Thank you Bill for this additional solution.
Andrija
On Tue, Mar 15, 2011 at 12:16 AM, wrote:
> It is possible to do it with numeric comparisons, as well, but to make life
> comfortable you need to turn off the warning system temporarily.
>
> df <- data.frame(q1 = c(0,0,33.33,"check"),
>
Hi all:
I have a question on sample size calculation of 2 groups of data. If 2
groups of data are all normal distribution, then I can use the function
"n.indep.t.test.eq" from samplesize package.But if 2 groups of data are all
skewed distribution, but not normal distribution,how can I calculate the
Hello Mr. Allan and Mr. David!
Thanks very much for your time and lessons.
Works pretty well, Mr. Allan, specially after Mr. David explanation (and the 2
days-bug correction).
Just a curious behavior of data tranformation: the R-transformation date was
2min and 20s late compared with excel 20
Hi Laura,
?is.vector
> is.vector(1:10)
[1] TRUE
> is.vector(matrix(1:10))
[1] FALSE
Cheers,
Josh
On Mon, Mar 14, 2011 at 9:58 PM, Laura Smith wrote:
> Hello:
>
> Here are some basic class items:
>
>> x <- 1:10
>> class(x)
> [1] "integer"
>> x.mat <- matrix(1:6,nrow=2)
>> class(x.mat)
> [1] "
Hello:
Here are some basic class items:
> x <- 1:10
> class(x)
[1] "integer"
> x.mat <- matrix(1:6,nrow=2)
> class(x.mat)
[1] "matrix"
> class(x>3)
[1] "logical"
> test <- function() { plot(1:10) }
> class(test)
[1] "function"
>
Is there something that says "vector", please? Or does it go to nu
I see what you are saying. The application I am working with is making
convenient single-call (user) interfaces for R functions, and I don't
want to write a wrapper for every possible combination (like Gene
suggested below). If we don't consider the ... arguments for a second,
and only consider fir
On Mon, Mar 14, 2011 at 9:06 PM, zhenjiang xu wrote:
> Hi,
>
> For example, the data.frame like:
>
> origdata.long <- read.table(header=T, con <- textConnection('
> subject sex condition measurement
> 1 M control 7.9
> 1 M first 12.3
> 1 M second
I just fix it afterward:
dta <- dta[,c("col1","col2","col3")]
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. wi
On Mar 14, 2011, at 10:24 PM, Henrique Dallazuanna wrote:
Try this:
reshape(origdata.long, direction = 'wide', timevar = 'condition',
idvar =
c('subject', 'sex'))
If you first set the order in the desired sequence:
origdata.long$condition <- factor(origdata.long$condition,
levels=c('fi
Try this:
reshape(origdata.long, direction = 'wide', timevar = 'condition', idvar =
c('subject', 'sex'))
On Mon, Mar 14, 2011 at 10:06 PM, zhenjiang xu wrote:
> Hi,
>
> For example, the data.frame like:
>
> origdata.long <- read.table(header=T, con <- textConnection('
> subject sex condition me
Hi:
This is straightforward with the reshape package:
library(reshape)
origdata.long$condition <- factor(origdata.long$condition,
levels = c('first', 'second', 'control'))
cast(origdata.long, subject + sex ~ condition)
Using measurement as value column. Use the value argument to cast to
ove
Hi,
It is indeed documented but I'd misinterpreted it, my mistake. I made
the wrong assumption that switch would have a factor method which
would use the string levels, but that just isn't the case.
Thanks for the clarification anyway,
baptiste
On 14 March 2011 22:22, Allan Engelhardt wrote:
>
On 11-03-14 8:12 PM, Gene Leynes wrote:
Yes, I understand. Normally I use Eclipse, which does what I want for
"save as..."
The bigger issue is that R can't tell the location of an open script,
which makes it harder to create new versions of existing work
But it can. If you open a script
Hi,
For example, the data.frame like:
origdata.long <- read.table(header=T, con <- textConnection('
subject sex condition measurement
1 M control 7.9
1 M first12.3
1 Msecond10.7
2 F control 6.3
2 F first
My preferred cycle with RGui is:
1) Open RGui
2) Save (empty) workspace in the new working directory
3) Quit RGui
4) Double-click on the saved workspace. This opens RGui and sets the working
directory.
5) Open Notepad++ and edit as desired
6) Select code to run
7) Press F8
I have also been playi
Yes, I understand. Normally I use Eclipse, which does what I want for "save
as..."
The bigger issue is that R can't tell the location of an open script, which
makes it harder to create new versions of existing work
Say you have some great analysis going in "Research 2011-01-01" with a
folder
Thank you, but I need only the area under the llm colored. What if there
were two lm lines? Is it possible to color only between them. I have
search and found shaded polygons, but not shades that incorporate lm lines.
Here is the updated data example. The color on this follows both the lm and
On 11-03-14 5:03 PM, Gene Leynes wrote:
As much as I love R, there are still the occasional shortcomings.
I would love to find a solution to the "save as..." problem.
Steps to reproduce the problem:
1. Open any version of he R GUI in Windows
2. Choose "File> Open" from the menu
3.
First off, pardon the simple question, but is the XXX.xls file in your
current working directory? The error message might suggest that R (or
more precisely, the perl script) never saw the input file.
Have you been able to load any other Excel file into R?
My only other guess is that for som
Hi:
Try this:
co <- coef(llm)
bord <- pmin(y, co[1] + co[2] * x)
plot(x,y)
polygon(c(x, x[length(x)]), c(bord, bord[1]), col = 'red')
abline(co)
HTH,
Dennis
On Mon, Mar 14, 2011 at 4:02 PM, Marlin Keith Cox wrote:
> Hi,
> I would like to add a color under a lm line and not the plotted line. I
It is possible to do it with numeric comparisons, as well, but to make life
comfortable you need to turn off the warning system temporarily.
df <- data.frame(q1 = c(0,0,33.33,"check"),
q2 = c(0,33.33,"check",9.156),
q3 = c("check","check",25,100),
Hi,
I would like to add a color under a lm line and not the plotted line. Is
this possible? In the example, I do not want the area under the curve red,
but rather under the llm line.
x=seq(0,5,len=100)
y=-(x-5)^2
llm<-lm(y~x)
plot(x,y)
polygon(c(x,x[length(x)]), c(y, y[1]), col='red')
abline(ll
The nnclust package will compute the minimum spanning tree, from which
you can extract hierarchical single-linkage clustering.
For N randomly-ordered observations it uses only NlogN memory, and
takes N^2 time in high dimensions (30 is high) but only NlogN in low
dimensions.
-thomas
On Tue,
Scott, thanks for the suggestion. I have already filtered genes from more than
3. Probably I should filter more. I will take a look at genefilter package.
John
From: "Ochsner, Scott A"
Sent: Mon, March 14, 2011 2:19:57 PM
Subject: RE: [R] hclust() memor
This might be off base, but would using --args help? It means "ignore
everything else, Mr. R executable". However you can still parse it within
the R environment.
I use something like this in my startup file (.site file):
if("--args" %in% commandArgs()){
i=grep("--args", commandArgs())
John,
First, why are you trying to cluster so many rows? Presumably, if this is a
gene expression array dataset, most of the array features are not going to
change across treatments/conditions and will be relatively uninformative. Try
using a filter which does not use treatment/condition info
I'm not 100% sure of your problem...
It seems that you may want to consider these functions
rapply
mapply
Also, making a separate wrapper function for what you want to do. E.g.:
ModelPrinterFun = function(dat){
model = glm(dist~speed, data=dat)
print(coef(model), digits=3)
}
Model
David,
thank you very much. I changed little bit my code and now it works. The
magic word was stringsASfactor=FALSE and i didn't realize at the first
time.
Andrija
On Mon, Mar 14, 2011 at 9:28 PM, David Winsemius wrote:
>
> On Mar 14, 2011, at 3:51 PM, andrija djurovic wrote:
>
> I would like
As much as I love R, there are still the occasional shortcomings.
I would love to find a solution to the "save as..." problem.
Steps to reproduce the problem:
1. Open any version of he R GUI in Windows
2. Choose "File > Open" from the menu
3. Open a script that is in a different directo
Hi, I have a microarray dataset of dimension 25000x30 and try to clustering
using hclust(). But the clustering on the rows failed due to the size:
> y<-hclust(dist(data),method='average')
Error: cannot allocate vector of size 1.9 Gb
I tried to increase the memory using memory.limit(size=3000), s
On Mon, Mar 14, 2011 at 1:06 PM, Pavan G wrote:
> Hello All,
> I have a histogram with values above and below 0. I would like to color the
> +ve bars green and -ve bars red. I am plotting data using:
>
> hist(a[,2],breaks=100,main="W3",xlab="Movement towards site (A)")
>
> Can someone please comme
Hi,
I have a bunch of R scripts which have the hash bang !/usr/bin/env Rscript
and I typically run these scripts by passing in some parameters like this:
./nameOfRScript arg1 arg2 ...
I know Rscript has the option to run in --vanilla. Where exactly do I
insert the --vanilla option? When I do s
On Tue, Mar 15, 2011 at 9:05 AM, Jeroen Ooms wrote:
> Hmmm I was hoping there would be a more natural way to do it. For example,
> if you actually try to call the first function with all arguments:
>
> lm(formula=dist~speed, digits=3, data=cars)
>
> R will match whatever it can, and give you a war
You mean like cumsum()?
> a<-1:10
> b<-cumsum(a)
> b
[1] 1 3 6 10 15 21 28 36 45 55
-Nick Larson
--
View this message in context:
http://r.789695.n4.nabble.com/code-for-permutative-operation-tp3354839p3354849.html
Sent from the R help mailing list archive at Nabble.com.
__
Hi R Users,
I commonly import multiple .csv files and then write loops to work with
those files like this:
>setwd('C://Nathaniel/R/allfiles')
>files<-list.files()
>allfiles<-lapply(t1,read.csv)
>for (i in 1:47) {
> t1<-allfiles[[i]]
etc. etc.
I have written a script that I would like to loo
On Mar 14, 2011, at 4:16 PM, Sascha Vieweg wrote:
On 11-03-14 21:09, David Winsemius wrote:
On Mar 14, 2011, at 3:52 PM, Sascha Vieweg wrote:
Hello, I need some form of a "permutative" operation on a numeric
vector x
x = (v1, v2, v3, ..., vN)
that produces
x.r = (v1, v1+2, v1+v2+v3, ... v
Thanks Martin!
From: Martin Morgan
Cc: r-help@r-project.org
Sent: Mon, March 14, 2011 12:33:40 PM
Subject: Re: [R] ideas on sorting
On 03/14/2011 10:32 AM, array chip wrote:
> Hi, I have a character vector as below:
>
> a<-c('10','3R','4','4R','5','5R','6','6
On Mar 14, 2011, at 16:32 , Allan Engelhardt wrote:
> You could try
>
> scan(what=character(0), sep=",", file=textConnection("first,second,third"))
>
>
> but better to put the strings in a file (say, strings.txt), one per line, and
> read it using
>
> scan("strings.txt", what=character(0), s
On Mar 14, 2011, at 3:51 PM, andrija djurovic wrote:
I would like to hide cells with values less the 10%, so "." or just
"" doesn't make me any difference. Also I used apply combined with
as.character:
apply(df, 2, function(x) ifelse(as.character(x) < 10,".",x))
This is, probably not a goo
On 11-03-14 21:09, David Winsemius wrote:
On Mar 14, 2011, at 3:52 PM, Sascha Vieweg wrote:
Hello, I need some form of a "permutative" operation on a numeric vector x
x = (v1, v2, v3, ..., vN)
that produces
x.r = (v1, v1+2, v1+v2+v3, ... v1+v2+...+vN)
If the operation is sum() I can run
x
Hello All,
I have a histogram with values above and below 0. I would like to color the
+ve bars green and -ve bars red. I am plotting data using:
hist(a[,2],breaks=100,main="W3",xlab="Movement towards site (A)")
Can someone please comment on how it can be done?
Thanks!
[[alternative HTML
On Mar 14, 2011, at 3:52 PM, Sascha Vieweg wrote:
Hello, I need some form of a "permutative" operation on a numeric
vector x
x = (v1, v2, v3, ..., vN)
that produces
x.r = (v1, v1+2, v1+v2+v3, ... v1+v2+...+vN)
If the operation is sum() I can run
x <- 5:8
m <- matrix(rep(x, length(x)), nc
Hmmm I was hoping there would be a more natural way to do it. For example,
if you actually try to call the first function with all arguments:
lm(formula=dist~speed, digits=3, data=cars)
R will match whatever it can, and give you a warning with the names of
remaining unmatched arguments. The only
Dear Duncan,
Rich Calaway helped me narrowing down the error. The problem is that I can't
execute the line
cl<- makeCluster(mpi.universe.size(), type ="MPI")
when I am connected via VPN. Rich finally suggested to post this on R-SIG-HPC,
which I did a couple of minutes ago.
Cheers,
Marius
On
On Mar 14, 2011, at 17:57 , Giles Crane wrote:
> Greetings,
>
> The aov() function may mislabel
> the random effects as in the example below:
> Has anybody else noticed this?
What's "mislabeled" about it??? Looks like you nave an unbalanced design (in
which case, aov() may be the wrong tool.)
Hello, I need some form of a "permutative" operation on a numeric
vector x
x = (v1, v2, v3, ..., vN)
that produces
x.r = (v1, v1+2, v1+v2+v3, ... v1+v2+...+vN)
If the operation is sum() I can run
x <- 5:8
m <- matrix(rep(x, length(x)), ncol=length(x))
(x.r <- rowsum(m * upper.tri(m, diag=TRU
I would like to hide cells with values less the 10%, so "." or just ""
doesn't make me any difference. Also I used apply combined with
as.character:
apply(df, 2, function(x) ifelse(as.character(x) < 10,".",x))
This is, probably not a good solution, but it works except that I lose row
names and
Thank you Ista. It worked
On 14 March 2011 14:36, Ista Zahn wrote:
> Hi Christain,
> You do not give us a reproducible example, nor do yo tell us how you
> are creating the table. The following works fine for me:
>
> dat <- data.frame(x1=1, x2=2, x3=3, x4=4)
> names(dat) <- c("$\\sum_{i}\\sum_
On 03/14/2011 10:32 AM, array chip wrote:
Hi, I have a character vector as below:
a<-c('10','3R','4','4R','5','5R','6','6R','7','8','9','7R','1','10R','11'
,'11R','12','12R','13','13R','14','14R','15','15R','1R','2','2R','3','8R'
,'9R')
Is there a clever way to sort this easily to return a vect
Hi Rex,
You're right... I missed the proper route of thought :) I should read in a
set of data that I can handle memory wise, calculate the covariance matrix
and add it to the total... Sorry I assessed your suggestion the wrong way.
Thanks a million,
Tsjerk
On Mar 14, 2011 7:58 PM, wrote:
Ts
On 3/13/2011 6:46 PM, David Winsemius wrote:
On Mar 13, 2011, at 8:51 PM, Mark Linderman wrote:
David, thank you for your quick reply. I spent a few minutes getting your
command to work with some sparse synthetic data, and then spent several
hours trying to figure out why my data didn't work (
Thank you Bill.
John
From: William Dunlap
Sent: Mon, March 14, 2011 10:47:39 AM
Subject: RE: [R] ideas on sorting
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of array chip
> Sent: Monda
On Mar 14, 2011, at 2:52 PM, andrija djurovic wrote:
Hi R users,
I have following data frame
df<-data.frame(q1=c(0,0,33.33,"check"),q2=c(0,33.33,"check",9.156),
q3=c("check","check",25,100),q4=c(7.123,35,100,"check"))
and i would like to replace every element that is less then 10
with . (d
That's a smart way to do this. Thanks!
John
From: Berend Hasselman
To: r-help@r-project.org
Sent: Mon, March 14, 2011 10:49:37 AM
Subject: Re: [R] ideas on sorting
array chip wrote:
>
> Hi, I have a character vector as below:
>
> a<-c('10','3R','4','4R','5
Hi Jim,
this may be barking up the wrong tree, but
create <-function(...) paste(substitute(list(...)))[-1]
createl <-function(...) {
tmp<-list(...)
names(tmp)<-create(...)
tmp
}
#eg
a<-1:4
b<-letters[2:6]
createl(a,b)
works. But I can't imagine that a named list is the one and only useful
d
Dear Andrija,
You could convert the factors to numeric class in order to test, get an index
of cells and then replace those. I wonder though if it wouldn't be easier to
do this at some step *before* the numbers are combined with strings?
At any rate, take a look at ?which ?factor ?as.numeric ?
On Mon, 14 Mar 2011, Jen wrote:
Hi Bill,
Thanks for your response and I'm sorry -- that was a misleading example of
what I was trying to show. This one should illustrate the point:
require(AER)
data_in = c(0,6,12,18,24,30,36,42,48,54,60,66,72,78)
data_in2 = data_in^2
data_in3 = data_in^3
data_o
On Mar 14, 2011, at 12:25 PM, Andreas Emanuelsson wrote:
Hi everyone, I have problems with a double for loop and the program
response: "Error in 1:mass[j] : NA/NaN argument"
I'm trying to create a very simple script to generate a vector full
of objects with the value [fuel] and in the amou
Tsjerk,
It seems to me that memory and not time is your big efficiency problem, and
I've showed you how to avoid storing your entire input.
If you want to avoid doing each multiplication twice, you can replace the
"outer" with a function that computes each product only once and accumulate
sums o
Hi R users,
I have following data frame
df<-data.frame(q1=c(0,0,33.33,"check"),q2=c(0,33.33,"check",9.156),
q3=c("check","check",25,100),q4=c(7.123,35,100,"check"))
and i would like to replace every element that is less then 10 with . (dot)
in order to obtain this:
q1q2q3q4
1
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Andreas Emanuelsson
> Sent: Monday, March 14, 2011 9:25 AM
> To: r-help@r-project.org
> Subject: [R] (no subject)
>
> Hi everyone, I have problems with a double for loop and the p
Hello,
I know that Mclust does the fitting on its own but I am trying to implement
an optimization with the aim to generate a the mixture gaussian with the
combine moments as closed as possible to the moment of my return
distribution.
The objective is to Min Abs((Mean Ret - MeanFit)/Mean Fit) +
Hi Bill,
Thanks for your response and I'm sorry -- that was a misleading example of
what I was trying to show. This one should illustrate the point:
require(AER)
data_in = c(0,6,12,18,24,30,36,42,48,54,60,66,72,78)
data_in2 = data_in^2
data_in3 = data_in^3
data_out =
c(139487.00,13.00,62500.00
Hi everyone, I have problems with a double for loop and the program response:
"Error in 1:mass[j] : NA/NaN argument"
I'm trying to create a very simple script to generate a vector full of objects
with the value [fuel] and in the amount [mass] for a hist plot.
H=1
for (j in 1:63)
Hi Jim
create <-function(...) paste(substitute(list(...)))[-1]
create(a,b,c)
should work.
hth.
Am 14.03.2011 10:29, schrieb Maas James Dr (MED):
> Is there a way to convince R to create a character vector without using the
> quotes?
>
> This works
>
> ex1 <- c("first","second")
>
> but w
I think you meant to use Depth as the grouping factor rather than s_name:
> ddply(my_data, .(s_name), function(x){
+ x$Im_looking <- x$Depth + as.numeric(factor(x$s_name)) / 100
+ x
+ })
Depth s_name index Im_looking
1 3852 Site_1 1443852.01
2 3852 Site_1 1443852
On Mon, 14 Mar 2011, Brian Smith wrote:
Hi,
I was trying to install the package Rmpi on a hpc cluster running SGE. The
Which is what? (If Sun Grid Engine, not relevant and it is your
unstated OS that we needed to know. Not just 'Linux', but the
precise distro.))
Please do note the posti
On Mon, Mar 14, 2011 at 9:59 AM, ONKELINX, Thierry
wrote:
> Something like this?
>
> my_data=read.table("clipboard", header=TRUE)
> my_data$s_name <- factor(my_data$s_name)
> library(plyr)
> ddply(my_data, .(s_name), function(x){
> x$Im_looking <- x$Depth + as.numeric(x$s_name) / 100
>
array chip wrote:
>
> Hi, I have a character vector as below:
>
> a<-c('10','3R','4','4R','5','5R','6','6R','7','8','9','7R','1','10R','11'
> ,'11R','12','12R','13','13R','14','14R','15','15R','1R','2','2R','3','8R'
> ,'9R')
>
> Is there a clever way to sort this easily to return a vector of in
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of array chip
> Sent: Monday, March 14, 2011 10:33 AM
> To: r-help@r-project.org
> Subject: [R] ideas on sorting
>
> Hi, I have a character vector as below:
>
> a<-c('10','3R','4',
Xiaoqi,
You need to specify the sizes. There are other search algorithms that
auotmatically pick the size (such as genetic algorithms), but I don't have
those in the package yet.
Another approach is to use univariate filtering (see the sbf function in caret).
Max
On Mar 13, 2011, at 8:49 PM,
Yes, shell("start") works. R does not wait for the command prompt to be closed,
but this can be worked around by running shell("start /WAIT") so it looks like
this should give a solution our guy's can use.
Many thanks,
Ralph
--- On Fri, 11/3/11, Gabor Grothendieck wrote:
From: Gabor Grothen
Hi, I have a character vector as below:
a<-c('10','3R','4','4R','5','5R','6','6R','7','8','9','7R','1','10R','11'
,'11R','12','12R','13','13R','14','14R','15','15R','1R','2','2R','3','8R'
,'9R')
Is there a clever way to sort this easily to return a vector of index that
would
produce a vector as
Hi:
Just a few additional comments to Spencer's.
1. There is an R-SIG-Robust group that you may wish to post your
question to if you have not already done so. There you **should** find
experts to help you, of which I'm also not one.
2. The situation regarding the effectiveness of robust techniqu
Greetings,
The aov() function may mislabel
the random effects as in the example below:
Has anybody else noticed this?
Cordially,
Giles Crane, MPH, ASA, NJPHA
gilescr...@verizon.net
> m2
Call:
aov(formula = y ~ ap + pe + Error(ju), data = d)
Grand Mean: 77.50667
Stratum 1: ju
Terms:
Tjerk,
This is just a pseudo code outline of what you need to do:
M = matrix(0, number of variables, number of variables)
V = rep(0, number of variables)
N = 0
While (more observations to read) {
X <- next observation
V <- V + X
M <- M + outer(X,X)
N <- N+1
}
Compute covariance matrix
There are better alternatives for big data than to revert to C.
http://code.google.com/p/pymc/
http://github.com/armstrtw/CppBugs (still alpha)
-Whit
On Mon, Mar 14, 2011 at 11:06 AM, nblarson wrote:
> Has anybody had issues running MCMC (either BUGS or JAGS) on data sets of
> this magnitude (
Hi,
I was trying to install the package Rmpi on a hpc cluster running SGE. The
command, and the sessionInfo() is as follows:
===
> install.packages("Rmpi",dependencies=TRUE)
also installing the dependency rsprng
trying URL '
http://www.ibiblio.org/pub/la
The reason this doesn't work is because R thinks that in the command
as.character(c(first,second)) that first and second are variables that
exist within the working environment. Since they don't (I assume), R
doesn't know what to do with the command. Using the quotes indicates to R
that you're s
Thanks, that is what I was trying to do.
From: Peter Ehlers [via R]
[mailto:ml-node+3350009-943873321-216...@n4.nabble.com]
Sent: Saturday, March 12, 2011 4:27 AM
To: Anthony Seeman
Subject: Re: Kendall Theil line as fit?
On 2011-03-11 17:10, jonbfish wrote:
> Th
Has anybody had issues running MCMC (either BUGS or JAGS) on data sets of
this magnitude (ie 30k x 20-30). I've been trying to run a hierarchical
random effects model on expression data but R completely stalls out on jobs
run on 32bit R on our server (doesn't respond...then eventually crashes out
On 03/13/2011 09:01 PM, Laura Smith wrote:
Hi
I'm trying to learn about S4 methods, classes, etc.
Is it better to use initialize or use a construction function, please?
Hi Laura -- partly a matter of taste; 'initialize' actually establishes
quite a complicated contract, and I find it easier
On Mar 14, 2011, at 6:36 AM, Allan Engelhardt wrote:
On 14/03/11 02:00, Raoni Rosa Rodrigues wrote:
Hello R Help,
I'm working in a project with a software that register date and
time data in serial time format. This format is used by excel, for
exemple. In this format, 40597.3911423958
I'm not an expert on robust modeling. However, as far as I know,
most robust regression procedures are based on heuristics, justified by
claims that "it seems to work" rather than reference to assumptions
about a probability model that makes the procedures "optimal". There
may be except
Hi Tierry,
Thanks for your answer, that is very close to I'm looking, but
there are this difference:
whit your code I get this:
Depth s_name indice Im_looking
3852Site_1 144 3852.01
3852Site_1 144 3852.01
3852Site_1 144 3852.01
3852site_A 145 3852.02
3852
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Jen
> Sent: Monday, March 14, 2011 3:27 AM
> To: r-help@r-project.org
> Subject: [R] different regression coeffs with different starting point
>
> Hi all,
> I have a question ab
Dear Abhijit,
Maybe am using the wrong test. But all i need is calculate non-parametrically
the accuracy ratios of my samples.
Ideally for that kind of an output then the answer would be between 0 and 1.
notice also one of my variable is a non-numeric.
Thank
Taby
An idea not coupled wit
You could try
scan(what=character(0), sep=",", file=textConnection("first,second,third"))
but better to put the strings in a file (say, strings.txt), one per
line, and read it using
scan("strings.txt", what=character(0), sep="\n")
Even better is to understand what you are really trying to
You need to read up on the Wilcoxon signed-rank test and the output from
wilcox.test!!!
The confidence interval is of the difference of medians, which can
certainly be negative. In fact, your estimate is -33, and the confidence
interval is (-68, 0) which is reasonable.
The value of W is a posi
You need to read up on the Wilcoxon signed-rank test and the output from
wilcox.test!!!
The confidence interval is of the difference of medians, which can
certainly be negative. In fact, your estimate is -33, and the confidence
interval is (-68, 0) which is reasonable.
The value of W is a posi
Something like this?
my_data=read.table("clipboard", header=TRUE)
my_data$s_name <- factor(my_data$s_name)
library(plyr)
ddply(my_data, .(s_name), function(x){
x$Im_looking <- x$Depth + as.numeric(x$s_name) / 100
x
})
Best regards,
Thierry
---
my output is as follows:
wilcox.test(main_samp$SCORE~main_samp$GENDER, conf.int = TRUE)
Wilcoxon rank sum test with continuity correction
data: main_samp$SCORE by main_samp$GENDER
W = 2780.5, p-value = 0.04829
alternative hypothesis: true location shift is not equal to 0
95 percent con
Thanks for the help! All the methods above worked well and cleared up some
misunderstandings.
Thanks!
Haakon =)
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Jim,
I don't know if this is the most elegant solution :
Copy the following to your clipboard (some editors will require also to take
last paragraph mark)
first
second
third
> ## now go to R and use
> dat <- read.table("clipboard") # will give data.frame
> dat
V1
1 first
2 second
3 thi
Hello,
we want to plot a proportional symbol map with ggplot. Symbols' area should
have the same proportions as the scaled variable.
Hereby an example we found on
http://www.r-bloggers.com/bubble-chart-by-using-ggplot2/ . In this example
we see the proportions of the symbols' area are differen
Can i ask a question> Do I need a good math for developing a web crawler ?
( I want to develop a simple web crawler to do something )
--
View this message in context:
http://r.789695.n4.nabble.com/Developing-a-web-crawler-tp3332993p3353291.html
Sent from the R help mailing list archive at Nabble
Hi all,
I have a question about the optimisation methods used in nonlinear
regression. I have some data that I would like to fit a tobit regression
model to (see code below). It seems that the solution is very sensitive to
the initial condition that I give it - is there any option to use a
differen
Dear community,
I have done model selection between candidate Cox models, using AICc
calculated with penalized log likelihoods. Then model averaging was done to
obtain model averaged parameter estimates. Is there a way to plot survival
curve from the averaged model, by estimating baseline hazard a
Hi Everyone,
I'm a pretty useless r-er but have data that SPSS etc doesn't like. I've
managed to do GLMs for my data, but now need to fit a thin plate spline for
my data (arcsine.success~date.num:clutch.size)
If anyone has a bit of spare time and could come up with a bit of code I'd
be very gratef
Hi guys
I have a basic question about S4 stuff
Here is how I do it:
setGenericVerif <-
function(x,y){if(!isGeneric(x)){setGeneric(x,y)}else{}}
# from the genolini intro to S4
# here is the class
setClass("RiskBuckets",
representation(
... xxx ...
)
)
# here
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