSA) created with GSA that contains
the value ngenes visible with GSA.obj$ngenes
And I have the list object (class list) created with GSA.listsets that
contains
Table of negative sets with sc
Erich Neuwirth wrote:
>
> Look for Rprof in the utils package.
>
This was already suggested- but the original poster clarified that he is
looking to profile the R interpreter it's self, not R scripts.
-
Charlie Sharpsteen
Undergraduate-- Environmental Resources Engineering
Humboldt State
Thank you so much for pointing on this obvious check of the MS Access
database! Inspired, I tried to import the csv-file directly into the MS
Access database and I encountered an Error saying (freely translated from
Danish) : "Cannot find search key".
The MS Access database is in MS Access-2000 for
Peter Ehlers wrote:
> > and then use the rest of the plotpc() code as is (except for
> > maybe having to use flip1=TRUE, etc).
Nikos:
> Hmm... I am _now_ working on it to understand how I could make this
> "automatic"!.
>
> If I give flip1, flip2 (=TRUE) the histograms are located where they shou
Peter Ehlers wrote:
> Nikos,
>
> I think you can just replace the line
>
> pc <- princomp(x[,1:2], scores=TRUE, na.action=na.fail)
>
> with
>
> pc <- prcomp(x[,1:2], retx=TRUE, center=pc.center,
> scale.=pc.scale, na.action=na.fail)
>
> and rename the components of
I could, but with close to 100 columns, its messy.
On 5/16/10 11:22 AM, Peter Ehlers wrote:
> On 2010-05-16 11:06, Noah Silverman wrote:
>> Update,
>>
>> I have it working, but now its producing really ugly labels. Must be a
>> small adjustment to the code. Any ideas??
>>
>> ##Create example da
On 2010-05-16 11:06, Noah Silverman wrote:
Update,
I have it working, but now its producing really ugly labels. Must be a
small adjustment to the code. Any ideas??
##Create example data.frame
group<- c("A", "B","B","C","C","C")
a<- c(1,4,3,4,5,6)
b<- c(5,4,5,3,4,5)
d<- data.frame(cbind(a,b,gr
Nikos,
I think you can just replace the line
pc <- princomp(x[,1:2], scores=TRUE, na.action=na.fail)
with
pc <- prcomp(x[,1:2], retx=TRUE, center=pc.center,
scale.=pc.scale, na.action=na.fail)
and rename the components of pc
names(pc) <- c('sdev', 'loadings', 'c
Nikos Alexandris:
> Among the (R-)tools, I've seen on the net, for (bivariate) Principal
> Component scatter plots (+histograms), "plotpc" [1] is the one I like
> most.
[...]
> I started the modification by attempting first to get a "prcomp" version of
> "plotpc()" (named it "plotpc.svd()") by a
Hello Hadley,
Thank you very much for your help! I have just received your book btw :)
On May 16, 2010, at 6:16 PM, Hadley Wickham wrote:
>Hi Giovanni,
>
>Have a look at the classifly package for an alternative approach that
>works for all classification algorithms. If you provided a small
>repr
Thank you for the helpful direction to the smoothing splines function, it was
very helpful and is exactly what i am trying to do. My data however is 3-D,
i.e. i have x and y values which are coordinates for different field sites
and z values which are really what I am interested in analysing with
Update,
I have it working, but now its producing really ugly labels. Must be a
small adjustment to the code. Any ideas??
##Create example data.frame
group <- c("A", "B","B","C","C","C")
a <- c(1,4,3,4,5,6)
b <- c(5,4,5,3,4,5)
d <- data.frame(cbind(a,b,group))
#create new frame with discretized
On 2010-05-16 6:22, Holger Steinmetz wrote:
Dear experts,
I tried to learn about Box-Cox-transformation but found the following thing:
When I had to add a constant to make all values of the original variable
positive, I found that
the lambda estimates (box.cox.powers-function) differed dramati
Thanks,
That gives me exactly what I'm looking for.
Two quick questions:
1) What would be the fastest way to do this if I have other continuous
data as well. For example, I have a data frame with 10 variable and
want to discretize one of them using this method. (Say, column 6 for
example.)
I
Look for Rprof in the utils package.
On 5/12/2010 9:22 PM, xiaoming gu wrote:
> Hi, all. Does anyone know how to profile R interpreter? I've tried gprof but
> it doesn't work. Thanks.
>
> Xiaoming
--
Erich Neuwirth, University of Vienna
Faculty of Computer Science
Center for Computer Science D
Use readBin:
readBin('/path/to/file', 'raw', n=100)
Here is an example of reading in a JPEG file:
> x <- readBin(fileName,'raw',n=100)
> str(x)
raw [1:801403] ff d8 ff e1 ...
> # convert to a HEX string (on the first 20 bytes)
> paste(sprintf("%s", x[1:20]), collapse='')
[1] "ffd8ffe1ff
Hi Giovanni,
Have a look at the classifly package for an alternative approach that
works for all classification algorithms. If you provided a small
reproducible example, I could step through it for you.
Hadley
On Sat, May 15, 2010 at 6:19 AM, Giovanni Azua wrote:
> Hello,
>
> I have a labelled
Dear R-list members,
About the parameter n of the function density() (Kernel Density
Estimation, package stats):
The R HTML documentation says about the parameter n: "the number
of equally spaced points at which the density is to be estimated.
When n > 512, it is rounded up to the next power of
runif(1)
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Soham
Sent: Saturday, May 15, 2010 9:05 AM
To: r-help@r-project.org
Subject: [R] p value
How to compute the p-value of a stati
Colleagues,
I am using R to assemble RTF documents (which are plain text). I need to embed
a JPEG graphic that was created with R. I presume that the steps need to be:
a. read the file into R
b. convert the object to HEX format
c. write the converted object to a textf
Dear experts,
I tried to learn about Box-Cox-transformation but found the following thing:
When I had to add a constant to make all values of the original variable
positive, I found that
the lambda estimates (box.cox.powers-function) differed dramatically
depending on the specific constant chos
When you combine zoo objects with arithmetic it merges them using all = FALSE:
> library(zoo)
> x <- data.frame(a=1:5*2, b=1:5*3)
> x <- zoo(x); x
a b
1 2 3
2 4 6
3 6 9
4 8 12
5 10 15
>
> # these two are the same
>
> x$a/x$a[1]
1
1
>
> m <- merge(x$a, x$a[1], all = FALSE)
> m
x$a x$a[
I started with the summarized data, and there are different ways to do
this. For this example, let there be four columns and a corresponding
sum of 1s.
library("ggplot2")
mydf <- data.frame(colname = c("A","B","C","D"),mycolsum=c(1:4))
p <- ggplot(mydf,aes(x=colname,y=mycolsum))
p <- p + geom_bar
Hi Steve,
I "think" what you want to do is get a unique time-date from the first two
columns.
Try something like this: (changing the file name obviously.
mydate should give you a time and date format that you can add to the existing
data.frame.
mydata <- read.table("C:/rdata/dates.junk.csv
In zoo the index= argument to read.zoo can be a vector of column
indices to indicate that the time is split across multiple columns and
the FUN= argument can be used to process the multiple columns. In
this example the resulting z uses chron:
L <- "Date,Time,Open,High,Low,Close,Up,Down
05/02/200
Let us see if it is a R issue.
Try this:
Read the CSV on Ms Access directly. It is an importation on MsAccess.
If you succeed we will check R then.
Caveman
On Sun, May 16, 2010 at 11:48 AM, Johan Lassen wrote:
> Dear R-community,
>
> After repeating the sqlSave-command 3 times on a dataframe
On May 16, 2010, at 2:24 AM, Agustín Muñoz M. (AMFOR) wrote:
hi everybody, a question, as I can know the location (number) of an
attribute with its name.
Ej.
X1 X2 X3 X4 X5 X6
1 3 5 2 1 7
6 7 4 5 2 9
as I can know that the attribute "X4" is in position 4
It is prob
Hi,
I am trying to load a data file that looks like this:
|Date,Time,Open,High,Low,Close,Up,Down
05/02/2001,0030,421.20,421.20,421.20,421.20,11,0
05/02/2001,0130,421.20,421.40,421.20,421.40,7,0
05/02/2001,0200,421.30,421.30,421.30,421.30,0,5
05/02/2001,0230,421.60,421.60,421.50,421.50,26,1|
etc.
On May 16, 2010, at 2:00 AM, Sean Carmody wrote:
I am a bit confused about the different approaches taken to
recycling in
plain data frames and zoo objects. When carrying out simple
arithmetic,
dataframe seem to recycle single arguments, zoo objects do not. Here
is an
example
x <- data.
On 05/16/2010 03:10 AM, rajesh j wrote:
Hi,
I have a plot whole tick values along the axis have a certain range 0 - x .
I need to normalize this range without changing my data files. for e.g.,
if my plot has tick values at 10,20,30,40,50... i have to make this 2,4,6,
etc. but without changing th
On 05/16/2010 12:03 AM, Giovanni Azua wrote:
Hello,
I managed to "linearize" my LDA decision boundaries now I would like to call abline three
times but be able to specify the exact x range. I was reading the doc but it doesn't seem to
support this use-case? are there alternatives. The reason w
On 2010-05-16 2:23, Laetitia Schmid wrote:
Hi,
I am sampling two random columns from females and two random columns
from males to produce tetraploid offspring. For every female I am
sampling a random male.
In the end I want to write out a a matrix with all the offspring, but
that does not work. I
Dear R-community,
After repeating the sqlSave-command 3 times on a dataframe (of size 13149
rows * 5 columns) to my MS-Access database I get the following error:
*Error in sqlSave(channel, eksport_transp_acc_2, "transp_acc_scenarier", :
unable to append to table transp_acc_scenarier*
**
This m
Hi,
On 16 May 2010 03:31, michael westphal wrote:
[ snipped ]
> Any suggestions?
>
i'd suggest you
- read the posting guide
- upgrade your R to the latest version
- don't post to two mailing lists
- make your example minimal, self-contained, reproducible
- show the result of sessionInfo()
HTH,
Hi Soham,
I don't feel your question is well defined.
But an equally ill defined answer would be:
Through a permutation test.
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Heb
On 16-May-10 06:11:52, Agustín Muñoz "M. (AMFOR) wrote:
> hi everybody, a question, as I can know the location (number) of an
> attribute with its name.
>
> Ej.
>
> X1 X2 X3 X4 X5 X6
> 1 3 5 2 1 7
> 6 7 4 5 2 9
>
> as I can know that the attribute "X4" is in position 4
>
Hi, if i just want a vector filled with names which has length(index) > 0.
For example if
nombreC <- c("Juan", "Carlos", "Ana", "María")
nombreL <- c("Juan Campo", "Carlos Gallardo", "Ana Iglesias", "María
Bacaldi", "Juan Grondona", "Dario Grandineti", "Jaime Acosta",
"Lourdes Serrano")
I would
Hi,
I am sampling two random columns from females and two random columns
from males to produce tetraploid offspring. For every female I am
sampling a random male.
In the end I want to write out a a matrix with all the offspring, but
that does not work. I get always only the offspring from th
On 2010-05-13 17:50, Phil Spector wrote:
Vincent -
I'm afraid there's no solution other than artificially modifying
the zeroes:
vec
[1] 26.58950617 5.73074074 5.9622 5.6478 20.95728395 0.
0.0700 12.8689
[9] 3.64543210 0.05049383 25.6089 3.53246914 0. 31.3904
hi everybody, a question, as I can know the location (number) of an
attribute with its name.
Ej.
X1 X2 X3 X4 X5 X6
1 3 5 2 1 7
6 7 4 5 2 9
as I can know that the attribute "X4" is in position 4
I hope you can help me
from already thank you very much to all
Agustín
__
Dear R Family,
I have an error message. I would like to learn how to deal with that.
The orginal series is as follows: I just pick up the first 10 observations.
> dif_transaud[1:10]
[1] 0.0065880493 -0.0065880490 -0.0131743570 0.0197745715 0.0065889175
[6] 0.0131813110 0.0065923924 -0.03955
hi everybody, a question, as I can know the location (number) of an
attribute with its name.
Ej.
X1 X2 X3 X4 X5 X6
1 3 5 2 1 7
6 7 4 5 2 9
as I can know that the attribute "X4" is in position 4
I hope you can help me
from already thank you very much to all
Agustín
__
How to compute the p-value of a statistic generally?
--
View this message in context:
http://r.789695.n4.nabble.com/p-value-tp2217867p2217867.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://st
hi everybody, a question, as I can know the location (number) of an
attribute with its name.
Ej.
X1 X2 X3 X4 X5 X6
1 3 5 2 1 7
6 7 4 5 2 9
as I can know that the attribute "X4" is in position 4
I hope you can help me
from already thank you very much to all
Agustín
__
I am a bit confused about the different approaches taken to recycling in
plain data frames and zoo objects. When carrying out simple arithmetic,
dataframe seem to recycle single arguments, zoo objects do not. Here is an
example
> x <- data.frame(a=1:5*2, b=1:5*3)
> x
a b
1 2 3
2 4 6
3 6
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