similar to: plotting

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, 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 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

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

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 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 Hanna, your function is essentially zero outside a short interval around 9. And the help page states: "If the function is approximately constant (in particular, zero) over nearly all its range it is possible that the result and error estimate may be seriously wrong." 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

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

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,

> 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 tried to work with par(mpg). > It does not

For a data frame, we can add an additional row or column easily. For example, we can add an additional row of zero and an additional row of column as follows. Is there an easy and similar way to add zeros in each dimension? For example, for array(1:12, dim=c(2,2,3))? Thanks for your help!! Hanna > x <- as.data.frame(matrix(1:20,4,5))> x[5,] <- 0> x[,6] <- 0> x V1 V2

With this data x <- c(0,0,1,1,2,2) y <- c(5,6,4,3,2,6) lwr <- y-1 upr <- y+1 xlab <- c("Low","Low","Med","Med","High","High") mydata <- data.frame(x,xlab,y,lwr,upr) I would like to make a dot plot and use lwr and upr as error bars. Above 0=Low. I would like there to be some space between the 5 and the 6 corresponding

Dear all, I got an error message when running the following code. Can anyone give any suggestions on fixing this type of error? Thank you very much in advance. Hanna > integrand <- function(x, rho, a, b, z){ + x1 <- x[1] + x2 <- x[2] + Sigma <- matrix(c(1, rho, rho, 1), 2,2) + mu <- rep(0,2) + f <-

Dear all, I want to fit a random effect model with only one random factor. I do not want to include the intercept term either. The model I using now is lmer(values ~ (1|lot), data=tmp) The results are as below. How do I take out the intercept term? Or if this is not possible for the lmer function, is it possible using lme function in the "nlme" package? Thank you very much in

Hi all, Is there a way to change the color of the boxplot plots outliers? Thanks. Hanna [[alternative HTML version deleted]]

Hi all, In my plotCI function, the argument x is chosen to be seq(0.05, 0.95, by=0.05). However, when I make the plot, the plot has the x coordinate goes 1:19. Does anyone know how to make the x coordinate to be (0,0.5, 0.1, ..., 0.95). Thank you. Hanna [[alternative HTML version deleted]]

Most likely, previous computations have ended up giving slightly different 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:

Dear all, The function f() below is a function of m1 and m2, both of which are matrices with 3 rows. The function works sequentially one row after another. So altogether there are three stages. I am trying to update the coding to write a generic function that will work for arbitrary k stages. I am hoping to get some suggestion and help. Thanks so much! Hanna ##x, y are two