Hello,
I want to create and save objects in a loop, but this is precluded by the
following obstacle:
this part of the script fails to work:
assign(x=paste(a, 1, sep=), value=1);
save(paste(a, 1, sep=), file=paste(paste(a, 1, sep=), .RData, sep=))
Do you know any workaround? I am already out
Martin Ivanov wrote
Hello,
I want to create and save objects in a loop, but this is precluded by the
following obstacle:
this part of the script fails to work:
assign(x=paste(a, 1, sep=), value=1);
save(paste(a, 1, sep=), file=paste(paste(a, 1, sep=), .RData,
sep=))
Do you know
I have managed to use the tripack package which is a bit simpler to use
(for me) and performs better (timely)...
##
# inputFile format: x y z
inputFile - d:\\inputfile.xyz.txt
# output file will contain a x3d scene embedded in a html file to be
You can do a transformation on your Y, and than use the normal tools.
Using something like a median test is not advisable due to low power, and a
wilcox.test is considered a better choice (although the meaning is more
complex when observations are not symmetrically distributed).
If you wish to
Hi all,
The past couple of days I have been trying to update my R installation
(Debian) and I am getting the following:
Get:9 http://http.us.debian.org squeeze/main amd64 Packages [6,542 kB]
Ign http://mirrors.nics.utk.edu squeeze-cran/ Release
Ign http://mirrors.nics.utk.edu squeeze-cran/
Hi,
This should make much more sense
onenew-data.frame(keyword=gsub((NA){0,1}\\|, ,one$keyword))
onenew1-data.frame(keyword=gsub((NA){0,1},,onenew$keyword))
onenew1
keyword
1 auto
2 auto insurance quote
3 auto insurance
4 insurance
5 auto
HI,
I guess you wanted to change the scale of Y-axis:
plot(X,Y,ylim=c(0,25))
A.K.
- Original Message -
From: Lekgatlhamang, lexi Setlhare lexisetlh...@yahoo.com
To: r-h...@stat.math.ethz.ch r-h...@stat.math.ethz.ch
Cc:
Sent: Thursday, July 19, 2012 5:31 PM
Subject: [R] A graph with
Hi all,
I'm trying to build a package 'Thiele', but run into problems with generic
functions.
I have a class Benefit, and defined the function print.Benefit.
When I try R CMD check Thiele in X11, this warning came out
---
Functions/methods with usage in documentation object 'Benefit' but
On 19/07/12 20:40, Erdal Karaca wrote:
Oh, I dont want anyone to do anything for me. I am just asking for
hints/infos to be able to do it myself...
I was hoping that someone has already done this...
But anyways, I will post my solution once I have succeeded...
If you haven't already looked at
Hi,
Check this link.
http://r.789695.n4.nabble.com/help-comparing-two-median-with-R-td822998.html
A.K.
- Original Message -
From: Data Analytics Corp. w...@dataanalyticscorp.com
To: r-help@r-project.org
Cc:
Sent: Thursday, July 19, 2012 4:14 PM
Subject: [R] median comparison tests
use function system in R
for (x in 1:100){
system (paste(rename file0.,x,_data.RData file,x,_data.Rdata,sep=))
}
-
Yasir Kaheil
--
View this message in context:
http://r.789695.n4.nabble.com/change-file-name-from-file0-1-data-RData-to-file1-data-Rdata-tp4637135p4637136.html
Sent from the R
I'm having a lot of trouble getting this done and nothing I've written has
been remotely successful. Basically I have about 64 binary data sets stored
as vectors, with 72900 entries in each. I am not very familiar with cross
correlations, but I was advised that I should create 1000 randomized date
Hi,
Just a variant of creating rounded down minutes vector:
minutes-ifelse(ceiling(as.numeric(substr(t[,1],15,16))30),30,00)
#or
minutes1-ifelse(floor(as.numeric(substr(dat1[,1],15,16))30),30,00)
identical(minutes,minutes1)
#[1] TRUE
minutes3-floor( as.numeric( substr( t[,1], 15, 16 ) ) / 30
Dear Sarah and A.K.,
Yes, both your suggestions give me exactly what I wanted. Thank you kindly.
Lexi
From: Sarah Goslee sarah.gos...@gmail.com
Cc: r-h...@stat.math.ethz.ch r-h...@stat.math.ethz.ch
Sent: Thursday, July 19, 2012 11:36 PM
Subject: Re: [R] A
I'm getting this error message:
nms-names(data)[grep(vars,names(data))]
Warning message:
In grep(vars, names(data)) :
argument 'pattern' has length 1 and only the first element will be used
Is there a way around this?
On Thu, Jul 19, 2012 at 6:17 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Really appreciate the discussion on outliers.
I come from an engineering signal processing background, and my thinking
has generally been that an outlier is outside a threshold of
- distance from the mean
- rarity
that we don't need/want to capture in whatever model we're building.
In my
I'm still getting the message (if this is what you were suggesting I try).
The data set I'm using has many more columns other than these variables;
could that be a problem? I didn't think it would affect it.
pattern - L[1-8][12]
nms-names(data)[grep(vars,names(data))]
Warning message:
In
Hi,
Try this:
dat1-read.table(text=
y A B C
0 1 1 2
0 1 2 1
1 1 1 2
0 1 1 2
1 1 1 2
1 1 2 1
0 1 2 2
,sep=,header=TRUE)
dat2-aggregate(y~A+B+C,data=dat1,sum)
dat2-dat2[,c(4,1:3)]
dat3-dat2[with(dat2,rev(order(y,A,B,C))),]
dat3
y A B C
2 2
Dear Sir,
I am using timereg package for analysing competing risk data.
There are two options in method: proportional and additive.
How to choose among the methods for analysis and how to check which of the
methods are appropiate for the specific data.
Tanks a lot.
Dr Suman Kumar
Hi,
Try this:
ans-function(){.Last.value}
2+3
[1] 5
ans()
[1] 5
A.K.
- Original Message -
From: darnold dwarnol...@suddenlink.net
To: r-help@r-project.org
Cc:
Sent: Friday, July 20, 2012 1:12 AM
Subject: [R] Last answer
Hi,
In Matlab, I can access the last computation as
Hi!
# toy data
toyData - data.frame(x = 1:4, y = 5:8, xy = 9:12, z = 13:16)
vars - c(x, z)
# pattern is an argument of grep
args(grep)
# pattern must only consist of a single element
# otherwise only the first element is used
grep(pattern =
On 19.07.2012 22:44, zgu9 wrote:
Hi all,
I'm trying to build a package 'Thiele', but run into problems with generic
functions.
I have a class Benefit, and defined the function print.Benefit.
When I try R CMD check Thiele in X11, this warning came out
---
Functions/methods with usage in
Hi,
What is the function for inverse normal transformation?
Thanks,
Carol
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Dear all,
Is it possible to create a pdf file with layers using the pdf() device in R?
Many thanks for your help!
Christoph
(using R 2.15.1 on Windows 7 64-Bit)
--
PD Dr Christoph Scherber
Georg-August University Goettingen
Department of Crop Science
Agroecology
Grisebachstrasse 6
D-37077
On 12-07-20 2:50 AM, Martin Ivanov wrote:
Hello,
I want to create and save objects in a loop, but this is precluded by the
following obstacle:
this part of the script fails to work:
assign(x=paste(a, 1, sep=), value=1);
save(paste(a, 1, sep=), file=paste(paste(a, 1, sep=), .RData, sep=))
On 12-07-20 6:21 AM, carol white wrote:
Hi,
What is the function for inverse normal transformation?
qnorm
Duncan Murdoch
Thanks,
Carol
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Thanks for your reply.
So to derive it from a given data set, is the following correct to do?
my_data.p =2*pnorm(abs(my_data),lower.tail=FALSE)
my_data.q = qnorm(my_data.p)
Cheers,
From: Duncan Murdoch murdoch.dun...@gmail.com
Cc: r-h...@stat.math.ethz.ch
Hi List
I have a dataframe (~1,200,000 rows deep) and I'd like to conditionally reorder
groups of rows in this dataframe.
I would like to reorder any rows where the Chr.Strand column contains a '-' but
reorder within subsets delineated by the Probe.Set.Name column.
# toy example
On 12-07-20 7:36 AM, carol white wrote:
Thanks for your reply.
So to derive it from a given data set, is the following correct to do?
my_data.p =2*pnorm(abs(my_data),lower.tail=FALSE)
my_data.q = qnorm(my_data.p)
I don't know what you're trying to do, but that doesn't look like it
does
Dear Peter,
This indeed resolves the problem.
Many thanks. My apologies for not starting a new thread. I am a new R user
and not yet fully integrated into the R community.
Kind regards,
Bart Ferket
--
View this message in context:
Dear prof. Harrell,
I'm not able to use the force option with fastbw, here an example of the error
I've got (dataset stagec rpart package):
fitstc - cph(Surv(stagec$pgtime,stagec$pgstat) ~ age + eet + g2 + grade +
gleason + ploidy, data=stagec)
fbwstc -
Hi,
I would like to evaluate a function, with 3 arguments, for instance,
myfunc-function(a,b,c) { sqrt(a)-exp(b)+4*c
}
How to execute myfunc(x,y,z), for all x, all y and all z, where x,y,z are
vectors?
Thank you very much in advance
--
View
On 20/07/12 12:07, Christoph Scherber wrote:
Dear all,
Is it possible to create a pdf file with layers using the pdf() device in R?
No. Is it possible to specify layers in the R graphics language or any
device? (From what I understand by 'layers', no.)
The author of pdf()
Many thanks
Thanks David for the solution, I paste here the code with a few changes:
par(xpd = NA, oma = c(5, 0, 0, 0))
matplot(t(buffersump[,1:6]), type=l, xaxt=n, lwd=3, ylab=Porcentajes de
respuestas afirmativas)
axis(1, 1:6, colnames(buffersump2))
legend(bottomright, legend=rownames(buffersump2),
Dear r-help members.
I would like to compare species numbers of moths between eight different
forests (each sampled for six nights). I would like to do a nested anova to
compare species numbers between forests and nights.
For more site specific details I wanted to do a Tukey test (TukeyHSD).
I have the following data:
x - as.factor(c(1,1,1,2,2,2,3,3,3))
y - as.factor(c(10,10,10,20,20,20,30,30,30))
z - c(100,100,NA,200,200,200,300,300,300)
I could create the cross tab of x and y with Sum of z as its elements using
the xtabs function as follows:
# X Vs. Y with Sum Z
xtabs(z ~ x + y)
Hello,
I have a file like this (just a snapshot) where I have numerical
values for various genes, I want a line plot with shading (may be
using smooth ?) using qplot or ggplot :
Gene1 10 14 12 23 11 11 33 1 ..(multiple columns)
Gene2 4 2 1 1 3 4 1 2 .
Gene3 2 5 7 5 6 89 7 3 ..
Gene4 1
HI,
Probably this might be helpful for you.
http://rgm2.lab.nig.ac.jp/RGM2/func.php?rd_id=fBasics:dist-nigFit
A.K.
- Original Message -
From: carol white wht_...@yahoo.com
To: r-h...@stat.math.ethz.ch r-h...@stat.math.ethz.ch
Cc:
Sent: Friday, July 20, 2012 6:21 AM
Subject: [R]
I have a continuous data. So to calculate the inverse normal transformation, I
thought that I should first calculate the Z-score normalized data and then,
calculate the p-value et the quantile transformation. Does this seem to be more
sensible
my_data.p
Hello,
No it's not correct, you are computing a what seems to be a two-tailed
probabiity, so the inverse should account for it. Look closely: you take
the absolute value, then the upper tail probability, then multiply 2
into it. Reverse these steps to get the correct value.
# Helper function
You mean executing the function for all combinations of values?
For example, if you have a-b-c-1:2
you would get back the values of
myfunc(1,1,1)
myfunc(1,1,2)
myfunc(1,2,1)
myfunc(1,2,2)
myfunc(2,1,1)
myfunc(2,1,2)
myfunc(2,2,1)
myfunc(2,2,2)
?
On 20.07.2012, at 13:05, carla moreira wrote:
Not quite sure what you are aiming at, but looking at ?mapply or ?expand.grid
could be helpful
Benno
On Jul 20, 2012, at 1:05 PM, carla moreira wrote:
Hi,
I would like to evaluate a function, with 3 arguments, for instance,
myfunc-function(a,b,c) { sqrt(a)-exp(b)+4*c
Hi Tjun Kat,
you can define variables outside the ode function, but normally NOT
state variables, because their values need to be updated by the solver
during the simulation process.
But, if you want to block this for any debugging purposes and want to
e.g. fix a derivative to a certain
See below for the complete mail to which I reply which was not sent to rhelp.
==
emptyexpandlist2-list(ne=0,l=array(NA, dim=c(1, 1000L)),len=1000L)
addexpandlist2-function(x,prev){
if(prev$len==prev$ne){
n2-prev$len*2
prev - list(ne=prev$ne, l=array(prev$l, dim=c(1, n2)),
Thanks Rui.
I changed my scripts to the followings and I think that it still is not
correct. See also the attached file.
Thanks for your help,
tmp
[1] 2.502519 1.828576 3.755778 17.415000 3.779296 2.956850 2.379663
[8] 1.103559 8.920316 2.744500 2.938480 7.522174 10.629200
General problem: I have 20 projects that can be invested in and I need to
decide which combinations meet a certain set of standards. The total
possible combinations comes out to 2^20. However I know for a fact that the
number of projects must be greater than 5 and less than 13. So far the the
code
Yes, I do.
But I need to control how the permutations are done.
Thank you.
2012/7/20 Jessica Streicher j.streic...@micromata.de
You mean executing the function for all combinations of values?
For example, if you have a-b-c-1:2
you would get back the values of
myfunc(1,1,1)
myfunc(1,1,2)
I think we need some raw data. Have a look at ?dput for a way to supply it.
It might help if you supplied some sample code of what you have tried.
I want a line plot is not particularly helpful.
John Kane
Kingston ON Canada
-Original Message-
From: anamika...@gmail.com
Sent:
Well, what do you want to control there?
Need a subset? Need an ordering?
On 20.07.2012, at 15:00, Carla Moreira wrote:
Yes, I do.
But I need to control how the permutations are done.
Thank you.
2012/7/20 Jessica Streicher j.streic...@micromata.de
You mean executing the function
Not quite what you asked for but would this do?
library(plyr)
x - as.factor(c(1,1,1,2,2,2,3,3,3))
y - as.factor(c(10,10,10,20,20,20,30,30,30))
z - c(100,100,NA,200,200,200,300,300,300)
df1 - data.frame(x, y, z)
(dsum - ddply(df1, .(x, y), summarise, sum = sum(z, na.rm = TRUE),
Dear R users,
I have the following problem. I plot a SpatialPixelsDataFrame object with
spplot and I need to add a contourplot
of another spatial variable. To be more precise, I plot the
SpatialPixelsDataFrame of the mean precipitation over Germany with coloured
regions and I want to overlay
On Jul 20, 2012, at 5:30 AM, vioravis wrote:
I have the following data:
x - as.factor(c(1,1,1,2,2,2,3,3,3))
y - as.factor(c(10,10,10,20,20,20,30,30,30))
z - c(100,100,NA,200,200,200,300,300,300)
I could create the cross tab of x and y with Sum of z as its elements using
the xtabs
There are lots of possibilities. Here's one using only xtabs():
dframe - na.omit(data.frame(x, y, z))
zsum - xtabs(z~x+y, dframe)
zcount - xtabs(~x+y, dframe)
zmean - ifelse(is.nan(zsum/zcount), 0, zsum/zcount)
--
David L Carlson
Associate Professor of
It would help if you supplied some code for us to see what you have tried.
However something like this would presumably give you the sample data sets
--note only n=100 in the example.
mymat - matrix(NA, ncol= 100, nrow= 100)
for (n in 1: 100) {
mymat[,n] - sample(c(0,1), 100, replace =
On 2012-07-20 04:05, carla moreira wrote:
Hi,
I would like to evaluate a function, with 3 arguments, for instance,
myfunc-function(a,b,c) { sqrt(a)-exp(b)+4*c
}
How to execute myfunc(x,y,z), for all x, all y and all z, where x,y,z are
I've had to do something similar, so I wrote a small function to help.
This runs in about 1/4 the time of your code on my machine.
Others may have a more efficient approach.
all.combs - function(num, from=0, to=num) {
# create a matrix of all possible combinations of num items
#
Inline.
-- Bert
On Fri, Jul 20, 2012 at 6:59 AM, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2012-07-20 04:05, carla moreira wrote:
Hi,
I would like to evaluate a function, with 3 arguments, for instance,
myfunc-function(a,b,c) { sqrt(a)-exp(b)+4*c
Rui's example included z-score data (drawn from rnorm). You converted your
data to z-scores so you need to compare your results to the z-scores not
the original data.
Change these lines:
tmp.qnorm = qnorm(tmp.p/2,lower.tail=FALSE)*sign(scale(tmp))
# sign is of scale(tmp) not tmp
On 20/07/2012 8:28 AM, carol white wrote:
Thanks Rui.
I changed my scripts to the followings and I think that it still is not correct.
You haven't told us clearly what you are trying to achieve, and you
don't tell us what is wrong with what you have. How do you expect
anyone to help?
Michael,
Use dput() to output your data (or perhaps a small subset). Then paste
the result of that call and your R code (just those lines of code that are
needed to reproduce the problem) right in your message to R-help. That
makes it easier for the R-help list readers to help you
On Fri, Jul 20, 2012 at 05:45:30AM -0700, wwreith wrote:
General problem: I have 20 projects that can be invested in and I need to
decide which combinations meet a certain set of standards. The total
possible combinations comes out to 2^20. However I know for a fact that the
number of projects
Dear Prof Ripley,
Many thanks for your response. In fact, latticeExtra and ggplot2 both state
they work with layers,
but essentially this is nothing more than sequentially adding graphical output
to an existing device.
But given that PDF can include layers since version 1.5 it would be
Hello,
I would like to convert date as a factor to represent time in a repeated
measure situation in the following code. How would I do that?
d - read.csv(file.path(dataDir,data.csv), as.is=T,stringsAsFactors =
FALSE)
d[1:2,]
id date ab c y
1 1
Petr,
This is great.
MUCH faster than the code I provided.
And much more elegant code.
Thanks for posting!
Jean
Petr Savicky savi...@cs.cas.cz wrote on 07/20/2012 09:26:34 AM:
On Fri, Jul 20, 2012 at 05:45:30AM -0700, wwreith wrote:
General problem: I have 20 projects that can be invested
On Fri, Jul 20, 2012 at 04:26:34PM +0200, Petr Savicky wrote:
On Fri, Jul 20, 2012 at 05:45:30AM -0700, wwreith wrote:
General problem: I have 20 projects that can be invested in and I need to
decide which combinations meet a certain set of standards. The total
possible combinations comes
Bert,
The only thing wrong is that I'm still 75% asleep! Yikes!!
Thanks for the heads-up.
Carla: See Bert's solution.
Peter Ehlers
On 2012-07-20 07:10, Bert Gunter wrote:
Inline.
-- Bert
On Fri, Jul 20, 2012 at 6:59 AM, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2012-07-20 04:05, carla
-Original Message-
Subject: [RsR] How does rlm in R decide its w weights for
each IRLS iteration?
I am also confused about the manual:
a. The input arguments:
wt.method are the weights case weights (giving the relative
importance of case, so a weight of 2 means
Duncan Murdoch-2 wrote
On 12-07-20 2:50 AM, Martin Ivanov wrote:
Hello,
I want to create and save objects in a loop, but this is precluded by the
following obstacle:
this part of the script fails to work:
assign(x=paste(a, 1, sep=), value=1);
save(paste(a, 1, sep=),
Michael Eisenring eimic...@ethz.ch wrote on 07/20/2012 09:35:03 AM:
Dear Jean,
thanks for this email. I'm grateful for this instruction Just to
make sure that I understood you properly: something like this?:
Michael,
Yes, this is perfect. Very helpful.
My data:
Reith, William [USA] reith_will...@bah.com wrote on 07/20/2012
09:52:02 AM:
Would this matrix eat up memory making the rest of my program
slower? Each x needs to be multiplied by a matrix and the results
checked against a set of thresholds. Doing them one at a time takes
at least 24 hours
Reith, William [USA] reith_will...@bah.com wrote on 07/20/2012
09:52:02 AM:
Would this matrix eat up memory making the rest of my program
slower? Each x needs to be multiplied by a matrix and the results
checked against a set of thresholds. Doing them one at a time takes
at least 24 hours
Thanks the aggregate() command is what I was looking for.
Chris
On Thu, Jul 19, 2012 at 9:03 PM, David L Carlson dcarl...@tamu.edu wrote:
dtf - read.table(text=y A B C
+ 0 11 2
+ 0 12 1
+ 1 11 2
+ 0 11 2
+ 1 11 2
+ 1 12
d$date - factor(d$date)
Best,
Michael
On Fri, Jul 20, 2012 at 9:44 AM, Yolande Tra yolande@gmail.com wrote:
Hello,
I would like to convert date as a factor to represent time in a repeated
measure situation in the following code. How would I do that?
d -
Dear R help list,
I have done a lot of searching but have not been able to find an answer to
my problem. I apologize in advance if this has been asked before.
I am applying a mixed model to my data using lmer. I will use sample data
to illustrate my question:
library(lme4)
library(arm)
Hi,
I am working with a SpatialPolygonsDataFrame of many islands. There are a
lot of polygons (islands) composing my SpatialPolygonsDataFrame.
I want to extract the elevation of each island.
I need to separate the different polygons (like dissolve function in
arcgis), to have the elevation of
Hi,
Just a doubt regarding the dataset.
Suppose, I include two more patients F and G with different missing values as
in this new dataset and run the code.
dat1-read.table(text=
Patient Cycle V1 V2 V3 V4 V5
A 1 0.4 0.1 0.5 1.5 NA
A 2 0.3 0.2 0.5 1.6 NA
A 3 0.3 NA 0.6 1.7
Hi,
Check this link.
http://stackoverflow.com/questions/5904513/tukeyhsd-after-within-factors-anova
A.K.
- Original Message -
From: Michael Eisenring eimic...@ethz.ch
To: r-help@r-project.org
Cc:
Sent: Friday, July 20, 2012 6:15 AM
Subject: [R] TukeyHSD not working
Dear r-help
Contact me with your full names, phone numbers and residential address if you
own this email r-h...@stat.math.ethz.ch.
I have a proposal for you
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Hi,
Try this:
dat1-data.frame(x,y,z)
xtabs(z~x+y,aggregate(z~x+y,dat1,mean))
y
x 10 20 30
1 100 0 0
2 0 200 0
3 0 0 300
table(dat1$z,dat1$y)
10 20 30
100 2 0 0
200 0 3 0
300 0 0 3
or,
table(dat1$x,dat1$z)
100 200 300
1 2 0 0
HI,
I don't have much experience with laercio package. If it works for you, it is
good.
A.K.
- Original Message -
From: Michael Eisenring eimic...@ethz.ch
To: arun smartpink...@yahoo.com
Cc:
Sent: Friday, July 20, 2012 8:38 AM
Subject: Re: [R] TukeyHSD not working
Hi
Thanks for
Yes,
that's this.
Thank you very much.
2012/7/20 Peter Ehlers [via R] ml-node+s789695n463721...@n4.nabble.com
On 2012-07-20 04:05, carla moreira wrote:
Hi,
I would like to evaluate a function, with 3 arguments, for instance,
myfunc-function(a,b,c) { sqrt(a)-exp(b)+4*c
Hi Rich,
I don't have a cenbox() function that I can find, but your solution will
(probably? hopefully?) be along the lines of:
foo - boxplot( y ~ x, data=sdf, plot=FALSE)
foo$names - ifelse(foo$names, Label for TRUE, Label for FALSE)
bxp(foo)
where sdf is a dataframe containing your data
Thanks,
Y
On Fri, Jul 20, 2012 at 11:47 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
d$date - factor(d$date)
Best,
Michael
On Fri, Jul 20, 2012 at 9:44 AM, Yolande Tra yolande@gmail.com
wrote:
Hello,
I would like to convert date as a factor to represent time in a
On Fri, 20 Jul 2012, MacQueen, Don wrote:
I don't have a cenbox() function that I can find, but your solution will
(probably? hopefully?) be along the lines of:
Don,
Well, I must have mis-typed that although I'm sure I read about it in the
NADA.pdf or Dennis' book. I'll look again. I don't
Thanks,
Y
On Fri, Jul 20, 2012 at 1:02 PM, arun smartpink...@yahoo.com wrote:
Hi,
Try this:
dat1-read.table(text=
iddateab cy
1 1 8/6/2008Red15B 22
2 1 8/6/2008 Green 15B 22
,sep=,header=TRUE)
Hi. Is there any package available to calculate the systematic root mean
square error (RMSEs) and unsystematic RMSE (RMSEu)? I could not find such a
package. I am too lazy to calculate them step by step.
The calculation could be found in equation 3 and 4 in the Appendix A of the
following link:
Hi,
Try this:
dat1-read.table(text=
id date a b c y
1 1 8/6/2008 Red 15 B 22
2 1 8/6/2008 Green 15 B 22
,sep=,header=TRUE)
dat1$date-with(dat1,as.factor(date))
dat1
id date a b c y
1 1 8/6/2008 Red 15 B
Hi
Here is some code to illustrate how the correlations are calculated.
data - c(word1, word1 word2,word1 word2 word3,word1 word2 word3
word4,word1 word2 word3 word4 word5)
frame - data.frame(data)
frame
data
1 word1
2
hi ,
last time I did not mention the specific version of linux and it was problem
with only syntax but with correction I got new error.
I found that some people has faced this problem before and some of them
have mentioned that it may be because of dependencies issue and
On 2012-07-20 09:41, Rich Shepard wrote:
On Fri, 20 Jul 2012, MacQueen, Don wrote:
I don't have a cenbox() function that I can find, but your solution will
(probably? hopefully?) be along the lines of:
Don,
Well, I must have mis-typed that although I'm sure I read about it in the
On Fri, Jul 20, 2012 at 4:51 PM, Celine bellard.cel...@gmail.com wrote:
Hi,
I am working with a SpatialPolygonsDataFrame of many islands. There are a
lot of polygons (islands) composing my SpatialPolygonsDataFrame.
I want to extract the elevation of each island.
Is each island a row in
Hello,
Yes, Carol is comparing what can't be compared. Your code should do it,
I hope.
Rui Barradas
Em 20-07-2012 15:12, David L Carlson escreveu:
Rui's example included z-score data (drawn from rnorm). You converted your
data to z-scores so you need to compare your results to the z-scores
That is faster than what I was doing and reducing 15% of my iterations it
still very helpful.
Next question.
I need to multiply each row x[i,] of the matrix x by another matrix A.
Specifically
for(i in 1:n)
{
If (x[i,]%*%A[,1].5 || x[i,]%*%A[,2]42 || x[i,]%*%A[,3]150)
{
x-x[-i,]
n-n-1
}. #In
This works to multiply the ith row of a by the ith value of b.
It might be what you can use
a - matrix(1:30, 6, 5)
b - 1:6
a
a*b
To simplify your code, I think you can do this---one multiplication
xA - x %*% A
Now you can do the tests on xA and not have any matrix multiplications
inside the
As a follow up is there any way to also get the count for each combination?
For example
y A B C
0 11 2
0 12 1
1 11 2
0 11 2
1 11 2
1 12 1
0 12 2
Should become:
y A B C count
2 1 1 24
1 1 2
Hi Michael, S Ellison,
I do not actually understand what you want to achieve with the M
estimates of rlm in MASS, but why you do not give a try of lmrob in
'robustbase'. Please have a llok in the references (?lmrob) about the
advantages of MM estimators over the M estimators.
Best regards,
Hello,
Still with aggregate, use length() not sum().
dtagroup - aggregate(y ~ A + B + C, data=dtf, sum)
dtalength - aggregate(y ~ A + B + C, data=dtf, length)
# Now merge the two
names(dtalength)[4] - count
mm - merge(dtagroup, dtalength)
# And make it pretty
mm - mm[, c(y, A, B, C, count)]
On 2012-07-20 12:09, Rui Barradas wrote:
Hello,
Still with aggregate, use length() not sum().
dtagroup - aggregate(y ~ A + B + C, data=dtf, sum)
dtalength - aggregate(y ~ A + B + C, data=dtf, length)
# Now merge the two
names(dtalength)[4] - count
mm - merge(dtagroup, dtalength)
# And make
On Fri, 20 Jul 2012, Peter Ehlers wrote:
Well, You didn't say (in your original request) that you were using
the NADA package. The function is cenboxplot() and it's just a
wrapper for boxplot() and hence passes arguments to boxplot().
Thus the solution to your problem is just to add the
Dear Carol,
-Original Message-
From: carol white [mailto:wht_...@yahoo.com]
Sent: July-20-12 2:45 PM
To: John Fox
Subject: Re: inverse normal transformation
Thanks John for your quick reply.
The purpose of applying inverse normal transformation is to reduce the
impact of
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