__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
I am having trouble selecting rows of a dataframe that will be included
in a regression. I am trying to select those rows for which the variable
Meno equals PRE. I have used the code below:
difffitPre<-lm(data[,"diff"]~data[,"Age"]+data[,"Race"],data=data[data[,"Meno"]=="PRE",])
summary(difffitPre
Hi all R users,
I want to draw a 3D plot, where the Z-axis will represent the normal
densities each with zero mean but different volatilities, Y-axis will
represent the SD [volatilities), and X-axis will represent time at which
these SD are calculated.
Can anyone give me any clue? Your help will
Thanks for the guidance. With the help of your explanation, I was able to
find this reference that provides a good explanation of why the definition
may be different in the "no intercept" or "regression through origin" case.
For future readers of this list, the reference is:
Joseph G. Eisenhaue
which.max()
b
On Jan 17, 2007, at 11:20 PM, Feng Qiu wrote:
> Hi all:
> A short question:
> For example, a=[3,4,6,2,3], obviously the 3rd entry of
> the array has the maxium value, what I want is index of the maxium
> value: 3. is there a neat expression to get this index?
Thanks Brian, that advice may help speed up my regexp operations in the
future. The computer science advice offered by those of you who are more
expert is appreciated by we biologists who are primarily working more at
the level of bioinformatics. Mark
Mark W. Kimpel MD
(317) 490-5129 Work, &
which.max
> a <- sample(1:10,10)
> a
[1] 3 4 5 7 2 8 9 6 10 1
> which.max(a)
[1] 9
>
On 1/17/07, Feng Qiu <[EMAIL PROTECTED]> wrote:
>
> Hi all:
> A short question:
> For example, a=[3,4,6,2,3], obviously the 3rd entry of the array
> has the maxium value, what I want
?which.max
?which.is.min
On Wed, 17 Jan 2007, Feng Qiu wrote:
> Hi all:
> A short question:
> For example, a=[3,4,6,2,3], obviously the 3rd entry of the array has
> the maxium value, what I want is index of the maxium value: 3. is there a
> neat expression to get this index?
>
One thing to watch with experiments like this is that the locale will
matter. Character operations will be faster in a single-byte locale (as
used here) than in a variable-byte locale (and I suspect Seth and Marc
used UTF-8), and the relative speeds may alter. Also, the PCRE regexps
are often
Or
which.max(a)
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of talepanda
Sent: Wednesday, January 17, 2007 11:45 PM
To: Feng Qiu
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] how to get the index of entry with max value in an array?
In R lang
In R language, one solution is:
a<-c(3,4,6,2,3)
which(a==max(a))
On 1/18/07, Feng Qiu <[EMAIL PROTECTED]> wrote:
> Hi all:
> A short question:
> For example, a=[3,4,6,2,3], obviously the 3rd entry of the array
> has the maxium value, what I want is index of the maxium value: 3
Hi all:
A short question:
For example, a=[3,4,6,2,3], obviously the 3rd entry of the array has
the maxium value, what I want is index of the maxium value: 3. is there a neat
expression to get this index?
Thank you!
Best,
Feng
[[alternative HTML version deleted]]
This is documented on ?summary.lm.
It is not that 'intercept is zero' or 'zero intercept', it is that there
is no intercept term in the model.
On Wed, 17 Jan 2007, endeitz wrote:
>
> I am curious as to the "lm" calculation of R2 (multiple coefficient of
> determination, I assume) when intercept
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
What does the FASTA header look like. You are using 'gene' to access things
in the array and if (for example) 'gene' is a character vector of 10, then
for every element of vectors that you are using (I count about 4-5 that use
this index) then you are going to have at least 550 * 6000 * 5 * 10 mor
Thanks for 6 ways to skin this cat! I am just beginning to learn about
the power of regular expressions and appreciate the many examples of how
they can be used in this context. This knowledge will come in handy the
next time the number of characters is variable both before and after the
dot. On my
Hi,
I'm about the head out of the office next 48 hours, but the short
answer is to override the finalize() method in your subclass of
Object. This will be called when R garbage collects the object. From
?finalize.Object:
setConstructorS3("MyClass", function() {
extend(Object(), "MyClass")
Here is one way of doing it by 'padding' all the elements to the same
length:
> x <- "Col1 Col2 Col3 Col159 Col160
+ Row1 0 0 LD 0 VD
+ Row2 HD0 0 0 MD
+ Row3 0 HDHD 0 LD
+ Row4 LDHDHD 0 0
+ LastRow
On Wed, 2007-01-17 at 16:46 -0800, Seth Falcon wrote:
> "Kimpel, Mark William" <[EMAIL PROTECTED]> writes:
>
> > I have a very long vector of character strings of the format
> > "GO:0008104.ISS" and need to strip off the dot and anything that follows
> > it. There are always 10 characters before t
"Kimpel, Mark William" <[EMAIL PROTECTED]> writes:
> I have a very long vector of character strings of the format
> "GO:0008104.ISS" and need to strip off the dot and anything that follows
> it. There are always 10 characters before the dot. The actual characters
> and the number of them after the
___
substr is vectorised, so it should work fine without needing an explicit
loop.
--
Hong Ooi
Senior Research Analyst, IAG Limited
388 George St, Sydney NSW 2000
+61 (2) 9292 1566
-Original Message-
Fr
Try this:
> gsub("[.].*", "", "GO:0008104.ISS")
[1] "GO:0008104"
On 1/17/07, Kimpel, Mark William <[EMAIL PROTECTED]> wrote:
> I have a very long vector of character strings of the format
> "GO:0008104.ISS" and need to strip off the dot and anything that follows
> it. There are always 10 characte
I have a very long vector of character strings of the format
"GO:0008104.ISS" and need to strip off the dot and anything that follows
it. There are always 10 characters before the dot. The actual characters
and the number of them after the dot is variable.
So, I would like to return in the format
Has anyone figured out how to create a destructor in R.oo?
How I'd like to use it: I have an object which opens a connection thru RODBC
(held as a private member) It would be nice if the connection closes
automatically
(inside the destructor) when an object gets gc()'ed.
Thanks in advance.
Re
Hi,
Working further on this dataframe : my_data
Col1 Col2 Col3 ... Col 159 Col 160
Row 1 0 0 LD ... 0 VD
Row 2 HD0 0 0 MD
Row 3 0 HDHD 0 LD
Row 4 LDHDHD 0 0
......
Las
I am curious as to the "lm" calculation of R2 (multiple coefficient of
determination, I assume) when intercept is zero. I have 18 data points, two
independent variables:
First, a model with an intercept:
> mod0=lm(Div~Rain+Evap,data=test)
> summary(mod0)$r.squared
[1] 0.6257541
> cor(predict(mo
try using strwidth & strheight
x<-c(0,1)
plot(x,x,type='l')
dimensions<-matrix(c(strwidth(expression(theta),cex=5),strheight(expression(theta),
cex=5)),nrow=1)
symbols(0.5,0.5
,rectangle=dimensions,bg='white',fg='white',add=TRUE,inches=FALSE)
text(0.5,0.5,expression(theta),cex=5)
On 1/17/07,
Hi -
When I'm trying to read in a text file into a labeled character array,
the memory stamp/footprint of R will exceed 4 gigs or more. I've seen
this behavior on Mac OS X, Linux for AMD_64 and X86_64., and the R
versions are 2.4, 2.4 and 2.2, respectively. So, it would seem that
this is pla
(I overlooked the reply).
Thanks, Gabor. That is neat and easy! (and I should have been able to
see it on my own :-(
Best,
R.
On 1/8/07, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> The S4 is not essential. You could do it in S3 too:
>
> > f.a <- function(x) x+1
> > f.b <- function(x) x+2
>
Lauri Nikkinen wrote:
> Hi R-users,
>
> I'm quite new to R and trying to learn the basics. I have a following
> problem concerning the convertion of array object into data frame. I have
> made following data sets
>
> tmp1 <- rnorm(100)
> tmp2 <- gl(10,2,length=100)
> tmp3 <- as.data.frame(cbind(t
Hi R-users,
I'm quite new to R and trying to learn the basics. I have a following
problem concerning the convertion of array object into data frame. I have
made following data sets
tmp1 <- rnorm(100)
tmp2 <- gl(10,2,length=100)
tmp3 <- as.data.frame(cbind(tmp1,tmp2))
tmp3.sum <- tapply(tmp3$tmp1,
In the package genefilter, from www.bioconductor.org there is a function
to do this (genefinder, if I recall correctly)
best wishes
Robert
Charles C. Berry wrote:
> On Wed, 17 Jan 2007, Damion Colin Nero wrote:
>
>> I am trying to find a way to perform pairwise correlations against one
>> gen
On Wed, 17 Jan 2007, Albrecht Kauffmann wrote:
> Hi all,
>
> I'm faced with a problem applying the sp package: The projection argument in
>
> readShapePoly(Shapefile,proj4string="CRS class argument")
>
> e.g.: CRS("+proj=aea +lat_1=46 +lat_2=73 +lat_0=60 +lon_0=84 +x_0=0
> +y_0=0 +ellps=clrk66
On Wed, 17 Jan 2007, Damion Colin Nero wrote:
> I am trying to find a way to perform pairwise correlations against one
> gene in a matrix rather than computing every pairwise correlation. I am
> interested in how 1 transcription factor correlates to every gene in a
> matrix of 55 experiments (col
Yes, you refer to
Cushny, A. R. and Peebles, A. R. The action of optical isomers: II
hyoscines. The Journal of Physiology, 1905, 32: 501.510.
which was used by 'Student' to illustrate the paired t-test.
This is indeed a crossover design.
On Wed, 17 Jan 2007, Tom Backer Johnsen
Dear All (apologies if some of you have received this twice)
Thanks very much for the rapid reply Prof Ripley. I had been looking at this
anaysis for my colleague (Prof Behnke) and suggested that he contact the R
mailing list because I couldn't answer his question. I think some of the detail
got
I suggest using permutation on each predictor and see how much the
accuracy drops, no matter what modeling approach you used.
HTH,
weiwei
On 1/17/07, Rupendra Chulyadyo <[EMAIL PROTECTED]> wrote:
> Hello all,
>
> I want to assign relative score to the predictor variables on the basis of
> its i
On 1/17/2007 12:24 PM, Duncan Murdoch wrote:
> On 1/17/2007 10:42 AM, Matt Sakals wrote:
>> I am trying to plot a non-linear trend line along with my data. I am
>> sure the solution is simple; any help greatly appreciated.
>>
>> Recent and historic attempts:
>> fit = nls((sd~a+b*exp(-c/time)+d*g
On 1/17/2007 10:42 AM, Matt Sakals wrote:
> I am trying to plot a non-linear trend line along with my data. I am
> sure the solution is simple; any help greatly appreciated.
>
> Recent and historic attempts:
> fit = nls((sd~a+b*exp(-c/time)+d*geology), start=list(a=1, b=1, c=10,
> d=-1), model
> In your case, read.table behaves as documented.
> The ' - character is one of the standard quoting characters. Some (but
> very few) of the entrys contain single ' chars, so sometimes more than
> ten thousand lines are just treated as a single entry. Try using
> quote="" to disable quoting, as
The problem is somewhere in the file, probably with tab characters, as removing
sep="" from your call does the job.
> dfr<-read.table("Tchange_rates_crawled.dat",header=TRUE)
> str(dfr)
'data.frame': 122271 obs. of 5 variables:
[skipped]
> dfr<-
read.table("Tchange_rates_crawled.dat",header=TR
Folks:
I think this and several other recent posts on ranking predictors are nice
illustrations of a fundamental conundrum: Empirical models are fit as good
*predictors*; "meaningful" interpretation of separate parameters/components
of the predictors may well be difficult or impossible, especially
Frank McCown schrieb:
> I have been trying to read in a large data set using read.table, but
> I've only been able to grab the first 50,871 rows of the total 122,269 rows.
>
> > f <-
> read.table("http://www.cs.odu.edu/~fmccown/R/Tchange_rates_crawled.dat";,
> header=TRUE, nrows=123000, comment
Frank McCown wrote:
> I have been trying to read in a large data set using read.table, but
> I've only been able to grab the first 50,871 rows of the total 122,269 rows.
>
> > f <-
> read.table("http://www.cs.odu.edu/~fmccown/R/Tchange_rates_crawled.dat";,
> header=TRUE, nrows=123000, comment.c
Hi all,
I'm faced with a problem applying the sp package: The projection argument in
readShapePoly(Shapefile,proj4string="CRS class argument")
e.g.: CRS("+proj=aea +lat_1=46 +lat_2=73 +lat_0=60 +lon_0=84 +x_0=0
+y_0=0 +ellps=clrk66 +units=m +no_defs")
doesn't have any impact on the plotted obj
say your data.frame is called "df"
df[order(df$evidence),]
or
df[order(df$evidence, decreasing=T),] # if you want the other way
around.
b
On Jan 17, 2007, at 11:21 AM, Milton Cezar Ribeiro wrote:
> Hi there,
>
> How can I sort (order?) a data.frame using a dataframe field
> (evidence) a
Hi,
I tried to do a model simplification in a repeated anova with 3
factors (population,treatment,site). I tried to get rid of some
interactions. Obviously it's not possible to use the function "anova"
straight away to compare the different models. I got the suggestion
to use the function
Hi there,
How can I sort (order?) a data.frame using a dataframe field (evidence) as
classifyer? My data.frame looks like:
record model evidence
1 areatotha 6638.32581
2 areatotha_ca000 8111.01860
3 areatotha_ca000_Pais 1721.41828
4 areatotha_ca020
I am trying to find a way to perform pairwise correlations against one
gene in a matrix rather than computing every pairwise correlation. I am
interested in how 1 transcription factor correlates to every gene in a
matrix of 55 experiments (columns) by 23,000 genes (rows), performing
the correlatio
I do not know the precise language to describe the situation. But here
it is what I want to do.
I need to annotate a graph that contains two or more curves with labels
that contain math symbols. The label must go on top of the curve.
The problem is that when I annotate the plot, I can see the cu
I am trying to plot a non-linear trend line along with my data. I am
sure the solution is simple; any help greatly appreciated.
Recent and historic attempts:
fit = nls((sd~a+b*exp(-c/time)+d*geology), start=list(a=1, b=1, c=10,
d=-1), model=TRUE)
plot(time, sd, col=3, main = "Regression", sub=
I have been trying to read in a large data set using read.table, but
I've only been able to grab the first 50,871 rows of the total 122,269 rows.
> f <-
read.table("http://www.cs.odu.edu/~fmccown/R/Tchange_rates_crawled.dat";,
header=TRUE, nrows=123000, comment.char="", sep="\t")
> length(f$c
Dear All
Thanks very much for the rapid reply Prof Ripley. I had been looking at
this anaysis for my colleague (Prof Behnke) and suggested that he
contact the R mailing list because I couldn't answer his question. I
think some of the detail got lost in translation (he grew up with the
GLIM package
Hi,
I tried to do a model simplification in repeated anovas. obviously
it's not possible to use the function "anova" straight away. I got
the suggestion to do it with proj (see below). In my data, however, I
have in "block" not the third order interaction (n:p:k), but to
factors and their
On Wed, 17 Jan 2007, Behnke Jerzy wrote:
> Dear All,
> I wonder if anyone can advise me as to whether there is a consensus as
> to how the effect size should be calculated from GLIM models in R for
> any specified significant main effect or interaction.
I think there is consensus that effect size
[Unstated, but a near repeat of a posting earlier today with the same
subject, in response to unacknowledged help.]
The problem is not with unlist (as in your subject line) but your input to
strptime.
strptime() is intended to convert character strings, and
> as.character(acc.period[16])
[1] "2
Rupendra Chulyadyo wrote:
> Hello all,
>
> I want to assign relative score to the predictor variables on the basis of
> its influence on the dependent variable. But I could not find any standard
> statistical approach appropriate for this purpose.
> Please suggest the possible approaches.
>
> Tha
Dear R-users,
I use unlist of POSIX dates to extract the year, hour etc. With that I
can search for files in my database which are in the form
'mmddhh_synops.txt'
However, I get stucked during midnight where unlist just gives NA's.
The script is given below; the problem accurs at acc.period[
Thanks for quick responses.
My dataset consists of 213 independent variables (disease related costs) and
the depedent variable is the total medical + pharmacy cost of the member.
As you had suggested, I tried to use hier.part function in R, but the
current implementation does not seem to allow mo
Dear All,
I wonder if anyone can advise me as to whether there is a consensus as
to how the effect size should be calculated from GLIM models in R for
any specified significant main effect or interaction.
In investigating the causes of variation in infection in wild animals,
we have fitted 4-way G
2007/1/17, Patrick Burns <[EMAIL PROTECTED]>:
> A logical reason for the phenomenon is that
> matrices are stored down their columns. For
> example:
>
> > matrix(1:15,5)
> [,1] [,2] [,3]
> [1,]16 11
> [2,]27 12
> [3,]38 13
> [4,]49 14
> [5,]5 10
Martin Keller-Ressel wrote:
> Thanks to Bettina, Dimitris and Gavin for their help.
> All their solutions work nicely.
> For future reference, here are three ways to draw a percent sign in R
> plots:
>
> plot(0:10, 0:10, type = "n")
> text(5,7,expression(paste(alpha == 5, "%", sep = "")))
> text(
Tom Backer Johnsen wrote:
> I am having a hard time understanding how to perform a "repeated
> measures" type of ANOVA with R. When reading the document found here:
>
> http://cran.r-project.org/doc/contrib/Lemon-kickstart/kr_repms.html
>
> I find that there is a reference to a function make.rm
A logical reason for the phenomenon is that
matrices are stored down their columns. For
example:
> matrix(1:15,5)
[,1] [,2] [,3]
[1,]16 11
[2,]27 12
[3,]38 13
[4,]49 14
[5,]5 10 15
When an 'apply' across rows is done, it will be
the values cor
I saw that R language has a cluster package which has in built PAM, CLARA and
Kmeans (and many more) Clustering Algorithms.
But, I couldnot find DBSCAN, ROCK, BIRCH algorithms (which I feel are standard
ones). Aren't these implemented as well?
Bhanu Kalyan K
B.Tech Final Year, CSE
Tel: +91-
> We are dealing with a variable (BA) which indicates the overlap
> between small mammal home ranges. It varies between 0 and 1 and it
> can be interpreted as "the probability of two home ranges to
> overlap", therefore we would have modelled it with the binomial
> family, also supported by the di
On Wed, 17 Jan 2007, Hanneke Schuurmans wrote:
> Dear R-users,
>
> I use unlist of POSIX dates to extract the year, hour etc. With that I
> can search for files in my database which are in the form
> 'mmddhh_synops.txt'
> However, I get stucked during midnight where unlist just gives NA's.
> T
Dear all,
We are dealing with a variable (BA) which indicates the overlap between
small mammal home ranges. It varies between 0 and 1 and it can be
interpreted as "the probability of two home ranges to overlap",
therefore we would have modelled it with the binomial family, also
supported by the di
I am having a hard time understanding how to perform a "repeated
measures" type of ANOVA with R. When reading the document found here:
http://cran.r-project.org/doc/contrib/Lemon-kickstart/kr_repms.html
I find that there is a reference to a function make.rm () that is
supposed to rearrange a "
Dear R-users,
I use unlist of POSIX dates to extract the year, hour etc. With that I
can search for files in my database which are in the form
'mmddhh_synops.txt'
However, I get stucked during midnight where unlist just gives NA's.
The script is given below, the problem accurs at acc.period[
> 1. The first I used to do in SPSS and I would like to be able
> to do it in R as well.
> This is the hierarchical model I would like to use: a continuous
> variable explained by factor A(fixed) + factor B(random)
> nested in A + factor C (random) nested in factor B (which is nested
> in A).
Y
When reading the documentation for the "sleep" data set in R, the
impression is clear, this is an "independent groups" kind of design
(two groups of 10 subjects each). However, when browsing the original
article (referred to in the help file), my impression is quite clear,
this is really a "re
Thanks to Bettina, Dimitris and Gavin for their help.
All their solutions work nicely.
For future reference, here are three ways to draw a percent sign in R
plots:
plot(0:10, 0:10, type = "n")
text(5,7,expression(paste(alpha == 5, "%", sep = "")))
text(5, 5, expression(paste(alpha, " = 5%")))
te
On Wed, 2007-01-17 at 09:57 +, Martin Keller-Ressel wrote:
> Hello,
>
> I would like to annotate a graph with the expression 'alpha = 5%' (the
> alpha should be displayed as the greek letter).
> I tried
>
> > text(1,1,expression(alpha == 5%))
>
> which gives a syntax error.
> escaping the
On Wed, Jan 17, 2007 at 09:57:49AM -, Martin Keller-Ressel wrote:
> I would like to annotate a graph with the expression 'alpha = 5%' (the
> alpha should be displayed as the greek letter).
> I tried
>
> > text(1,1,expression(alpha == 5%))
text(1,1, expression(paste(alpha == 5, '%')) )
cu
you can try
text(1, 1, expression(paste(alpha, " = 5%")))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Martin Keller-Ressel wrote:
> Hello,
>
> I would like to annotate a graph with the expression 'alpha = 5%' (the
> alpha should be displayed as the greek letter).
> I tried
>
>> text(1,1,expression(alpha == 5%))
Try
text(1,1,expression(alpha == 5*"%"))
Best,
Bettina
__
Marco Helbich wrote:
> Dear List,
>
> I want to make a parallel coordinates plot with the specific variables on
> the abscissa and the cases on the ordinate should be dyed dependent on
> another nominal variable from the data frame. I use the parcoord function.
>
Hi Marco,
If I understand your qu
Hello,
I would like to annotate a graph with the expression 'alpha = 5%' (the
alpha should be displayed as the greek letter).
I tried
> text(1,1,expression(alpha == 5%))
which gives a syntax error.
escaping the percent sign (\%) or doubling (%%) does not help.
What do I do?
Thanks,
Martin Ke
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