Wacek Kusnierczyk wrote:
Barry Rowlingson wrote:
On Wed, May 13, 2009 at 5:36 PM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
Barry Rowlingson wrote:
Soln - for loop:
z=list()
for(i in 1:1000){z[[i]]=rnorm(100,0,1)}
now inspect the individual bits:
hist(z[[1]])
Hi, postscript should do, *.eps specifically.
x=rnorm(100,0,1)
e=rnorm(100,0,1)
y=x+e
postscript(myeps.eps)
plot(y~x)
dev.off()
Cheers,
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org
I want generate R code to determine the real root of the polynomial
x^3-2*x^2+3*x-5. Using an initial guess of 1 with Newton's method.
Help please...
--
View this message in context:
http://www.nabble.com/Finding-root-using-Newton%27s-method-tp23534519p23534519.html
Sent from the R help mailing
Hi everybody.
I want to identify not only duplicate number but also the original number
that has been duplicated.
Example:
x=c(1,2,3,4,4,5,6,7,8,9)
y=duplicated(x)
rbind(x,y)
gives:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
x123445678 9
y0
On Wednesday 13 May 2009 10:54:01 pm Peverall Dubois wrote:
Hi,
Suppose I have a calculation that contain an estimation
of certain value and its confidence interval
(this is computed via Chi Square approximation).
Is there any approach in R in which we can
evaluate the confidence interval?
On Thu, May 14, 2009 at 2:16 PM, christiaan pauw cjp...@gmail.com wrote:
Hi everybody.
I want to identify not only duplicate number but also the original number
that has been duplicated.
Example:
x=c(1,2,3,4,4,5,6,7,8,9)
y=duplicated(x)
rbind(x,y)
gives:
[,1] [,2] [,3] [,4] [,5] [,6]
Barry Rowlingson wrote:
On Wed, May 13, 2009 at 9:56 PM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
Barry Rowlingson wrote:
n = 1000
benchmark(columns=c('test', 'elapsed'), order=NULL,
'for'={ l = list(); for (i in 1:n) l[[i]] = rnorm(i, 0, 1)
Homework? - see your instructor! Otherwise, provide minimal self-contained
code and show us where you are stuck.
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org
Barry Rowlingson wrote:
[...]
Yes, you can probably vectorize this with lapply or something, but I
prefer clarity over concision when dealing with beginners...
but where's the preferred clarity in the for loop solution?
Seriously? You think:
lapply(1:n, rnorm, 0, 1)
is
Yes, they don't seem to show ways to evaluate confidence interval though.
Only functions for computing confidence interval.
-G.V
On Thu, May 14, 2009 at 3:19 PM, J Dougherty j...@surewest.net wrote:
On Wednesday 13 May 2009 10:54:01 pm Peverall Dubois wrote:
Hi,
Suppose I have a calculation
Thanks for these suggestions. However I have one more question. Is there any
way to extract only numbers? For example I want to extract only 88 in my
example.
Regards,
MUHC-Research wrote:
Hi Ron,
Look up the grep() function.
Cheers,
--
*Luc Villandré*
/Biostatistician
McGill
Dimitris Rizopoulos wrote:
Wacek Kusnierczyk wrote:
Barry Rowlingson wrote:
On Wed, May 13, 2009 at 5:36 PM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
Barry Rowlingson wrote:
Soln - for loop:
z=list()
for(i in 1:1000){z[[i]]=rnorm(100,0,1)}
now inspect the
RON70 wrote:
Thanks for these suggestions. However I have one more question. Is there any
way to extract only numbers? For example I want to extract only 88 in my
example.
if you have just one integer number represented in a single string,
here's one way go:
strings = c('foo 1', '2
Dear Ravi:
Thanks for pointing out the homotopy methods. Coming from Mathematics I was
always considering SINGULAR for such a task which is also providing results
when the solution set is not isolated points, but an algebraic variety.
For single points, homotopy methods appear to be an
I want generate R code to determine the real root of the polynomial
x^3-2*x^2+3*x-5. Using an initial guess of 1 with Newton's method.
Homework? - see your instructor! Otherwise, provide minimal
self-contained
code and show us where you are stuck.
It gets a little suspicious when there are
Stats Wolf stats.wolf at gmail.com writes:
Saving a plot with pdf gives a very nice result:
pdf(myplot.pdf)
par(font=1,family='serif')
plot(pressure)
dev.off()
Doing the very same with other formats (png, jpeg, tiff) gives far
worse results. Is there anything to do to make a plot in
Thank you, Paul.
Postscript, however, does not have to be what I need for two reasons.
First, it does not accept some special characters from foreign
languages (exactly like PDF). Second, not too many journals accept it.
So from your -- and Daniel's -- reply I understand other formats will
not
Gabor Grothendieck ggrothendieck at gmail.com writes:
R interprets backslash to give special meaning to the next character, i.e.
it strips off the backslash and send the following character to gsub
possibly reinterpreting it specially (for example \n is newline). Thus
a backslash will
Try this
x%in%x[which(y)]
From your example
x=c(1,2,3,4,4,5,6,7,8,9)
y=duplicated(x)
rbind(x,y)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
x123445678 9
y000010000 0
which(y)
[1] 5
x[which(y)]
[1] 4
Nair, Murlidharan T wrote:
Does anyone have any suggestion on this? I tried par(new=T) It does not seem
to work with scatterplot3d. Any suggestion is appreciated.
Cheers../Murli
Please be patient for a day or two before expecting a reply. If you
expect some Europeans to answer: Not all of
Dear all,
I write this message because I have a problem in data importation. I hope that
you help me.
My data base is in an Excel spreasheet. I import this data base using the
following code:
library(RODBC)
db - C:/Users/Axel/Desktop/estimation/data.xls
channel - odbcConnectExcel(xls.file =
Dear R-community,
I was using SVM regression (svm {e1071}) for predictions of single
soil properties of a huge data set (3000 samples). There are for the
eps-regression using the radial basis kernel three optimization
parameters needed.
To make things easier (using only two optimization
I am trying to use the kernel function. To understand how it works I tried
out some of the examples.
None of them works as shown in the following:
kernel(daniell, 50) #
Error in kernel(daniell, 50) : unused argument(s) (50)
kernel(daniell, 10) #
Error in kernel(daniell, 10) : unused
Wacek Kusnierczyk waclaw.marcin.kusnierc...@idi.ntnu.no wrote
Seriously? You think:
lapply(1:n, rnorm, 0, 1)
is 'clearer' than:
x=list()
for(i in 1:n){
x[[i]]=rnorm(i,0,1)
}
for beginners?
Firstly, using 'lapply' introduces a function (lapply) that doesn't
have an intuitive
The operator %in% is very good! And that can be simpler like this:
x %in% x[duplicated(x)]
[1] FALSE FALSE FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE
On Thu, May 14, 2009 at 4:43 PM, Andrej Blejec andrej.ble...@nib.si wrote:
Try this
x%in%x[which(y)]
From your example
Peter Flom wrote:
Seriously? You think:
lapply(1:n, rnorm, 0, 1)
is 'clearer' than:
x=list()
for(i in 1:n){
x[[i]]=rnorm(i,0,1)
}
for beginners?
Firstly, using 'lapply' introduces a function (lapply) that doesn't
have an intuitive name. Also, it takes a function as an
On 13/05/2009 2:38 PM, Rebecca Sela wrote:
I am creating an R package. I ran R CMD check on the package, and everything
passed until it tried to run the examples. Then, the result was:
* checking examples ... ERROR
Running examples in REEMtree-Ex.R failed.
The error most likely occurred in:
I wrote
As a beginner, I agree the for loop is much clearer to me.
Wacek Kusnierczyk waclaw.marcin.kusnierc...@idi.ntnu.no replied
well, that's quite likely. especially given that typical courses in
programming, afaik, include for looping but not necessarily functional
stuff -- are
As a beginner, I agree the for loop is much clearer to me.
[Warning: Contains mostly philosophy]
To me, the world and how I interact with it is procedural. When I want
to break six eggs I do 'get six eggs, repeat break egg until all
eggs broken'. I don't apply an instance of the break
Peter Flom wrote:
As a beginner, I agree the for loop is much clearer to me.
well, that's quite likely. especially given that typical courses in
programming, afaik, include for looping but not necessarily functional
stuff -- are you an r beginner, or a programming beginner?
Barry Rowlingson wrote:
As a beginner, I agree the for loop is much clearer to me.
[Warning: Contains mostly philosophy]
maybe quasi ;)
To me, the world and how I interact with it is procedural. When I want
to break six eggs I do 'get six eggs, repeat break egg until all
I have 3 columns: flow, month, and monthname, where month is 1-12, and
monthname is name of month. I can't get the plot to replace the 1-12
with monthname using ticks.lab. What am I doing wrong?
plot(flow~factor(month),xlab=Month,ylab=Total Flow per Month,
ylim=c(0,55000), ticks.lab=monthname)
Hi,
I was wondering how to specify the number of decimal numbers in my
computation using R? I have too many decimal numbers for my result, when I
convert them to string with as.character, the string will be too long.
Thanks and regards!
--
View this message in context:
On 14-May-09 11:28:17, Wacek Kusnierczyk wrote:
Barry Rowlingson wrote:
As a beginner, I agree the for loop is much clearer to me.
[Warning: Contains mostly philosophy]
maybe quasi ;)
To me, the world and how I interact with it is procedural. When I want
to break six eggs I do
Barry Rowlingson wrote:
Computer scientists will write their beautiful manuscripts, but how
many people who come to R because they want to do a t-test or fit a
GLM will read them? That's the R-help audience now.
don't forget that r seems to take, maybe undeserved, the pride of being
a
Depending on what you want to do, use 'sprintf':
x - 1.23456789
x
[1] 1.234568
as.character(x)
[1] 1.23456789
sprintf(%.1f %.3f %.5f, x,x,x)
[1] 1.2 1.235 1.23457
On Thu, May 14, 2009 at 7:40 AM, lehe timlee...@yahoo.com wrote:
Hi,
I was wondering how to specify the number of
Stats Wolf wrote:
Dear colleagues,
Saving a plot with pdf gives a very nice result:
pdf(myplot.pdf)
par(font=1,family='serif')
plot(pressure)
dev.off()
Doing the very same with other formats (png, jpeg, tiff) gives far
worse results. Is there anything to do to make a plot in some other
format
S. Nunes wrote:
Hi all,
I am doing some explorations using a dataset with the following
structure (id, value, flag).
For instance:
a, 2.2, 1
b, 3.0, 1
c, 2.9, 0
d, 3.1, 1
...
I have plotted a standard histogram using a simple command like:
hist(data$value)
My question:
I would like to
Alternatively,
signif(c(pi,exp(1)), 3)
?signif # and others in that page
HTH,
baptiste
On 14 May 2009, at 13:47, jim holtman wrote:
Depending on what you want to do, use 'sprintf':
x - 1.23456789
x
[1] 1.234568
as.character(x)
[1] 1.23456789
sprintf(%.1f %.3f %.5f, x,x,x)
[1] 1.2
On 14-May-09 11:40:21, lehe wrote:
Hi,
I was wondering how to specify the number of decimal numbers in my
computation using R? I have too many decimal numbers for my result,
when I convert them to string with as.character, the string will
be too long.
Thanks and regards!
Since you say you
Thanks!
In my case, I need to deal with a lot of such results, e.g. elements in a
matrix. If using sprintf, does it mean I have to apply to each result
individually? Is it possible to do it in a single command?
jholtman wrote:
Depending on what you want to do, use 'sprintf':
x -
On 14/05/2009 7:31 AM, Graves, Gregory wrote:
I have 3 columns: flow, month, and monthname, where month is 1-12, and
monthname is name of month. I can't get the plot to replace the 1-12
with monthname using ticks.lab. What am I doing wrong?
plot(flow~factor(month),xlab=Month,ylab=Total Flow
As I do not thoroughly understand the way 'lsoda' works, I face some
difficulties to 'get' myself into the function(), though I changed the code
as follows:
--
require(deSolve)
qm-0.36
y0-c(0)
parms-c(K1,K2)
times-seq(0,1,1)
kinetic.model-function(t,y,parms){
It all depends on what you want to do with the result. Here are some
variations:
x - matrix(runif(16), 4)
x
[,1] [,2] [,3] [,4]
[1,] 0.2655087 0.2016819 0.62911404 0.6870228
[2,] 0.3721239 0.8983897 0.06178627 0.3841037
[3,] 0.5728534 0.9446753 0.20597457 0.7698414
jim holtman wrote:
Depending on what you want to do, use 'sprintf':
x - 1.23456789
x
[1] 1.234568
as.character(x)
[1] 1.23456789
sprintf(%.1f %.3f %.5f, x,x,x)
[1] 1.2 1.235 1.23457
... but remember that sprintf introduces excel bugs into r
Thanks for everyone.
I think the approach below is most suitable for me, as a beginner.
x=list()
for(i in 1:n){
x[[i]]=rnorm(i,0,1)
}
Now, I am trying to obtain the sample variance (S^2) of the 1000 samples that I
have generated before.
I am wondering what command I should use in
To make things easier (using only two optimization parameters and not
loosing performance) I wanted to use LS SVM regression (lssvm{kernlab}). But
it looks to me that it is not yet implemented. At least I got error
messages, which I could not find a solution for (Error in if (n !_dim(y)[1]
If the input is gdfsa-sdhchc99-88 then
assuming you only want 88 but not
99 then if s is the vector of words
that we already computed:
s[regexpr(^[0-9]+, s) 0]
or that could be combined with the strapply solution
into one line:
strapply(gdfsa-sdhchc99-88, \\w+, ~ if (regexpr(^[0-9]*$, x) 0)
Thanks to all of you!
Yes, I am generating a latex table in my report.
jholtman wrote:
It all depends on what you want to do with the result. Here are some
variations:
x - matrix(runif(16), 4)
x
[,1] [,2] [,3] [,4]
[1,] 0.2655087 0.2016819 0.62911404
Hi,
I am using latex.table to write my results into a latex table. If my results
is like a matrix except that some column has strings and others have
numbers. Is it possible to feed my results into latex.table?
Thanks and regards!
--
View this message in context:
On 14-May-09 12:03:43, lehe wrote:
Thanks!
In my case, I need to deal with a lot of such results, e.g. elements
in a matrix. If using sprintf, does it mean I have to apply to each
result individually? Is it possible to do it in a single command?
Yes, in various ways depending on how you
Debbie Zhang schrieb:
Now, I am trying to obtain the sample variance (S^2) of the 1000 samples that
I have generated before.
I am wondering what command I should use in order to get the sample variance
for all the 1000 samples.
What I am capable of doing now is just typing in
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Thursday, May 14, 2009 4:44 AM
To: Nair, Murlidharan T
Cc: r-help@r-project.org
Subject: Re: [R] scatterplot3d
Nair, Murlidharan T wrote:
Does anyone have any suggestion on this? I tried par(new=T)
Another way of making functions visible to examples is to list the
package name as Depends in the package DESCRIPTION file.
Alternatively, if you precede use in \examples with
library(packageName), you should also list it as Suggests in
DESCRIPTION.
Hope this helps.
Spencer
Since you got the most suitable way to get x, why can't you get the
variances in the same way? Just like:
v = vector()
for (i in 1:length(x)) v[i] = var(x[[i]])
BTW, it is much better to use lapply, like this:
lapply(x, var)
On Thu, May 14, 2009 at 8:26 PM, Debbie Zhang
Thanks. But ... Still don't get it.
Referring explicitly to the monthname column which had many entries
across many years did not work. So taking your clue I created a new
variable list of names of month
month.name-list(c(Jan, Feb, Mar, Apr, May, Jun, Jul,
Aug, Sep, Oct, Nov, Dec))
then
lehe wrote:
I am using latex.table to write my results into a latex table. If my
results is like a matrix except that some column has strings and others
have numbers. Is it possible to feed my results into latex.table?
Such a matrix-like structure is called a dataframe. See latex in
On 14-May-09 12:27:40, Wacek Kusnierczyk wrote:
jim holtman wrote:
Depending on what you want to do, use 'sprintf':
x - 1.23456789
x
[1] 1.234568
as.character(x)
[1] 1.23456789
sprintf(%.1f %.3f %.5f, x,x,x)
[1] 1.2 1.235 1.23457
... but remember that sprintf introduces
Graves, Gregory wrote:
Thanks. But ... Still don't get it.
Referring explicitly to the monthname column which had many entries
across many years did not work. So taking your clue I created a new
variable list of names of month
month.name-list(c(Jan, Feb, Mar, Apr, May, Jun, Jul,
Aug, Sep,
Axel Leroix wrote:
Then I perform an lm regression using the following code:
reg1 -lm(data$prod~data$pri+data$cli)
summary(reg1)
Use
reg1 -lm(prod~pri+cli, data=data)
instead. It is not necessary to call the data frame you read your stuff into
data, any more useful name, such as
Hi,
I am using rpart for my decision stump. I am trying to extract the learned
stump from the output of rpart.
For example:
cntrl - rpart.control(maxdepth = 1, minsplit = learn-1,
maxsurrogate = 0, usesurrogate=0, maxcompete = 1,
cp = 0,
threshold wrote:
Hi, do you have any suggestions how to make 3D scatterplot, BUT under linux.
Worth mentioning is the fact that 'scatterplot3d' does not load under Ubuntu
8.10.
Why not? It is in pure R code, hence it is just as OS dependent as R itself.
Uwe Ligges
Do you know any
Stefan Grosse wrote:
Debbie Zhang schrieb:
Now, I am trying to obtain the sample variance (S^2) of the 1000 samples
that I have generated before.
I am wondering what command I should use in order to get the sample variance
for all the 1000 samples.
What I am capable of doing now
I have genetic data, and I would like to do an dissimilariry table of the
factorial analysis.
What package should I use and the command line?
Lassana TOURE
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
mau...@alice.it wrote:
I am trying to use the kernel function. To understand how it works I tried out some of the examples.
None of them works as shown in the following:
kernel(daniell, 50) #
Error in kernel(daniell, 50) : unused argument(s) (50)
which works for me. Have you
Erin Hodgess wrote:
Hi R People:
I have a question about setClass please. I'm working thru R
Programming for Bioinformatics.
Actually, I was wondering if there is such a thing as an updateClass,
in order to change a contains option, please?
that is, if I had
setClass(dog,
Noting that:
ave(x, x, FUN = length) 1
[1] FALSE FALSE FALSE TRUE TRUE FALSE FALSE FALSE FALSE FALSE
try this:
rbind(x, dup = ave(x, x, FUN = length) 1)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
x 123445678 9
dup0001
(Ted Harding) wrote:
On 14-May-09 12:27:40, Wacek Kusnierczyk wrote:
... but remember that sprintf introduces excel bugs into r (i.e.,
rounding is not done according to the IEC 60559 standard, see ?round):
ns = c(0.05, 0.15)
round(ns, 1)
# 0.0 0.2
Wacek Kusnierczyk wrote:
(Ted Harding) wrote:
On 14-May-09 12:27:40, Wacek Kusnierczyk wrote:
... but remember that sprintf introduces excel bugs into r (i.e.,
rounding is not done according to the IEC 60559 standard, see ?round):
ns = c(0.05, 0.15)
round(ns, 1)
# 0.0 0.2
It is not recommended to plot variables on different scales on the same graph.
You can use something like par(mfrow=...) (see ?par) to stack graphs on top of
each other (or next to each other, or both) for easy comparison, but each with
their own scales/axes.
If you have considered all that
James W. MacDonald wrote:
Wacek Kusnierczyk wrote:
(Ted Harding) wrote:
On 14-May-09 12:27:40, Wacek Kusnierczyk wrote:
... but remember that sprintf introduces excel bugs into r (i.e.,
rounding is not done according to the IEC 60559 standard, see
?round):
ns = c(0.05, 0.15)
Good morning.
I want to estimate a mixture of two Weibull distribution by using the EM
algorithm .
Does it exist an available package to do that ?
Or if not, Is it some avaible EM Algorithm ?
Thanks
Cordially
Abdoul Aziz Junior NDOYE
GREQAM Marseille.
[[alternative
I have a database like this:
kol1;kol2;kol3 ...
2;5;9
9;6;6
4;6;5...
I looking for a kod in R which let mi aviod in column unnecessary value, for
example number 9.
So, if I have:
Kol1
2
9
4
4...
after loop in R I would like to get my column like this:
Kol1
2
4
4...
If you have any idea
If I want to install RWeka I get an error:
library(RWeka)
Error occurred during initialization of VM
java.lang.OutOfMemoryError: unable to create new native thread
Could you have any idea to help me?
--
View this message in context:
Hi all,
I have a 36x14 matrix of hexadecimal coded colors (see attached file) that
was created with the following code:
cellcol[is.na(cellcol)] - #00
#Aliceblue
cellcol[cellcol 5] - #F0F8FF
#Skyblue1
cellcol[cellcol = 5 cellcol 20] - #87CEFF
#Blue
cellcol[cellcol = 20 cellcol 35] -
Wacek Kusnierczyk wrote:
do they make pompous claims about their software and disregarding claims
about others' as well?
My mistake. I thought your concern was for the quality of the software
(quality of course being defined by a certain committee of one). But it
appears this is of a more
Dear list,
I'd like to test the null hypothesis of independence in a 2*2
contingency table in a Bayesian way.
As far as I can see, this can be done by the function ctable in library
LearnBayes. (More precisely, a Bayes factor can be computed.)
Two questions:
1) Is there any other
Dear R-list,
another question trying to build consensus cluster ;-)
Using the package clue I have found a method of building consensus
clusters the following way from one distance matrix:
clust1 - c(ward, single, complete, average, mcquitty,
median, centroid)
clust_res - lapply(clust1,
James W. MacDonald wrote:
Wacek Kusnierczyk wrote:
do they make pompous claims about their software and disregarding claims
about others' as well?
My mistake. I thought your concern was for the quality of the software
(quality of course being defined by a certain committee of one). But
it
I am not sure what you are trying to achive. How about:
df-data.frame(kol1=c(2,9,4,3),kol2=c(5,6,6,2),kol3=c(9,6,5,1))
(df)
new.df-df[(!(df[[1]] %in% 9)!(df[[2]] %in% 9)!(df[[3]] %in% 9)),]
(new.df)
But why would you want to delete all the valid values in all the other
columns if you have your
... or, similar in character to Gabor's solution:
tbl - table(x)
(tbl[as.character(sort(x))]1)+0
Bert Gunter
Nonclinical Biostatistics
467-7374
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Gabor Grothendieck
Sent: Thursday,
Dear all,
I'm writing my own method to be used in Rpart by defining the list of
functions named init, split and eval. I'm following the example given in the
file 'tests/usersplits.R' in the sources.
By now I'm able to define the split function (and it works correctly in the
tree construction)
On 14-May-09 15:15:16, James W. MacDonald wrote:
Wacek Kusnierczyk wrote:
(Ted Harding) wrote:
On 14-May-09 12:27:40, Wacek Kusnierczyk wrote:
... but remember that sprintf introduces excel bugs into r (i.e.,
rounding is not done according to the IEC 60559 standard, see
?round):
(Ted Harding) wrote:
This happens also when you use C's fprintf and sprintf (at any rate
in my gcc):
r's printing routines (e.g., print, sprintf, cat, anything else?) seem
to rely on the underlying c sprintf, with no prior r-implemented
rounding. hence they import into r whatever
On 5/13/09, Liati liat...@hotmail.com wrote:
Hi,
Thank you for your reply.
This is exactly what I have done.
However, I would like to get the two plots next to each other (to use as one
figure).
Look at ?print.trellis and its examples.
-Deepayan
Dear all -
We perform some measurements with a machine that needs to be
recalibrated. The best calibration we get with polynomial regression.
The data might look like follows:
true_y - c(1:50)*.8
# the real values
m_y - c((1:21)*1.1, 21.1, 22.2, 23.3 ,c(25:50)*.9)/0.3-5.2
# the measured data
On 5/9/09, John Maindonald john.maindon...@anu.edu.au wrote:
The following tinkers with the strip labels, where the
different panels are for different levelf of a conditioning
factor.
tau - (0:5)/2.5; m - length(tau); n - 200; SD - 2
x0 - rnorm(n, mean=12.5, sd=SD)
matdf - data.frame(
Dieter Wirz wrote:
Dear all -
We perform some measurements with a machine that needs to be
recalibrated. The best calibration we get with polynomial regression.
The data might look like follows:
true_y - c(1:50)*.8
# the real values
m_y - c((1:21)*1.1, 21.1, 22.2, 23.3 ,c(25:50)*.9)/0.3-5.2
-- but it is preferable to use the appropriate access functions:
coef(mylm)
?coef
Bert Gunter
Nonclinical Biostatistics
467-7374
## Now, what I believe you're looking for ;
mylm$coefficients ;
Cheers,
--
*Luc Villandré*
/Biostatistician
McGill University Health Center -
Montreal
Don't think I have seen this one come across:
x - c(1,2,3,2,4,4,6,1)
duplicated(x) | duplicated(x, fromLast=TRUE)
[1] TRUE TRUE FALSE TRUE TRUE TRUE FALSE TRUE
On Thu, May 14, 2009 at 12:09 PM, Bert Gunter gunter.ber...@gene.comwrote:
... or, similar in character to Gabor's solution:
is there a way to label points in a graph using text(locator(1),text)
after ggplot() or qplot() ?
qplot(date, psavert, data = economics, geom = line,main=jhdjd)-p
p+opts(text(locator(1),),new=T)
does not work.
--
View this message in context:
Hello
I have to import 2 txt files into R. 1 file contains the data and the other
contains the header, column headings, datatypes and labels for the data.
I have 2 problems:
1) my data file has mixed type of data e.g. 1 2 3 4 5 3-5 02/04/06 3 4 5 and
so on, the data file is tab separated. when
What have you tried? Check the Intro manual for hints.
?read.table probably using sep='\t'
On Thu, May 14, 2009 at 1:30 PM, Sunita22 sunita...@gmail.com wrote:
Hello
I have to import 2 txt files into R. 1 file contains the data and the other
contains the header, column headings,
Hello
Yes I have used read.table(file name, sep=\t) for reading the text file
Thank you
On Thu, May 14, 2009 at 11:07 PM, jim holtman jholt...@gmail.com wrote:
What have you tried? Check the Intro manual for hints.
?read.table probably using sep='\t'
On Thu, May 14, 2009 at 1:30 PM,
Hi,
Thanks for the contribution. I tested the code but I got a couple or errors.
I'll try to debug these in the next days.
Maybe there is a simpler approach to what I am trying to do.
I have a bunch of measures and a column with the success of the case (1/0).
I simply want to evaluate my
The table()-based solution can have problems when there are
very closely spaced floating point numbers in x, as in
x1-c(1, 1-.Machine$double.eps, 1+2*.Machine$double.eps)[c(1,2,3,2,3)]
It also relies on table(x) turning x into a factor with the default
levels=as.character(sort(x)) and that
I am writing a custom function that uses an R-function from the
reshape package: cast. However, my question could be applicable to
any R function.
Normally one writes the arguments directly into a function, e.g.:
result=cast(table1, column1 + column2 + column3 ~column4,
mean)
I am currently doing some prediction work using the knn script in the
'class' package. Does anyone know a way of having R return the IDs (sample
IDs, or column IDs of the training matrix) of the 'k' samples that are
chosen by the algorithm as being nearest to a given test sample?
I have
On Thu, May 14, 2009 at 12:16 PM, Lori Simpson
lori.simp...@dc-energy.com wrote:
I am writing a custom function that uses an R-function from the
reshape package: cast. However, my question could be applicable to
any R function.
Normally one writes the arguments directly into a function,
I don't know the book but I doubt that it is a good way to learn R.
I'd suggest having a look at some of the documentation available on the R site.
Click on Other (in left column of page) have a look there and then select the
contributed documentation link to get more documentation. Have
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