similar to: for loop

Dear all, I have a function f(x) which return a list as result. $T1 [1] 0.03376190 $T2 [1] 0.04725 $T3 [1] 0.3796071 $T4 [1] 0.3713452 $T5 [1] 0.4523651 $T6 [1] 0.4575873 I now find the result for a vector of x values at one time. I want to store the reuslt for each xi value in a column of a matrix x <- seq(0,1, by=0.1) result <- matrix(0, nrow=6, ncol=length(x)) for (i in

I used RJS extensively in my last web app with no problem. I''m just beginning a new one and having a very odd problem. The called method, get_names, retrieves some values from the database, then ends: render :layout => false the get_names.rjs file is rendered into Javascript, but then is just dumped as Javascript (text) into the browser: e.g: try {

All, I''ve written a simple job/opportunity tracking database app via RoR. I have about 10 internal users. Sometimes 2 people will want to update the same record at more or less the same time and they collide. In particular user 1 may edit a record, then get distracted, and finally an hour later they click save. During that hour user 2 may have pulled up the record, edited it, and

Dear all, For two sets of random variables, say, x <- rnorm(1000, 10, 10) and y <- rnorm(1000. 3, 20). Is there any way to overlay the histograms (and density curves) of x and y on the plot of y vs. x? The histogram of x is on the x axis and that of y is on the y axis. The density curve here is to approximate the shape of the distribution and does not have to have area 1. Thank you

Hello, How much would you charge, got a contact email I can get you on. Thanks Hannah Sent from my iPhone > On 15 Nov 2015, at 04:54, Scott Winterstein <0turn1 at gmail.com> wrote: > > I can help... is this a free need or can you pay a little for the programming... im good and Im cheap > > Sehr geehrte Damen und Herren, > > Hallo, > > > Vielen Dank

Also, it is possible to change "ylim" also? 2012/7/1 li li <hannah.hlx@gmail.com> > Dear all, > I have a few questions regarding the boxplot output from the > "geom_boxplot" function. > Attached is the output I get. Below are my questions: > > 1. How can I define the xlab and ylab myself? > Also I would like to remove

Hello, Im new to ice cast and was wondering how I would go about setting it up to allow multiple DJ's for different locations.I would like to have an auto dj playing and then Dj's want to broastcast the auto dj stops, and when the dos stop the auto dj kicks back in straight away. If anyone has any idea on how to do this that would be great! Thanks Hannah -------------- next

Is there an easy way to make graphs for the following data. I have pretest and posttest scores for men and women. I would like to form a 'titlted segment' plot for the data. That is, make segments joining the scores, with different types of segments for men and women. Example data: menpre <- c(43,42,26,39,60,60,46) menpost <- c(40,41,36,42,54,58,43) womenpre <-

Dear all, I need some help on plotting multiple boxplots on one figure. I have three matrix A, B and C. Each of them is a 1000 by 10 matrix. The 10 columns of all three matrix correspond to the 10 values of the same parameter, say k=1, ..., 10. I want to make a plot where x axis represents different values of k. For each k value, I want to plot three boxplots, one on top of another. For

We used jplayer, on the web site side AMR.fm On January 4, 2016 11:00:15 AM "Marvin Scholz" <epirat07 at gmail.com> wrote: > On 4 Jan 2016, at 17:23, Hannah Carroll wrote: > >> Hello,I would like to add different listening options to my website >> but I having a bit of difficulty doing this.I would like the options >> of iTunes, windows media player and

Hannah, thanks so much - - -how do I log into administration. Just go the icecast website and type in my url? Regards Gary Hudson Home: 817-710-6367 Cell: 0959476691 From: icecast-bounces at xiph.org [mailto:icecast-bounces at xiph.org] On Behalf Of Hannah Carroll Sent: Tuesday, February 16, 2016 9:39 AM To: Icecast streaming server user discussions Subject: Re: [Icecast] listener

On 4 Jan 2016, at 23:41, Hannah Carroll wrote: > Hello,can you expand a bit more on how its not possible.How have those > websites managed it then? They simply have not. When I click those, it does not open in the player it says there. And it makes no sense imo to have this kind of links, it's better to have a playlist link like the m3u Icecast provides and the user will have the

Hi all, I want to add 6 plots in the format of 2 columns and 3 rows as one figure in latex. The plots are in .eps file. I know how to add 2 plots side by side, but could not figure out how to do multiple rows. I know this may not be the right place to ask such a question. But I do not know who to ask, so just try my luck here. Thank you in advance.

Hi, Here are three plots: library(ggplot2) data(diamonds) randind <- sample(nrow(diamonds),1000,replace=FALSE) dsmall <- diamonds[randind,] qplot(carat, data=dsmall, geom="histogram",binwidth=1) qplot(carat, data=dsmall, geom="histogram",binwidth=.1) qplot(carat, data=dsmall, geom="histogram",binwidth=.01) What are ways to put these three plots on a single

Hi all, I am have some difficulty with the legend function. I need to add a legend to describe the different line types in a plot. The legend box is small. It did not include sufficient length of each line type to help distinguish the differnt line types. Is there a way to fix this. Thank you Hannah [[alternative HTML version deleted]]

All, For my understanding, I wanted to see if I can get glmnet predictions using both the predict function and also by multiplying coefficients by the variable matrix. This is not worked out. Could anyone suggest where I am going wrong? I understand that I may not have the mean/intercept correct, but the scaling is also off, which suggests a bigger mistake. Thanks for your help. Juliet Hannah

Using this data as an example dat <- read.table(textConnection("Id myvar 12 1 12 2 12 6 34 9 34 4 34 8 65 15 65 23"), header = TRUE) closeAllConnections() how can I create another data set that does not have duplicate entries for 'Id', but the included values are randomly selected from the available ones. Thanks! Juliet

Hi all, I have a function rho.f which gives a list of estimators. I have the following problems. rho.f(0.3) gives me the right answer. However, if I use rho.f(corr[4]) give me a different answer, even though corr[4]==0.3. This prevents me from using a for loop. Can someone give me some help? Thank you very much in advance. Hannah >

Have you got a messenger or email? I live in the uk and it will cost a lot to call. Sent from my iPhone > On 15 Nov 2015, at 21:48, Scott Winterstein <0turn1 at gmail.com> wrote: > > well its unclear what all is needed, I however am very open and fair. I work hard to earn my money... I can call you and we can chat. > > > Sehr geehrte Damen und Herren, > >

Dear all, I have the following problem with the uniroot function. I want to find roots for the fucntion "Fp2" which is defined as below. Fz <- function(z){0.8*pnorm(z)+p1*pnorm(z-u1)+(0.2-p1)*pnorm(z-u2)} Fp <- function(t){(1-Fz(abs(qnorm(1-(t/2)))))+(Fz(-abs(qnorm(1-(t/2)))))} Fp2 <- function(t) {Fp(t)-0.8*t/alpha} th <- uniroot(Fp2, lower =0, upper =1,