[R] matrix of higher order differences

Is there an easy way to turn a vector of length n into an n by n matrix, in which the diagonal equals the vector, the first off diagonal equals the first order differences, the second... etc. I.e. to do this more efficiently: diffmatrix <- function(x){ n <- length(x); M <- diag(x);

Re: [R] Barplot for degree distribution

Thanks for the info. I have 2 degree distributions that have different degrees. I want both these barplots to have the same axes. Is this possible? I have used xlim and ylim. ylim works fine for both plots But xlim I am not getting the values till 60. And if I give names(dd) <- 0:60 it gives an e

[R] Problem about step and stepAIC

Hello, I am now running a multiple linear regression program, but I do not know the difference between the command step and stepAIC. Thanks. Maggie [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/

Re: [R] sub-matrix block size

Hi: Maybe this can help get you started. Reading your data into a matrix m, m <- structure(c(1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0,

Re: [R] Problems saving an Object called by get

Sorry David, I understand what you mean but could you help with how it would be done more specifically. Thanks On Wed, Apr 27, 2011 at 11:27 AM, David Winsemius wrote: > > On Apr 26, 2011, at 7:37 PM, Luis Felipe Parra wrote: > > Hello. I am trying to save an object which I created using assign

Re: [R] Problems saving an Object called by get

On Apr 26, 2011, at 7:37 PM, Luis Felipe Parra wrote: Hello. I am trying to save an object which I created using assign as following: assign(paste(NombreAlgoritmo,"_Portafolio",sep=""),PortafolioInicial) save(get(paste(Algoritmo,"_Portafolio",sep="")),file=paste(camino,"\ \Libreria\\Portafoli

Re: [R] Barplot for degree distribution

Hi: Firstly, you should have mentioned that you were using the igraph package; with over 2700 R packages available on CRAN, it's unreasonable to expect folks to know to which package a particular function resides, but you may not have been aware of that fact. It turns out that dd1 is a numeric vec

Re: [R] sub-matrix block size

Tena koe Santosh It is not clear to me precisely what your blocking rules are (e.g., where should matrix[4,4] go, is matrix[5:7,5:7] to be considered as a different block to matrix[8:10,8:10]). Also, what happens if there is an isolated 1 (e.g., in location matrix[9,6])? However, I imagine so

Re: [R] density plot of simulated exponential distributed data

Hi: Try this (and note the use of vectorization rather than a loop): rate <- 3 dta <- -log(1 - runif(1000))/rate hist(dta, nclass = 30, probability = TRUE) x <- c(0.001, seq(0, 3, by = 0.01)) lines(x, dexp(x, rate = 3)) This is the difference in timings between the vectorized and iterative metho

Re: [R] How can I extract information from list which class is nls

On Tue, Apr 26, 2011 at 2:21 PM, Schatzi wrote: > How do I extract the standard error of the parameter estimates? > > Also, if I would like to add two parameters together (x+y), can I use this > equation to calculate the new standard error?: > x = parameter 1 > y = parameter 2 > xSE = SE parameter

Re: [R] Help

Is this what you were looking for as output. You did not show what the output would look like: > x var1 var2 X. varN 1 122 nnn1 …1 2 213 nnn2 …2 3 422 nnn4 …2 4 432… …3 5 441… …4 6 500… …4 7 550… …4 > str(x) 'data.frame': 7 obs. of 4

Re: [R] How can I extract information from list which class is nls

> I would generally use the coef() extractor function if > it's available (and it is for nls()). ?nls has an example: > >  coef(summary(fm1DNase1)) I knew about coef() on model objects, but I did not know it had methods for their summaries. What wonderful information! Josh > > which is a matrix

[R] Help

Hey Everyone! I´m a quite new R user .. I found a problem that I'd like to share with you and help me find a solution. I have a large txt. file which I opened with read.table command, and what I understood from many R manuals is that I have a kind of matrix readed with read.table, I've used order

[R] logistic regression: wls and unbalanced samples

Greetings from Rio de Janeiro, Brazil. I am looking for advice / references on binary logistic regression with weighted least squares (using lrm & weights), on the following context: 1) unbalanced sample (n0=1, n1=700); 2) sampling weights used to rebalance the sample (w0=1, w1=14.29); e 3) a

[R] Barplot for degree distribution

In barplot for degree distribution x-axis is not seen. See the example below > g = barabasi.game(500, 0.4) > dd1 = degree.distribution(g) > plot(dd1, xlab="degree", ylab = "frequency") whereas barplot doesnot have any x-axis > barplot(dd1, xlab = "degree", ylab = "frequency") Please see the fig

[R] density plot of simulated exponential distributed data

Hi all, I tried to plot the density curve using the data from simulation. I am sure that the data should be exponentially distributed, but the plot of density curve always starts from (0,0) which is not the case for exponential distribution. Is there any way around this, to keep the curve

Re: [R] Undestanding return()

Yes, if there are no control statements such as "if" that prevent that return function from being activated. --- Jeff Newmiller The . . Go Live... DCN: Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Res

[R] sub-matrix block size

Dear Rxperts Below is a small vector of values of zeros and non-zeros... was wondering if there is an efficient way to get the block sizes of submatrices of a big matrix similar to the one shown below? diagonal elements can be zero too. Rows with only a diagonal element may be considered as a unit

[R] Problems saving an Object called by get

Hello. I am trying to save an object which I created using assign as following: assign(paste(NombreAlgoritmo,"_Portafolio",sep=""),PortafolioInicial) save(get(paste(Algoritmo,"_Portafolio",sep="")),file=paste(camino,"\\Libreria\\Portafolio\\Port_",NombreAlgoritmo,"\\",NombreAlgoritmo,"_Portafolio.

Re: [R] My code is too "loopy"

Hi: One approach is to remove the top two observations from each group. Here's one way: ddply(mydata, .(group), function(d) tail(d, -2)) Now apply the previous procedure to this data subset. HTH, Dennis On Tue, Apr 26, 2011 at 7:18 AM, Dimitri Liakhovitski wrote: > Dennis, this is really grea

[R] calculate probability circles

Hi, All, I want to have the 1%, 2%, 3%, ... contours for Dirichlet distribution. I need the exact contour circles (mathematically) instead of contour plots. Can anyone help me with this? Many thanks, Cindy [[alternative HTML version deleted]] ___

Re: [R] How can I extract information from list which class is nls

On 2011-04-26 15:11, Joshua Wiley wrote: Hi, One way would be: summary(nls.object)[["coefficients"]][, "Std. Error"] If you have a hankering to do it yourself rather than go through the summary formula, the code here will get you there: getAnywhere("summary.nls") If you are going to be doing

Re: [R] How can I extract information from list which class is nls

Hi, One way would be: summary(nls.object)[["coefficients"]][, "Std. Error"] If you have a hankering to do it yourself rather than go through the summary formula, the code here will get you there: getAnywhere("summary.nls") If you are going to be doing it a lot, creating a little function might

Re: [R] Least Squares Method

Hi Vana, I am not sure what package gls() is in off hand, but many model fitting functions have a subset argument. If not, supposing your data is in "dat", and the variable with the zeros in it that are concerning you is "X", then something like: newdat <- dat[dat[, "X"] != 0, ] and now fit gls

[R] Least Squares Method

Hi everyone, I am running the 'gls' command (least squares method) for a number of data out of which many are zeros. I strongly believe that the output is wrong and I think that this is due to the large number of zero values included in my dataset. I would like to ask if there is a command that w

Re: [R] matrix

On Tue, Apr 26, 2011 at 2:28 PM, Val wrote: > Hi all, > > Assume I have a matrix > xv=  [1 0 0 0 0 12, >      0 1 0 0 0 10, >  *    0 0 1 0 0 -9,* >      0 0 0 1 0 20, >    *  0 0 0 0 1 -5]* > > if the last column of "xv"  less than  0 then I want to set zero the entire > row. > The desired output

Re: [R] matrix

Try this: replace(m, m[,ncol(m)] < 0, 0) On Tue, Apr 26, 2011 at 6:28 PM, Val wrote: > Hi all, > > Assume I have a matrix > xv=  [1 0 0 0 0 12, >      0 1 0 0 0 10, >  *    0 0 1 0 0 -9,* >      0 0 0 1 0 20, >    *  0 0 0 0 1 -5]* > > if the last column of "xv"  less than  0 then I want to set

Re: [R] Multi-dimensional non-linear fitting - advice on best method?

On Sun, Apr 24, 2011 at 07:02:48PM -0400, Ravi Varadhan wrote: > Julian, > > You have not specified your problem fully. What is the nature of f? Is f a > scalar function or is it a vector function (2-dim)? It's something like this (only a bit worse): given x, work out alpha from cos(alpha)

[R] matrix

Hi all, Assume I have a matrix xv= [1 0 0 0 0 12, 0 1 0 0 0 10, *0 0 1 0 0 -9,* 0 0 0 1 0 20, * 0 0 0 0 1 -5]* if the last column of "xv" less than 0 then I want to set zero the entire row. The desired output looks like the following. In this case row 3 and row 5 are set

Re: [R] Multi-dimensional non-linear fitting - advice on best method?

On Mon, Apr 25, 2011 at 12:57:46AM +0200, peter dalgaard wrote: > > I have a set of data of the form (x, y1, y2) where x is the > > independent variable and (y1, y2) is the response pair. The model is > > some messy non-linear function: > > > > (y1, y2) = f(x; param1, param2, ..., paramk) + (y1e

Re: [R] Undestanding return()

As soon as you execute the 'return' , the value is returned. What you did not show is did the code have if-then-else to go down separate paths. On Tue, Apr 26, 2011 at 5:29 PM, Bogaso Christofer wrote: > Here, I have following generic function: > > > > Fn1 <- function(x) { > >                ...

[R] Undestanding return()

Here, I have following generic function: Fn1 <- function(x) { return(x) # assume x is calculated in previous steps . return(y) # assume y is calculated in previous steps ..

Re: [R] Normality tests

Hi Bruce, One way is via apply() # some data set.seed(123) X <- matrix(rnorm(100), ncol = 5) X # tests t(apply(X, 2, function(x){ sw <- shapiro.test(x) c(sw$statistic, P = sw$p.value) })) See ?apply and ?str and ?shapiro.test for more information. HTH, Jorge

Re: [R] Normality tests

On Tue, 2011-04-26 at 16:15 -0400, Bruce Kindseth wrote: > I have a large amount of data which I break down into a collection of > vectors of 100-125 values each. I would like to test the normality of the > vectors and compare them. In the interactive mode I can test any one vector > using the Sh

Re: [R] Normality tests

On 2011-04-26 13:15, Bruce Kindseth wrote: I have a large amount of data which I break down into a collection of vectors of 100-125 values each. I would like to test the normality of the vectors and compare them. In the interactive mode I can test any one vector using the Shapiro-Wilk test or t

Re: [R] Tell the difference between characters

Yes. That is what I want. Thank you very much. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Tell-the-difference-between-characters-tp3476130p3476288.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-proje

[R] Normality tests

I have a large amount of data which I break down into a collection of vectors of 100-125 values each. I would like to test the normality of the vectors and compare them. In the interactive mode I can test any one vector using the Shapiro-Wilk test or the Kolmogorov-Smirnov test. My problem is th

Re: [R] Tell the difference between characters

Thanks a lot. Lisa -- View this message in context: http://r.789695.n4.nabble.com/Tell-the-difference-between-characters-tp3476130p3476352.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.

[R] plot() under Rserve

Hi, Rserve allows one to render plots into jpeg with R code such as: jpeg(filename, ...) plot(...) dev.off() However, substituting png, tiff, or bmp device redirection fails on Rserve with "eval error" on OSX. png(filename, ...) plot(...) dev.off() The above in an eval prod

[R] Segmented Error Message

Hi All, I am trying to use the segmented package to determine the break point in a simple quadratic regression. I am receiving the error "(Some) estimated psi out of range" I searched the message board and found others that have had the same issue, but no clear solutions have been provid

[R] Questions about 'heatmap.2' function

I am trying a heatmap. I have 30 individuals and 1,000 genes. Some components I want for this heatmap plot: (1) the 'color-key'; (2) a banner across the top to emphasize how the hierarchical clustering has correctly split the data into the two groups; I use the 'heatmap.2' function in the 'gpl

[R] About snow packages

Dear Luke ! Thanh you for the lovely packages. I have used it, it woks fine for Linux, XP, but concerning about windows 7 - 64 bits. I have problem with function makeCluster(). with R 13.0 version. I wonder it may caused a update problem or ??? Do you have any hint ??? Best Truc ___

Re: [R] Hook into Coercion Framework for data.frame

statconnDCOM, availagble from rcom.univie.ac.at is a (D)COM server accessing R. rcom (available from CRAN) is an R package which can turn R into a COM server or into a COM client. These tools have their own mailing list which can be subscribed at rcom.univie.ac.at. On 4/26/2011 6:20 PM, Von De

Re: [R] separate and arrange outputs

On Tue, 2011-04-26 at 19:46 +0200, ivan wrote: > Hello, > > given that a self made function produces multiple outputs, is there a > possibility to separate latter by stars or simply by blank lines? Are there > funtions for the arrangement of the output in general? You can create a class for your

Re: [R] Using Java methods in R

Thanks Icn for pointing that out, but I don't understand it. My use of .jcall to return a type double scalar or String worked. My use of .jfield to get a one dimensional static array worked. My use of .jfield to get a static scalar constant did not work. My use of .jfield to get a two dimensional s

Re: [R] Tell the difference between characters

Try this: sapply(apply(sapply(temp, charToRaw), 2, unique), length) == 1 On Tue, Apr 26, 2011 at 3:09 PM, Lisa wrote: > Dear all, > > I just want to determine if the characters in a character string are the > same or not. For example, > > temp <- c("aa", "aA", "ab") > > How do I determine the fi

Re: [R] Tell the difference between characters

This will handle varying length strings if you want to test for the same character: > temp <- c("aa", "aA", "ab") > x <- strsplit(temp, '') > x [[1]] [1] "a" "a" [[2]] [1] "a" "A" [[3]] [1] "a" "b" > sapply(x, function(z) all(z[1] == z)) [1] TRUE FALSE FALSE > On Tue, Apr 26, 2011 at 2:09 P

Re: [R] Tell the difference between characters

Hi Lisa, On Tue, Apr 26, 2011 at 2:09 PM, Lisa wrote: > Dear all, > > I just want to determine if the characters in a character string are the > same or not. For example, > > temp <- c("aa", "aA", "ab") > > How do I determine the first one have the two same “a”, and the second and > third have th

Re: [R] Tell the difference between characters

Hi Lisa, Is this what you have in mind? > temp <- c("aa", "aA", "ab") > temp == temp[1] [1] TRUE FALSE FALSE HTH, Jorge On Tue, Apr 26, 2011 at 2:09 PM, Lisa <> wrote: > Dear all, > > I just want to determine if the characters in a character string are the > same or not. For example, > > tem

Re: [R] Predicting with a principal component regression model: "non-conformable arguments" error

Hello again all, I am responding to my own earlier post about a "non-conformable arguments" error with the predict() function of the pls package ( http://cran.r-project.org/web/packages/pls/) in R 2.13.0 (running in Ubuntu 10.10). I believe I have narrowed down the cause of the error. My new unde

[R] Tell the difference between characters

Dear all, I just want to determine if the characters in a character string are the same or not. For example, temp <- c("aa", "aA", "ab") How do I determine the first one have the two same “a”, and the second and third have the different characters? Thanks in advance. Lisa -- View this messa

Re: [R] How can I extract information from list which class is nls

How do I extract the standard error of the parameter estimates? Also, if I would like to add two parameters together (x+y), can I use this equation to calculate the new standard error?: x = parameter 1 y = parameter 2 xSE = SE parameter 1 ySE = SE parameter 2 NewSE=(x+y)*sqrt((xSE/x)^2+(ySE/y)^2)

Re: [R] Factor function

On Apr 26, 2011, at 18:52 , Petr Savicky wrote: > On Tue, Apr 26, 2011 at 10:51:33AM +0200, Petr PIKAL wrote: >> Hi >> >> >> d<-data.frame(matrix(c("ww","ww","xx","yy","ww","yy","xx","yy","NA"), >> ncol=3, byrow=TRUE)) >> >> Change character value "NA" to missing value >> d[d[,3]=="NA",3]<-N

[R] separate and arrange outputs

Hello, given that a self made function produces multiple outputs, is there a possibility to separate latter by stars or simply by blank lines? Are there funtions for the arrangement of the output in general? Thank you. Best Regards [[alternative HTML version deleted]] _

[R] Hook into Coercion Framework for data.frame

Hi, I am looking into a way to hook into the R coercion framework to allow me to convert table-like data stored within a COM object into a data.frame. Some of our COM objects have their own table-like data storage, and from R's point of view it's an object (EXTPTRSXP) decoarated with a sepcial sy

Re: [R] Factor function

On Tue, Apr 26, 2011 at 10:51:33AM +0200, Petr PIKAL wrote: > Hi > > > d<-data.frame(matrix(c("ww","ww","xx","yy","ww","yy","xx","yy","NA"), > ncol=3, byrow=TRUE)) > > Change character value "NA" to missing value > d[d[,3]=="NA",3]<-NA > > If you want drop any unused levels of a factor just u

Re: [R] graphics: 3D regression plane

On Tue, Apr 26, 2011 at 11:18 AM, agent dunham wrote: > Dear community, > > As it's explained I've tried the following: > > model<- lm(log(v1)~log(v2)+v3, data=dat) > newax<- expand.grid( >v2 = seq(min(log(dat$v2)), max(log(dat$v2)), length=100), >v3= seq(min(dat$v3), max(dat$v3), length=1

Re: [R] Second largest element from each matrix row

On 2011-04-26 09:11, William Dunlap wrote: A different approach is to use order() to sort first by row number and then break the ties by value. It is quick when there are lots of short rows. Bill, That's what Dmitris did, too; his solution falls between f3 and f4. Unless speed were paramount,

Re: [R] splitting and reorganising a data.frame

try this: > x nr height age Seed 1 1 4.51 3 301 2 15 10.89 5 301 3 29 28.72 10 301 4 43 41.74 15 301 5 57 52.70 20 301 6 71 60.92 25 301 7 2 4.55 3 303 8 16 10.92 5 303 9 30 29.07 10 303 10 44 42.83 15 303 11 58 53.88 20 303 12 72 63.39 25 30

Re: [R] Second largest element from each matrix row

And I hit the send button before adding the timings for when there were lots of columns and few rows. f3 changes from the best to the worst in this case. There is rarely one most efficient function for all datasets. > x <- t(x) > benchmark(r1 <- f1(x), r2 <- f2(x), r3 <- f3(x), r4 <- f4(x), repl

Re: [R] Second largest element from each matrix row

A different approach is to use order() to sort first by row number and then break the ties by value. It is quick when there are lots of short rows. > f1 <- function (x) +apply(x, 1, function(row) sort(row, decreasing = TRUE)[2]) > f2 <- function (x) + -apply(-x, 1, function(row) sort.in

[R] splitting and reorganising a data.frame

Hey, i have a question about how to reorganize a data frame in the easiest way. my example: what would be the easiest way to bring a data.frame such like this: nr height age Seed 1 1 4.51 3 301 2 15 10.89 5 301 3 29 28.72 10 301 4 43 41.74 15 301 5 57 52.70 20 30

Re: [R] Factor function

Thank you so much! Lisa -- View this message in context: http://r.789695.n4.nabble.com/Factor-function-tp3473984p3475549.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/li

Re: [R] graphics: 3D regression plane

Dear community, As it's explained I've tried the following: model<- lm(log(v1)~log(v2)+v3, data=dat) newax<- expand.grid( v2 = seq(min(log(dat$v2)), max(log(dat$v2)), length=100), v3= seq(min(dat$v3), max(dat$v3), length=100) ) fit <- predict(model,newax) graph <- persp(x=seq(m

[R] Problem with lapply and splitted variables

Dear Community, I have the following two variables, which I have split according to a factor: *y1* [1] 1 2 3 [2] 2 3 4 and *y2* A B [1] 1 4 2 5 3 6 [2] 2 5 3 6 4 7 Now I need the following Vector Autoregressive Models: VAR(cbind(y1[1],y2[1]$A)), VAR(cbind(y1[1],y2[1]$B)), VAR(cbind(

Re: [R] Second largest element from each matrix row

On Apr 26, 2011, at 14:36 , David Winsemius wrote: > > On Apr 26, 2011, at 8:01 AM, Lars Bishop wrote: > >> Hi, >> >> I need to extract the second largest element from each row of a >> matrix. Below is my solution, but I think there should be a more efficient >> way to accomplish the same, or

Re: [R] My code is too "loopy"

I would probably still have to "manually" specify the values for row 1 and row 2 - and then loop through groups. Something like: mydata[!is.na(mydata$myvalue),"new.value"]<-mydata[!is.na(mydata$myvalue),"myvalue"]*0.5 # this calculates the values for row 1 h <- function(x) embed(x, 3) %*% c(0.5,

Re: [R] Second largest element from each matrix row

Thanks a lot to all of you! On Tue, Apr 26, 2011 at 10:34 AM, Peter Ehlers wrote: > On 2011-04-26 05:26, Dimitris Rizopoulos wrote: > >> one way is the following: >> >> a<- matrix(rnorm(9), 3 ,3) >> >> aa<- a[order(row(a), -a)] >> matrix(aa, nrow(a), byrow = TRUE)[, 2] >> > > That's clever, Dmit

Re: [R] Second largest element from each matrix row

On 2011-04-26 05:26, Dimitris Rizopoulos wrote: one way is the following: a<- matrix(rnorm(9), 3 ,3) aa<- a[order(row(a), -a)] matrix(aa, nrow(a), byrow = TRUE)[, 2] That's clever, Dmitris. And very fast! Lars: my apologies; I didn't read your request carefully. You asked for more _efficient

Re: [R] Running Fortran code from R

On 26/04/2011 8:19 AM, A.Noufaily wrote: Dear R users, I have a Fortran code that I would like to compile and call from R later. I have never worked with Fortran before. Does anyone know the steps to create Fortran DLLs for R on a Windows PC. Is anyone aware of a manual (or does anyone know how

Re: [R] R plot : hat symbol and cex.lab

On 26/04/2011 8:01 AM, BMichel wrote: Hello, Does anybody know how to make the "hat" correctly appears in the label of this plot (with this cex.lab coefficient) : plot(1:10, 1:10,ylab = expression(hat(h)),cex.lab = 1.5) The "hat" does not completely appear on my graph, it is like cut on the le

Re: [R] My code is too "loopy"

Dennis, this is really great, thanks a lot! Do you know how to prevent the result from omitting the first 2 values. I mean - it starts (within each group) with the 3rd row but omits the first 2... Dimitri On Mon, Apr 25, 2011 at 5:31 PM, Dennis Murphy wrote: > Hi: > > I think the embed() function

Re: [R] R plot : hat symbol and cex.lab

plot(1:10, 1:10, ylab="") mtext(side=2, expression(hat(h)),cex = 1.5, line=2.5) Rich On Tue, Apr 26, 2011 at 7:45 AM, BMichel wrote: > Hello, > > Does anybody know how to make the "hat" correctly appears in the label of > this plot (with this cex.lab coefficient) : > > plot(1:10, 1:10,ylab = expr

[R] VIP extraction

Hi, Does anyone know how to extract the VIP (Variable influence on Projection) from a Partial Least Squares model? I'm using Package PLS. Thanks in Advance Tommy -- View this message in context: http://r.789695.n4.nabble.com/VIP-extraction-tp3475099p3475099.html Sent from the R help mailing

[R] R plot : hat symbol and cex.lab

Hello, Does anybody know how to make the "hat" correctly appears in the label of this plot (with this cex.lab coefficient) : plot(1:10, 1:10,ylab = expression(hat(h)),cex.lab = 1.5) The "hat" does not completely appear on my graph, it is like cut on the left side. It tried to change the margin :

Re: [R] RStudio -manipulate command

require(quantmod) getSymbols("GLD") library(manipulate) manipulate( plot(Cl(GLD), xlim=c(0,x.max)), x.max=slider(10,100)) Why the graph is not plotting? thanks veepsirtt -- View this message in context: http://r.789695.n4.nabble.com/RStudio-manipulate-command-tp3474947p3474976.html Sent f

[R] RStudio -manipulate command

Hello, How to plot the time series values of YHOO symbol?, getSymbols("YHOO",src="google") library(manipulate) manipulate( plot(YHOO, xlim=c(0,x.max)), x.max=slider(10,250)) while running this code I am getting this message Warning message: In plot.xts(YHOO, xlim = c(0, x.max)) : only t

[R] R plot : hat symbol and cex.lab

Hello, Does anybody know how to make the "hat" correctly appears in the label of this plot (with this cex.lab coefficient) : plot(1:10, 1:10,ylab = expression(hat(h)),cex.lab = 1.5) The "hat" does not completely appear on my graph, it is like cut on the left side. It tried to change the margin :

[R] Running Fortran code from R

Dear R users, I have a Fortran code that I would like to compile and call from R later. I have never worked with Fortran before. Does anyone know the steps to create Fortran DLLs for R on a Windows PC. Is anyone aware of a manual (or does anyone know how to) that explains: - What tools

Re: [R] what's wrong with plot(..., type="p")?

Dear Bob, Thanks for your answer! I didn't think factors would be a problem for plot() Ivan Le 4/26/2011 12:21, Bob O'Hara a écrit : On 26 April 2011 12:08, Ivan Calandra > wrote: Dear users, I'm trying to get a dot plot but always end up with a

[R] Immutable ticks using ts.plot?

Hello all-   What I'm going for here is a stack of four time series plots that use a common X axis on the bottom plot. And setting up customized tick marks on each plot that help illustrate the respective trends. Here's my code:   Start fig_data <- ts(read.csv("F:/mydata.csv"), start=20

Re: [R] Second largest element from each matrix row

On Apr 26, 2011, at 8:01 AM, Lars Bishop wrote: Hi, I need to extract the second largest element from each row of a matrix. Below is my solution, but I think there should be a more efficient way to accomplish the same, or not? set.seed(1) a <- matrix(rnorm(9), 3 ,3) sec.large <- as.vector

Re: [R] Second largest element from each matrix row

On 2011-04-26 05:01, Lars Bishop wrote: Hi, I need to extract the second largest element from each row of a matrix. Below is my solution, but I think there should be a more efficient way to accomplish the same, or not? set.seed(1) a<- matrix(rnorm(9), 3 ,3) sec.large<- as.vector(apply(a,

Re: [R] Second largest element from each matrix row

would this work (shorter)? apply(a, 1, function(x) x[order(x)[2]]) On Tue, Apr 26, 2011 at 5:31 PM, Lars Bishop wrote: > Hi, > > I need to extract the second largest element from each row of a > matrix. Below is my solution, but I think there should be a more efficient > way to accomplish the s

Re: [R] Second largest element from each matrix row

one way is the following: a <- matrix(rnorm(9), 3 ,3) aa <- a[order(row(a), -a)] matrix(aa, nrow(a), byrow = TRUE)[, 2] I hope it helps. Best, Dimitris On 4/26/2011 2:01 PM, Lars Bishop wrote: Hi, I need to extract the second largest element from each row of a matrix. Below is my solution

[R] Second largest element from each matrix row

Hi, I need to extract the second largest element from each row of a matrix. Below is my solution, but I think there should be a more efficient way to accomplish the same, or not? set.seed(1) a <- matrix(rnorm(9), 3 ,3) sec.large <- as.vector(apply(a, 1, order, decreasing=T)[2,]) ans <- sappl

Re: [R] what's wrong with plot(..., type="p")?

On 2011-04-26 03:08, Ivan Calandra wrote: Dear users, I'm trying to get a dot plot but always end up with a boxplot. Can someone please tell me what I am doing wrong? df<- structure(list(FACETTE = structure(c(1L, 1L, 1L, 1L, 2L, 2L, + 2L, 2L), .Label = c("base", "tip"), class = "factor"), Sq =

Re: [R] what's wrong with plot(..., type="p")?

On 26 April 2011 12:08, Ivan Calandra wrote: > Dear users, > > I'm trying to get a dot plot but always end up with a boxplot. Can someone > please tell me what I am doing wrong? > > df <- structure(list(FACETTE = structure(c(1L, 1L, 1L, 1L, 2L, 2L, > + 2L, 2L), .Label = c("base", "tip"), class =

[R] what's wrong with plot(..., type="p")?

Dear users, I'm trying to get a dot plot but always end up with a boxplot. Can someone please tell me what I am doing wrong? df <- structure(list(FACETTE = structure(c(1L, 1L, 1L, 1L, 2L, 2L, + 2L, 2L), .Label = c("base", "tip"), class = "factor"), Sq = c(274836, + 0.74182, 0.709205, 0.984552,

Re: [R] Factor function

Hi d<-data.frame(matrix(c("ww","ww","xx","yy","ww","yy","xx","yy","NA"), ncol=3, byrow=TRUE)) Change character value "NA" to missing value d[d[,3]=="NA",3]<-NA If you want drop any unused levels of a factor just use factor(d[,3]) [1] xx yy Levels: xx yy Regards Petr r-help-boun...@r-