similar to: Adding info from summary(lm(...)) to plot

... and the mailing list is picky about attachments... whatever you attached did not conform to the stringent requirements mentioned in the Posting Guide. Pasting the code right into the email is usually safest, though you DO have to post using plain text (as the Posting Guide indicates) or your code may get mangled by the automatic html format removal. On May 15, 2018 7:04:31 AM PDT, Bert Gunter

Unless there is good reason not to, always cc the list -- there are lots of smarter folks than I on it who can help. I may or may not have time to look at this. Hopefully someone else will. -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip

OK, Let's try this again! Here is the reproducible script; it is long because I had to copy the panel dataset here. My question is related to systemfit; I don't know how to get the result for the entire panel. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) install.packages("systemfit")

Sadly you failed to set your email program to send plain text and the data is corrupted at my end. I also think you need to reduce the size of the data set... the intent here is to increase your understanding, not debug your particular analysis. I will say that I am having a very challenging time understanding what you are trying to accomplish though. What are the equations that you think need

I can't get my simultaneous equations to work using system fit. Please help. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) str(Empdata) Empdata$gnipc<-as.numeric(Empdata$gnipc) install.packages("systemfit") library("systemfit") pdata <- plm.data(Empdata,

Hi r-users,   I would like to extract the data that match.  Attached is my data: I'm interested in matchind the value in column 'intg' with value in column 'rand_no' > cbind(z=z,intg=dd,rand_no = rr)             z  intg rand_no    [1,]  0.00 0.000   0.001    [2,]  0.01 0.000   0.002    [3,]  0.02 0.000   0.002    [4,]  0.03 0.000   0.003    [5,]  0.04 0.000   0.003    [6,] 

Dear all, I have a table with a column named "type" and I ''d like to access the type field of the correspondant model object but I get a deprecation warning when I use the following code : modelObject.type "type deprecated use class instead" How can I access such a named field in rails ? Johan Johan Duflost Analyst Programmer Belgian Biodiversity Platform (

All, I want to run automated unit tests as part of my build. I am building to my test environment. Upon successful completion of all of my unit tests, I would then like to load some data (using a fixture, I imagine) into my test database that will act as fresh "system test" data for my users to play with. What is the best way to "load the standard system test/user acceptance

I would like to validate a number within a certain range. To first test that a entry is a valid number i use validates_numericality_of :my_number, :message =>''sorry, not a number'' to test if the number is within a certain range i use validates_inclusion of :my_number, :in => min..max, message => ''outside range!'' If the numericality test fails, I

Hi Folks, I'm trying to find a neat solution to an apparently simple problem, but one which turns out to be a bit more intricate and tricky than one might expect. Suppose I have numbers given to a large number of digits. For example 1234567021 where (though I don't know this beforehand) only the last 3 digits will be varying (and all 3 will vary). What I want is, give a vector x of

Hi Hadley and everyone, here's a patch for ggplot2 that fixes the behavior of opts(legend.position={left,top,bottom}). If you try the following code in an unmodified ggplot2 options(warn = -1) suppressPackageStartupMessages(library("ggplot2")) data <- data.frame( x = c(1, 2, 3, 4, 5, 6), y = c(2, 3, 4, 3, 4, 5), colour = c(TRUE, TRUE, TRUE, FALSE, FALSE, FALSE))

I know that this maybe a trivial question. I am not able to plot pedices in graph axes. Instead I am able to plot different math symbols : XLABEL <- expression(paste(cmH,lim(f(x), x %->% 0),"O PEEP")) works well XLABEL <- expression(paste(cmH,[2],"O PEEP")) is considered a wrong expression. Thanks

Dear all, I'm working with Cox-regression, because data could be censored. But in this particular case not. Now I have a simple example: PRO and PRE are (0,1) coded. The response is not normal distributed. We are interested in a model which could describe interaction. But my results are depending strongly in the choose of the contrast option. It is clear that there is some dependence in

Good day all, I'm trying to plot a continuous line plot, which is followed by two forecast points eg. one forecast point is 12 months out, and another 24 months out from the last date of the line plot. In my attempts so far, the second plot (the forecast points) is scaled against a new axis scale, thus the two plots are not directly comparable (I need the forecast points to be scaled

Dear all, I am trying to plot a matrix I have? as an image str(matrixToPlot) ?num [1:21, 1:66] 0 0 0 0 0 0 0 0 0 0 . ?that contains only 0s and 1s, where the xlabel will be Labeled as str(xLabel) ?num [1:66] 1e+09 1e+09 1e+09 1e+09 1e+09 ... and the yLabels will be labeled as str(yLabel) ?num [1:21] -88 -87 -86 -85 -84 -83 -82 -81 -80 -79 ... I have found on the internet that I can do

Hi Kenneth, > Well...strictly as LLVM IR I find externally visible incorrect behavior > unlikely, it's just a "different definition". For C and C++, I'd be > looking at more complicated variations of > > int main() > { > volatile int i = 1; > return 0; > } > > It's clear that the LLVM IR representation of i cannot be simply >

Dear users, I'm trying to produce a 3d bar plot but the x and y dimensions have categorical data -- so I only want 3 points on each axis. So I try: require(scatterplot3d) mymat<-data.frame( x=c(1,1,1,2,2,2,3,3,3), y=c(1,2,3,1,2,3,1,2,3), z=c(1,2,3,4,5,6,7,8,9)) scatterplot3d(mymat, type="h", lwd=5, pch=" ", xlab="xlabel",

Hi! How to find out which plot function is used when i call plot(hist.default(1:10,plot=F)) and all works fine ? The reason why I would like to know it is that after loading some self written R functions > plot(hist.default(1:10,plot=F)) Error in xy.coords(x, y, xlabel, ylabel, log) : x and y lengths differ > traceback() 5: stop("x and y lengths differ") 4: xy.coords(x, y,

Dear All: good morning I am trying to graph the function y=f(x)=1/x over the interval (-5,5). But I am getting an error message. Please see below. I am getting the error message: *Error in xy.coords(x, y, xlabel, ylabel, log) : * * 'x' and 'y' lengths differ* x x <- seq(-5, 5, 0.01) y < 1/x plot(x,y, type='l', xlim=c(-5, 5), ylim=c(-5, 5), xlab = "x",

I don't usually do much with graphs in R, and this is my first time adding a line of best fit. Hopefully this is an easy problem to solve. I'm looking at a variable called soloKills along the range 5:28. Here are all my commands, in script form: range=5:28 graph=soloKills title="Solo kill/death data" xlabel="Number of deaths/1 game" ylabel="Mean number of kills/1