Dear R helpers
I am trying to fit the Gumbel distribution to a data. I am using lmom package.
I am getting problem in Cumulative Distribution Function of Gumbel distribution
as I am getting it as a series of 0's and 1's thereby affecting the P P Plot.
My R code is as follows.
library(quantr
Altough it depends on what crit you keep your variables, but maybe you should
take a look at ?step.
Bart
Paul Hermes wrote:
>
> ok,
> i think i have to be more precise of what we are doing.
> first thing: this code is not from me, and Im new to R (and never touched
> anything like this)
> I
Hi all,
I have the following data
AAA 1e+06 1222.312
AAC 100.2 0.33
However the following command
dat <- read.table("mydat.txt");
print(dat)
n1 <- sum(as.matrix(dat$V1));
gives:
Error in sum(as.matrix(dat$V1)) : invalid 'type' (character) of argument
What's wrong with the data type so t
I hope this will be an easy question to resolve, but I haven't been able to
find a way to hide a legend in a filled.contour plot. I've looked at the
filled.contour help file and Googled, but to no avail so far.
I'd appreciate any solutions to the problem.
Thanks,
-John
[[alternative HTM
On Fri, 13 Mar 2009, Tal Galili wrote:
Thanks Thomas.
Assuming I want to change the "k" factor (used in AIC type procedures), is
there a way to do that ?
There is no "k" factor in the leaps algorithm. It always reports the best
model with one predictor, the best model with two predictors, th
Hi,
I wonder if the following is possible in R:
Suppose a function takes an argument, and wants to modify this argument
so that the change is visible _at the call side_. It would be what is
for example known as pass-by-reference-to-non-const in C++.
test <- function(x)
{
x <- 10
...
retur
Thomas Mang fiwi.at> writes:
> I wonder if the following is possible in R:
>
> Suppose a function takes an argument, and wants to modify this argument
> so that the change is visible _at the call side_. It would be what is
> for example known as pass-by-reference-to-non-const in C++.
>
> test
The package does not install under Windows: see
http://www.r-project.org/nosvn/R.check/r-release-windows-ix86/DPpackage-00check.html
The maisntainer has been informed (probably many times via the
automated notification system). With 1700 packages, the volunteers
have no time to spend correcti
Debabrata Midya commerce.nsw.gov.au> writes:
> This has reference to the package âDPpackageâ. The binary version is
> available on Mac OS, but I am using Windows XP.
>
> May I request you to assist me in the followings:
> How can I prepare Windows binary of DPpackage from source?
>
>
On Fri, 13 Mar 2009, Thomas Mang wrote:
Hi,
I wonder if the following is possible in R:
Suppose a function takes an argument, and wants to modify this argument so >that the change is visible _at the call side_. It would be what is for
example known as pass-by-reference-to-non-const in C++.
Dieter Menne wrote:
Thomas Mang fiwi.at> writes:
I wonder if the following is possible in R:
Suppose a function takes an argument, and wants to modify this argument
so that the change is visible _at the call side_. It would be what is
for example known as pass-by-reference-to-non-const in C
Do an 'str(dat)' and see what the structure is. When I read in your data I get:
> x
V1V2 V3
1 AAA 100.0 1222.312
2 AAC 100.20.330
> str(x)
'data.frame': 2 obs. of 3 variables:
$ V1: Factor w/ 2 levels "AAA","AAC": 1 2
$ V2: num 100 100
$ V3: num 1222.31 0.
Dear Paul,
Could you be more specific about what you mean here?
I don't know the Runger paper so it's hard to tell what it is that
you're looking for.
Blatant plug: I developed a package for hidden Markov models
called depmixS4 that in some sense does what you want: clustering
taking dependencie
Prof Brian Ripley stats.ox.ac.uk> writes:
> The maintainer has been informed (probably many times via the
> automated notification system). With 1700 packages, the volunteers
> have no time to spend correcting packages for non-responsive
> maintainers (we do quite a lot for responsive ones).
Hi,
I want to run the R-function ctree (package party) from Java over Rserve
with the following Java-Code:
try{
RConnection v = new RConnection();
v.voidEval("library(party)");
v.voidEval("try(load(\"C:\\Documents and Settings\\daten2.rda\"))");
v.voidEval("try(pdf(\"C:\\Documents and Settings\\t
Dear Emma,
Have you tried a simpler model? False convergence can be due to an
overcomplex model. Can you give a brief outline of your data? E.g. how
many sites, how many data per site, ... Cross tabulations of all pairs
of factor variables are usefull too.
HTH,
Thierry
---
Hi,
I am not sure which test is applied to the data if you use cor.test(x, y) ?
Is it an unpaired t-Test?
Regards
--
View this message in context:
http://www.nabble.com/cor.test%28x%2Cy%29-tp22492993p22492993.html
Sent from the R help mailing list archive at Nabble.com.
_
Hi Thierry,
Thanks, my data are pasted in below: n = 48, with different numbers if site
replicates.
Site passes length width height ohwidth ohheight style shape nogap pgaps
1 1 32 38.21 7.18 2.600.00 0.00 1 1 0 0.00
2 1 7 154.38 2.55 2.430.00
I think what you are missing is that the default map database is "world".
If you use:
library(maps)
require("mapproj")
longlatLimit<-c(-107,-93,40,52)
par(plt=c(0,1,0,1),cex=1,cex.main=1) #Set plotting parameters
map(regions="USA", projection="azequalarea",
type="n",xlim=long
Hi Thierry,
That's great thanks!
I have done as you have said but I keep getting a warning message here is
my code:
G1Hvol<-glmer(passes~hvolume+style+habitat(1|Site),family = poisson)
And this is the message i get:
Warning message:
In mer_finalize(ans) : false convergence (8)
any ideas??
Dear all,
This seems like a simple problem but i've searched the help files and tried
various options but failed, so apologies in advance for asking what i'm sure
is an easy thing to do!
In short, I have displayed behavioural data using the TraMineR package such
that there is a colour change bet
Hello,
I'm trying to put a dynamic table and a dynamic graph side by side in a pdf
document using Sweave.
The data.frame used to generate the table is called rg (rg.txt):
"Date"; "Code"; "Data1"; "Data2"
"2009-03-10";"1";1958;147
"2009-03-10";"2";302;144
"2009-03-10";"3";4;141
"2009-03-10";"4";4
Dear Emma,
First of all make shure that style and habitat are factors (assuming
that they are categorical).
glmer(passes~hvolume+style+habitat(1|Site),family = poisson)
Style has 3 levels, habitat 5 levels. So your model needs to estimate 1
parameter for hvolume, 2 for style, 4 for habitat and 1
Hi,
I'm working with custom slides(Cy5) and working in the normalization of the
arrays.
I have three arrays (technical replicates).
I have sucesfully normalized the data using vsn, however i would like to
normalize using spike in controls.
My controls are annotated as CTL-1 to x and i would like t
Hello, everybody
Say I have
nm1 <- c(rep(1,10), rep(0,10))
then I can do:
diff(nm1)
to see where I have shift in value
but what if I have
nm2 <- c(rep("SPZ8", 10), rep("SPX9", 10))
how can I produce the same ouput as diff(nm1) does, that is zeros
everywhere except for one place where SPZ8 change
Sean Zhang wrote:
> Dear Jens and Wacek:
>
> I appreciate your answers very much.
>
> I came up an example based on your comments.
> I feel the example helped me to understand...(I could be missing your points
> though :( )
> If so, please let me know.
> Simon pointed out the following link:
> http
2009/3/13 Sergey Goriatchev :
> Say I have
> nm1 <- c(rep(1,10), rep(0,10))
> then I can do:
> diff(nm1)
> to see where I have shift in value
>
> but what if I have
> nm2 <- c(rep("SPZ8", 10), rep("SPX9", 10))
>
> how can I produce the same ouput as diff(nm1) does, that is zeros
> everywhere excep
Hi Thierry,
Thanks again! You are a great help!!
I have taken habitat out, and then run it with style but still the same
problem exists, so I have taken both style and habitat out. The problem
here is it leaves me with only 3 parameters and because they are all
correlated I cant use them in t
Hi:
For A, you can use head and tail but you have to add a zero the front.
For B, you can use the same function, but put it inside an sapply and
run over the columns and then cbind it back with the original dataframe.
A)
nm2 <- c(rep("SPZ8", 10), rep("SPX9", 10))
-1.0*c(0,as.numeric((head(nm2,
==
Course: 1 day introduction to R and S-PLUS
Location: Basingstoke, UK
Date: Wednesday 29 April 2009
Cost: 500 GBP
==
Details:
SP411: An introduction to R and TIBCO Spotfire S-PLUS
This course will give an intr
Ptit_Bleu yahoo.fr> writes:
> I'm trying to put a dynamic table and a dynamic graph side by side in a pdf
> document using Sweave.
> The data.frame used to generate the table is called rg (rg.txt):
>
> "Date"; "Code"; "Data1"; "Data2"
> "2009-03-10";"1";1958;147
> "2009-03-10";"2";302;144
...
>
Hi,
I have two lists of numbers which are both 1,2,3,4.
I would like to combine pairs so that I have:
1,2
1,3
1,4
2,3
2,4
3,4.
I know that expand.grid() can give me all combinations of pairs.
Any suggestions would be much appreciated.
Emma
--
View this message in context:
http://www.nabble.
I have solved my problem using:
> x<-1:4
> x
[1] 1 2 3 4
> combn(x,2)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]111223
[2,]234344
Thanks Emma
emj83 wrote:
>
> Hi,
>
> I have two lists of numbers which are both 1,2,3,4.
> I would like to combine pai
On Thu, Mar 12, 2009 at 01:25:40PM -0700, Atul Joshi wrote:
> When I run my .r script using source (myfilename) from console the plots
> appear.
>
> However, when I execute the same file via terminal command line using R CMD
> BATCH myfilename, the plots do not appear.
>
> I am working with Mac
Fabulous, guys!
Let me try your suggestions.
Thanks a lot!
Best,
SErgey
On Fri, Mar 13, 2009 at 11:53, wrote:
> Hi:
> For A, you can use head and tail but you have to add a zero the front.
> For B, you can use the same function, but put it inside an sapply and run
> over the columns and then
Where do you want to print it? Is it the console (if so, try
'print(median)') or if it is the plot, use 'text' with the appropriate
parameters. It would help if you listed the " I tried many things but
without success..." and what you were expecting vs. what you got.
Reproducible code would be use
Try
#install.packages('RColorBrewer')
example(brewer.pal, pack='RColorBrewer')
hth,
Kingsford Jones
On Fri, Mar 13, 2009 at 3:20 AM, Ross Culloch wrote:
>
> Dear all,
>
> This seems like a simple problem but i've searched the help files and tried
> various options but failed, so apologies in a
Hi Kingsford,
Thanks for the reply - some of the sets/palettes in the RColorBrewer are
ideal, but the problem with the problem i have is that they only go up to 12
colours, and i need 15 colours - so i assume the only thing i can do is
create my own palette, but i'm having limited success in tryi
I'm doing a permutation test and need to efficiently generate all
distinct subsets of size r from a set of size n. P 138 of MASS (4th
ed) notes that "The code to generate this efficiently is in the
scripts". I was unable to find this code on quick inspection of the
\library\MASS\scripts file for
Dear all.
After much grief I have finally found the source of some weird
discrepancies in results generated using R. It turns out that this is
due to the way R handles multi-line expressions. Here is an example
with R version 2.8.1:
# R-script.
> From: Christopher David Desjardins
> To: r-help@r-project.org
> Date: Thu, 12 Mar 2009 17:37:26 -0500
> Subject: [R] Unable to run smoother in qplot() or ggplot() - complains about
> knots
> I get the following error when I run qplot()
>
> qplot(grade, read,data = hhm.long.m, geom = c("point",
One option for creating your own palette is
#install.packages('epitools')
mycols <- colors.plot(locator = TRUE)
then left-click on 15 colors of your liking and then right-click 'Stop'.
mycols will be a data.frame with the third column containing the color names.
Kingsford
On Fri, Mar 13, 2009
This is a perfectly legal expression:
f <- a
+ b
+ c;
Type it in a the console, and it will assign a to f and then print out
the values of b and c. In parsing 'f <- a' that is a complete
expression. You may be confused since you think that semicolons
terminate an expression; that is not t
On Fri, 13 Mar 2009, Paul Suckling wrote:
Dear all.
After much grief I have finally found the source of some weird
discrepancies in results generated using R. It turns out that this is
due to the way R handles multi-line expressions. Here is an example
with R version 2.8.1:
---
Did you read the help for cor.test? Test statistics, references
looks pretty complete to me. If the descriptions are too terse,
then the references given would be the next step.
Sarah
Excerpted from ?cor.test
If 'method' is '"pearson"', the test statistic is based on
Pearson's pro
Many thanks yet again for your reply, thanks for that method, i gave it a go
and i checked 'mycols' and sure enough it had selected the chosen colours
and listed their names, but when i used it for making the graph warnigs
informed me that the supplied colour in not numeric or character.
Ross
Paul Suckling wrote:
<...>
>
> # R-script...
>
> r_parse_error <- function ()
> {
>
<...>
> f <- a
> + b
> + c;
> }
>
>
<...>
> f 1
>
>
> As far as I am concerned f should hav
> Thanks for the reply - some of the sets/palettes in the RColorBrewer are
> ideal, but the problem with the problem i have is that they only go up to 12
> colours, and i need 15 colours - so i assume the only thing i can do is
> create my own palette, but i'm having limited success in trying to wo
For permutations a couple of options are 'permutations' in package
gtools, and 'urnsamples' in package prob
hth,
Kingsford Jones
On Fri, Mar 13, 2009 at 6:35 AM, Dale Steele wrote:
> I'm doing a permutation test and need to efficiently generate all
> distinct subsets of size r from a set of size
I get it. Thanks everyone for the feedback.
Now that I understand how it works, my comment would be that this
system is dangerous since it makes it difficult to read the code and
easy to make errors when typing it. I recognise that this is something
so fundamental that it is unlikely to be changed
Hi Oliver,
Thanks very much for your reply. I have tried your script, but when the
script for the graph runs it comes up with several error messages repeating
that 'supplied colour is not numeric or character'
Olivier Delaigue wrote:
>
>> library(colorRamps)
>> image(matrix(1:150, 10), col =
jim holtman wrote:
>
>
>> if (1 == 1) {print (TRUE)
>>
> + } else {print (FALSE)}
> [1] TRUE
>
>
> so the parse knows that the initial 'if' is not complete on the single line.
>
... and likewise the original code could be rewritten as
f <- { a
+ b
+ c }
vQ
__
If all your code has semicolons you could write a program that
puts each statement on one line based on the semicolons and
then passing it through R will reformat it in a standard way.
See Rtidy.bat in the batchfiles distribution for the reformatting part:
http://batchfiles.googlecode.com
On Fri,
On Thu, Mar 12, 2009 at 5:37 PM, Christopher David Desjardins
wrote:
> I get the following error when I run qplot()
>
> qplot(grade, read,data = hhm.long.m, geom = c("point", "smooth"))
>
> Error in smooth.construct.cr.smooth.spec(object, data, knots) :
> x has insufficient unique values to suppo
On Fri, Mar 13, 2009 at 7:19 AM, Ross Culloch wrote:
>
> Many thanks yet again for your reply, thanks for that method, i gave it a go
> and i checked 'mycols' and sure enough it had selected the chosen colours
> and listed their names, but when i used it for making the graph warnigs
> informed me
Hello Dieter,
And thank you for the corrections.
Concerning the point 3, I'm a bit lost. Is it a problem of place to put the
table and the graph side by side (my english is quite as low as my skills in
Latex) ?
I tried with \begin{minipage}{0.45\textwidth} instead of 0.7 and I put
"//tiny" but no
Hi Hadley,
Many thanks for your post. You're not wrong - i'm certainly finding it
challenging, but i assumed it was because i was making some basic errors. My
data are 15 types of behaviour, e.g. resting, alert, locomotion, etc. so i
need to use 15 colours to tell each appart in a barplot which
Hi Kingsford,
Thanks yet again for your help! I have tried this, and once again i have
failed! I have put the code that i've used below (i'm sure you'll note some
bad practice) if that is any use to help explain where i'm going wrong, it
seems to run fine and feeds back just what you noted it wou
Gabor Grothendieck wrote:
> If all your code has semicolons you could write a program that
> puts each statement on one line based on the semicolons and
> then passing it through R will reformat it in a standard way.
> See Rtidy.bat in the batchfiles distribution for the reformatting part:
> http:/
On 13-Mar-09 12:55:35, Paul Suckling wrote:
> Dear all.
> After much grief I have finally found the source of some weird
> discrepancies in results generated using R. It turns out that this is
> due to the way R handles multi-line expressions. Here is an example
> with R version 2.8.1:
>
> ---
On Fri, Mar 13, 2009 at 10:11 AM, Wacek Kusnierczyk
wrote:
> Gabor Grothendieck wrote:
>> If all your code has semicolons you could write a program that
>> puts each statement on one line based on the semicolons and
>> then passing it through R will reformat it in a standard way.
>> See Rtidy.bat
I would argue that this is a matter of preference and the arguments on
"principle" for one side or another are not particularly compelling.
When the "=" was introduced for assignment, an argument was made that
name=value function arguments are also implicitly a kind of assignment.
While Duncan
Gabor Grothendieck wrote:
> On Fri, Mar 13, 2009 at 10:11 AM, Wacek Kusnierczyk
> wrote:
>
>> Gabor Grothendieck wrote:
>>
>>> If all your code has semicolons you could write a program that
>>> puts each statement on one line based on the semicolons and
>>> then passing it through R will r
Dear Ingmar,
Thank you for your reply, I hope I answer your question ---
A couple specific applications I have in mind:
* We work with customers to reduce energy consumption from use of hot
water. Baseline data was gathered at several locations by attaching a
temperature sensor downstream from
Alan Zaslavsky wrote:
> I would argue that this is a matter of preference and the arguments on
> "principle" for one side or another are not particularly compelling.
indeed; i have argued (i think...) for treating them as equals, the
vhoice being a matter of taste.
> When the "=" was introduce
I am trying to update the packages that I have installed but I get the
following warning messages:
package 'tseries' successfully unpacked and MD5 sums checked
Warning: cannot remove prior installation of package 'tseries'
bundle 'forecasting' successfully unpacked and MD5 sums checked
Warning: c
I am trying to update the packages that I have installed but I get the
following warning messages:
package 'tseries' successfully unpacked and MD5 sums checked
Warning: cannot remove prior installation of package 'tseries'
bundle 'forecasting' successfully unpacked and MD5 sums checked
Warning: c
If your subsets are to be taken from the rows of a dataframe, df, and
the size is r, then something like this could satisfy:
apply(combn(nrow(df), r), 2, function(x) df[x, ])
(Although these are not really permutations as I understand that term.)
--
David Winsemius
On Mar 13, 2009, at 8:35 A
Maybe they are loaded, use search() to see if they are. If yes, then
use detach() to unload them first.
Best
2009/3/13 :
> I am trying to update the packages that I have installed but I get the
> following warning messages:
>
> package 'tseries' successfully unpacked and MD5 sums checked
> Warn
thanks for your supportive comments!
by that time r programs will be scanned directly from your head, i
suppose, and the intelligent scanner will as gladly take <- as it will
=, so the problem will rather vanish.
Yes, and maybe the scanner will be more intelligent than the programmer so
when th
Hii Jholtman,
I will make a graph of the median values and not to print to the console.
I tried to plot with:
plot(V3 ~ grup, data = median) ??
but I get an error message.
I also tried to print the output of median<-with(x, tapply(V3, grup,
median)) to a text file with the X and Y Koordinates
> library(ROCR)
> library(randomForest)
> plot(results,significant)
Error in as.double(x) :
cannot coerce type 'S4' to vector of type 'double'
After try with all the possibilities, it shows same error in simpleaffy data
plot for the results.
please someone guide me
wiener30 wrote:
>
> I
Hii,
I will save the elements of the vector median<-with(x, tapply(V3, grup,
median)). The output of this vector is:
25 50 75 100 125 150 175 200 225 250 275 300 325 350 375
400 425 450 475 500
17.8 17.8 17.5 17.8 17.7 17.6 17.7 17.6 17.8 17.7 17.6 17.7 17.8 17.7 17.8
17.
> library(colorRamps)
> image(matrix(1:150, 10), col = blue2green2red(15))
Regards,
Olivier Delaigue
Ross Culloch wrote:
>
> Dear all,
>
> This seems like a simple problem but i've searched the help files and
> tried various options but failed, so apologies in advance for asking what
> i'm s
Sergey,
Sorry I missed your post earlier. I had this same problem and I reinstalled
the DDE Server and Excel Add-In, rebooted and it fixed it. I thought they were
already installed as I could pull down data in Excel from Bloomberg, but
reinstalling them fixed my R problem. You can find the d
Can you provide a reproducible example with the data so that we
understand what you are working with. Need to see what the structure
of 'x' and 'median' are.
On Fri, Mar 13, 2009 at 8:35 AM, johnhj wrote:
>
> Hii Jholtman,
>
> I will make a graph of the median values and not to print to the cons
'str(x)' would help. I assume that this is a named list, so this
might plot it for you:
plot(names(x), x)
On Fri, Mar 13, 2009 at 10:59 AM, johnhj wrote:
>
> Hii,
>
> I will save the elements of the vector median<-with(x, tapply(V3, grup,
> median)). The output of this vector is:
>
> 25 50
Hi,
take a look to
?write.table
regards
2009/3/13 johnhj
>
> Hii,
>
> I will save the elements of the vector median<-with(x, tapply(V3, grup,
> median)). The output of this vector is:
>
> 25 50 75 100 125 150 175 200 225 250 275 300 325 350 375
> 400 425 450 475 500
> 17
Ptit_Bleu yahoo.fr> writes:
> Concerning the point 3, I'm a bit lost. Is it a problem of place to put the
> table and the graph side by side (my english is quite as low as my skills in
> Latex) ?
> I tried with \begin{minipage}{0.45\textwidth} instead of 0.7 and I put
> "//tiny" but no success.
?cat
On Mar 13, 2009, at 10:59 AM, johnhj wrote:
Hii,
I will save the elements of the vector median<-with(x, tapply(V3,
grup,
median)). The output of this vector is:
25 50 75 100 125 150 175 200 225 250 275 300 325 350
375
400 425 450 475 500
17.8 17.8 17.5 17.8 17
Hi, Kent
Thank you for that!
I'll pass the info to our systems administrator, he is the only one
who is allowed to do things like that.
Have a nice weekend!
Best,
Serge
On Fri, Mar 13, 2009 at 15:32, Voss, Kent wrote:
> Sergey,
>
> Sorry I missed your post earlier. I had this same problem and
I did similar things with polygon().
Le mar. 10 mars à 13:30, g...@ucalgary.ca a écrit :
For a given random variable rv, for instance, rv = rnorm(1000),
I plot its density curve and calculate some quantiles:
plot(density(rv))
P10P50P90 = = quantile(rv,probs = c(10,50,90)/100)
I would like to co
Thanks Dieter for the link.
In fact it may be a problem with R.
The .tex created with R for the table put \begin{table}[ht] and \end{table}\
between \begin{minipage} and \end{minipage} (see below)
If I manually change these positions, according to your link, there is no
more error ... but the ta
This may be somewhat useful, but I might have more later.
http://florence.acadiau.ca/collab/hugh_public/index.php?title=R:CheckBinFit
(the code below is copied from the URL above)
CheckBinFit <- function(y,phat,nq=20,new=T,...) {
if(is.factor(y)) y <- as.double(y)
y <- y-mean(y)
y[y>0
Hi, sorry if it is a too stupid question, but how do I a string search
in R:
I have a dataframe A with A$test like:
test1
bcdtestblabla2.1bla
cdtestblablabla3.88blabla
and I want to search for string that start with 'dtest' and ends with
number and return the location of that substring and th
Hi Ross,
If you really need 15 colors, maybe you can use the Set3 palette
provided by RColorBrewer (this is the one used by TraMineR up to 12
states) and add yourself 3 more colors ?
For example (you can mix the hexadecimal color numbers from the
RColorBrewer palette and real color names in
Hi Alexis,
In my opinion you are nothing short of genius! That worked a treat, i
thought it would be something simple, but i could not find the script or an
example anywhere!!!
That was a massive help and has salvaged my day! :clap:
Thanks very much to everyone else that helped, it was much app
Yes, a random variable, discrete or continuous one, should associate with
a probability space and a measurable space.
I thought that graph of density(rv) below could give us an example of a
density function. I am very sorry for confusing you.
My question is how to find/estimate maximum values of a
This is a simple question, but I'm going on the supposition that the
only stupid question is the one not asked.
1. I have many sets of 5 proportions that are different from each
other (prop.test), and want to know which proportions are different
from each other. In other words, I want t
Hi Ross,
If you really need 15 colors, maybe you can use the Set3 palette
provided by RColorBrewer (this is the one used by TraMineR up to 12
states) and add yourself 3 more colors ?
For example (you can mix the hexadecimal color numbers from the
RColorBrewer palette and real color names in
Ok, sorry...
here is my code so far...
x <-read.table(file="D:/output.txt")
x$grup <- 25*rep(1:144, each=5)
median<-with(x, tapply(V3, grup, median))
sd<-with(x, tapply(V3, grup, sd))
I will plot to types of graph. One graph with the median values for example
with the "plot()" function,
Oh, this seemed so simple (and I'm sure the answer will be, as usual, so
thanks in advance for enlightening me). I need to sort each row of a matrix
independent of the others. For example,
> test <- matrix(c(8,7,1,2,6,5,9,4,3),nrow=3)
> test
[,1] [,2] [,3]
[1,]829
[2,]76
Dear:
I am trying to plot x against y for a particular subset of data, say z=1,
and labelling data by another variable say, k.
My data are:
y<-c(69.7, 82.3, 66.3, 107.3, 90.1, 63.7, 82, 74.4, 61.7, 93.4,
73.4, NA, NA, 70.7, 67.7, NA, NA)
x<-c(71.2, 82.6, 67.4, 107.1, 90.5, 66.7, 83.9, 73.9, 61
Thank you all,
I did it with "write.table"...
greetings,
johnh
johnhj wrote:
>
> Hii,
>
> I will save the elements of the vector median<-with(x, tapply(V3, grup,
> median)). The output of this vector is:
>
> 25 50 75 100 125 150 175 200 225 250 275 300 325 350 375
> 40
If you are trying to build your own function then presumably you do
not want the global maximum, since that is trivially returned by max.
So what do you really want? Is this a programming question or just a
general statistics question?
If you want to search along a (specific) sequence for l
On 3/13/2009 12:07 PM, Ptit_Bleu wrote:
Thanks Dieter for the link.
In fact it may be a problem with R.
The .tex created with R for the table put \begin{table}[ht] and \end{table}\
between \begin{minipage} and \end{minipage} (see below)
If I manually change these positions, according to your li
Kevski wrote:
> Oh, this seemed so simple (and I'm sure the answer will be, as usual, so
> thanks in advance for enlightening me). I need to sort each row of a matrix
> independent of the others. For example,
>
>
apply(matrix, 1, sort)
vQ
__
R-he
Now I just need the resulting matrix:
2 8 9
4 6 7
1 3 5
On Mar 13, 2009, at 1:26 PM, Wacek Kusnierczyk wrote:
Kevski wrote:
Oh, this seemed so simple (and I'm sure the answer will be, as
usual, so
thanks in advance for enlightening me). I need to sort each row of
a matrix
independent o
On Fri, Mar 13, 2009 at 5:26 PM, Wacek Kusnierczyk
wrote:
>> Oh, this seemed so simple (and I'm sure the answer will be, as usual, so
>> thanks in advance for enlightening me). I need to sort each row of a matrix
>> independent of the others. For example,
>>
> apply(matrix, 1, sort)
t(apply(te
Thanks for your ideas.
I would like to find all possible maximums - "mountains" on a graph of a
given density function but I have no ideas. In calculus, there is a
general approach for a given function f(x): Find derivative of f(x) and
estimate all zeros of f'(x). These zeros give us locations of m
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