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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 "fact...

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

Hi alL, I have a data frame with 4 columns: "value", "time", "group" and "id". I would like to plot "value" vs. "time" with different colors for different levels of "group" and different symbols for different values of "id". I think I could do this but I would like to see what is an easier way to plot the

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Hi all, I am trying to download the Rcolorbrewer package from Cran http://cran.r-project.org/web/packages/RColorBrewer//index.html It seems the files have been removed. Does anyone know where can I download the package? Thanks. Hannah [[alternative HTML version deleted]]

Have you tried P2["20", "10", "0"] ? Jean On Thu, Jun 1, 2017 at 3:10 PM, li li <hannah.hlx at gmail.com> wrote: > Hi all, > I have a three dimensional array with the corresponding dimension names. > I would like to subset the array according to the dimension names. For > example, > suppose I want to extract the values corresponding to A=20, B=10, C=0. I > know I...

Sorry. I meant in the previous email that the function h() is a monotone decreasing function. Thanks very much. 2018-02-06 13:32 GMT-05:00 li li <hannah.hlx at gmail.com>: > Hi all, > The function h below is a function of c and it should be a monotone > increasing function since the integrand is nonnegative and integral is > taken from c to infinity. However, as we can see from the plot, it is not > shown to be monotone. Something...

Hi all, I have a three dimensional array with the corresponding dimension names. I would like to subset the array according to the dimension names. For example, suppose I want to extract the values corresponding to A=20, B=10, C=0. I know I can do: P2[dimnames(P2)$A==20, dimnames(P2)$B==10, dimnames(P2)$C==0] But is there a better way for doing this? Thanks for your help! Hanna >

Hi all, When making a bunch of plots using a loop function, how to add title to reflect different plots. Specifically, for the code below, I generated 9 plots. I would like to add a title to each plot. For example, the titles will be respectively, plot1, plot 2, … plot 9. Thank you very much! Hanna par(mfrow=c(3,3), pty="s", pch=16, col="blue") for ( i in 1:9) { lm1

Hi all, The function h below is a function of c and it should be a monotone increasing function since the integrand is nonnegative and integral is taken from c to infinity. However, as we can see from the plot, it is not shown to be monotone. Something wrong with the usage of integrate function? Thanks so much for your help. Hanna h <- function(c){ g <- function(x){pnorm(x-8.8,

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.

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

..." You could try to integrate over a finite interval, say (7, 12). G?ran Brostr?m On 2018-02-06 19:40, li li wrote: > Sorry. I meant in the previous email that the function h() is a monotone > decreasing function. Thanks very much. > > 2018-02-06 13:32 GMT-05:00 li li <hannah.hlx at gmail.com>: > >> Hi all, >> The function h below is a function of c and it should be a monotone >> increasing function since the integrand is nonnegative and integral is >> taken from c to infinity. However, as we can see from the plot, it is not >> shown...

> On 1 Jun 2017, at 03:41, li li <hannah.hlx at gmail.com> wrote: > > Hi all, > I have a question with regard to making plots using function > "scatterplot3d". > Please see the example below. It looks like, for y axis, the tickmark text > was cutoff. > The number "10" does not show up completely....

I have the joint distribution of three discrete random variables z1, z2 and z3 which is captured by "z" and "prob" as described below. For example, the probability for z1=0.46667, z2=-1 and z3=-1 is 2.752e-13. Also, the probability adds up to 1. > head(z) z1 z2 z3 [1,] -0.46667 -1.0000 -1.0000 [2,] -0.33333 -0.9333 -0.9333 [3,] -0.20000 -0.8667 -0.8667

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 >

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,

Hi all, I have a question with regard to making plots using function "scatterplot3d". Please see the example below. It looks like, for y axis, the tickmark text was cutoff. The number "10" does not show up completely. I tried to work with par(mpg). It does not seem to work. Hope to get some advice here. Thanks much! Hanna C <- runif(30) B <- rep(1:3, each=10) A

Hi all. I was having some trouble with a for loop and I found the problem is the following. Does anyone have some idea why I got the following R result? Since mone is equal to 3, why mu1 only have 2 components? library(MASS) > p0 <- seq(0.1, 0.9,by=0.1) > m <- 10 > > > p0 <- p0[7] > > ## data generation > > mzero <- p0*m > mone <- m-mzero >

...rent values of say 0.13333. A pragmatic way out is to round to, say, 5 digits before applying unique. In this particular case, it seems like all numbers are multiples of 1/30, so another idea could be to multiply by 30, round, and divide by 30. -pd > On 28 Jul 2017, at 17:17 , li li <hannah.hlx at gmail.com> wrote: > > I have the joint distribution of three discrete random variables z1, z2 and > z3 which is captured by "z" > and "prob" as described below. > > For example, the probability for z1=0.46667, z2=-1 and z3=-1 is 2.752e-13. > Also, th...