similar to: help with plotting lines linking two sets of points?

Hello, I'd have a time series, where I am plotting the means and sd of a distance for a variety of positions along a bird's bill. I'd like to set each line (represented by "point") to start at zero, so that I can look at the absolute change along the series. At the moment I only know how to do that in Excel, by subtracting the value of time 1, point 1 from all other times

hello everyone. please bear with me if this is very easy... I have a data set with many trials, and frames within each trial. I would like to pull out one random frame from each trial. here is an example. i have 4 unique trials (file), and various frames within each (time_pred). I would like to randomly sample 4 rows, but 1 from each trial (file). this sample data is called "h"

Hello I have data of body length and body weight of people of different skin colors. I tried to write a code to plot body length and body weight according to the skin colors. (Thanks for Petr's advice so far.) A loop is used but an error shows up in the following code. It says: unexpected '}' in "

Hello R experts I have two linear regressions for sexes (Male, Female, Unknown). All have a good correlation between body length (response variable) and head length (explanatory variable). I know it is not recommended, but for a good practical reason (the purpose of study is to find a single conversion factor from head length to body length), the regressions need to go through the origin (0

Hello I want to draw a plot using the code below. Further, I want to add labels (White, Yellow, Red) to x-axis. Please kindly advise how to add the categorical labels, by modifying the code. Thank you. Elaine Code dataN <-read.csv("H:/skin_color.csv",header=T, row.names=1) dim(dataN) dataN[1,] str(dataN) obs.group<-dataN$skin_color

Dear All, I want to create a surface plot from the data. My data set is consists of x, y and z data. I plotted in very easy way by Excel worksheet as shown in the attached picture. I did some steps in R, but I cannot have the same plot as in Excel worksheet's figure. the R code is x <- c(-1,-0.75,-0.5,-0.25,0,0.25,0.5,0.75,1) y<- c(-1,-0.75,-0.5,-0.25,0,0.25,0.5,0.75,1) z<-

Hi I have two processes which take with a certain probability (p1 and p2) x number of years to complete (age1 and age2). As soon as thge first process is completed, the second one begins. I want to calculate the time it takes for the both processes to be completed. I have the following script which gives me the answer, butI think there must be a more elegant way of doing the calculations

Hello, Simply put, I'm trying to call a function "testme" with value "age=NA". I wish to use "dotlist<-list(...)" inside the function and have dotlist become: $age [1] NA I'm modifying existing code and need to minimize changing that code so it's easiest to conform how I call the existing function. My sample code fragment, results, and R.version

Hi everyone, I have posted a similar question to this list, but I don't get a reply. I really want to solve this problem, so I post it again... I am trying to use R to fit some mixed-effects models for a nested data. The data is a simulated data with 111 subjects. Each subject has 6 waves' data. Below are the first two subjects' data : > simu1[1:12,] Grouped Data: gf ~ age |

Dear list I'm using the reshape command and want to reshape a wide data set to a long one e.g. I have the variables y1,y2,y3,age1,age2,age3,sex,ethnic I want my new long data set to consist of the variables y (which has been created from y1,y2,y3), age (which has been created from age1,age2,age3), sex and ethnic I have tried to use the command:

Dear R-Users, I used the box.tidwell () function of the car Package. When I used the following formula: semi.sub.in.mi1.boxtidwell_h<-box.tidwell(RENT_LG ~ I(age+1)+I(age2+1)+X06A + I(X08B+1) + I(X22+1) + I(X24+1) + X31A, ~B_YEAR + C_X01 + C_X14 + C_X19 + C_X29A +C_X21 + C_X23 + D_X12 + D_X17 + D_X18 + D_X25 + D_X27 + D_X30 + D_X32 + D_X35, data = semi.sub.in.mi1) everything is

I'm trying to run an ordered logistic regression model. I've run the following code, but the output does not provide the p-values. Is there some command to include the p-values in the output. reg2 <- polr(trade1 ~ age2 + education2 + personal2 + economy2 + partisan2 + employment2 + union2 + home2 + market2 + race2 + income2) summary(reg2) Re-fitting to get Hessian# Call:

R-helpers; I used Proc Model in SAS to fit the following model to data: proc model data = dbsmv; a = a1*F**2; b = b1*F + b2*T + b3*F*T; tph2 = tph1 *((1 - exp(-a*age2)) / (1 - exp(-a*age)))**-b; fit tph2; and yielded the following estimated parameters after iterations: a1 = -0.15943, a2 = -1.8177, b1 = -0.01911, b2

Below you will find the output from a failed multi-level model run. I am trying to estimate the following model: Pr(PLFP=1)= logistic regression -> B1_j * bm + B2_j * wm + B3_j bf + B4_j wf + B5 yrsed+ B6 age+ B7 age^2+e_ij B1_j = G01 + G11 bmxd + d1 B2_j = G02 + G12 wmxd + d2 B3_j = G03 + G13 bfxd + d3 B4_j = G04 + G14 wfxd + d4 d1-d4 freely correlated Note that there is no

Dear all, I would like to know whether it is possible to unlist elements and keep the original format of the data. To make it more clear, let me give an exemple: I have a list l of dataframes that I created with apply but which looks like this: x1=data.frame(Name=LETTERS[1:2],Age=1:2) x2=data.frame(Name=LETTERS[3:4],Age=3:4) l=list(x1,x2) l [[1]] Name Age 1 A 1 2 B 2 [[2]] Name

Greetings to you all, I am performing a semi parametric bootstrap in R on a Gamma Distributed data and a Binomial distributed data. The main challenge am facing is the fact that the residual variance depends on the mean (if I am correct). I strongly feel that the script below may be wrong due to mean-variance relationship #####R code####### fit1s

Hi, Is there a simple command for computing the standard error of a combination of parameter estimates in a GLM? For example: riskdata$age1 <- riskdata$age riskdata$age2 <- ifelse(riskdata$age<67,0,riskdata$age-67) model <- glm(death~age1+age2+ldl, data=riskdata,family=binomial(link=logit)) And we want to find the standard error of the partial linear predictor for

Hi, i'am with Wine 1.3.15 and i had install Dragon Age 2 succesful, but the game didn't want start, Wine crash. I make some search and i found that, the folder "actvation.exe" tell :Runtime Error! Program: C:Program Files/Core/activation.exe R6034 An application has made an attempt to load the C runtime library incorrectly. I want to know if i can do something with this?

Hi, I am just wondering if there is a test available for testing if a linear fit of an independent variable in a Cox regression is enough? Thanks for any suggestions. John Zhang ____________________________________________________________________________________ [[elided Yahoo spam]]

Hi all, I am hoping someone can help me out with this: If I have dataframe of years and ages and the first column and first row are filled with leading values: Df<- age1 age2 age3 Yr1 1 0.4 0.16 Yr2 1.5 0 0 Yr3 0.9 0 0 Yr4 1 0 0 Yr5 1.2 0 0 Yr6 1.4 0 0 Yr7 0.8 0 0 Yr8 0.6 0 0 Yr9 1.1 0 0 Now the rest of the cells need to be filled according to the previous year and age