similar to: Ylim

Hello all, I have this simple barplot code: ifn <- "id.dat" dat <- read.table(ifn) ofn <- "id.png" bitmap(ofn, type = "png256", width = 30, height = 30, pointsize = 30, bg = "white",res=50) par(mar=c(5, 5, 3, 2),lwd=5) par(cex.main=1.6,cex.lab=1.6,cex.axis=1.6) names(dat)<-c("NumberOfPeople","Average")

Hello All: Could anybody tell me please what is the meaning of this error message: ESP Ghostscript 815.02: Unrecoverable error, exit code 1 Error in plot.xy(xy, type, ...) : ignoring SIGPIPE signal Calls: plot -> plot.default -> plot.xy Execution halted My code has to go to 18 loops for three times and in each loop it has to read a file which is 100-300 Megs. It always crashes at loop 14

Hello All: I am having the following error message when I increase the resolution ("res=30" to "res=350") of my graph: Error in plot.xy(xy, type, ...) : ignoring SIGPIPE signal Calls: plot -> plot.default -> plot.xy Execution halted Here is the code: ofn <- "MAY11.png" bitmap(ofn, type = "png256", width = 21, height = 30, pointsize = 30, bg =

Hello All: I am having the following data file named "data.dat" Number of people Number of pets Age of trees ifn <- "data.dat" dat <- read.table(ifn) colnames(dat)<-c("Variables") I want R to read all these in a string but when I ask R to read these, it gives me error because there more than one word in each line. Any comments to solve this problem?

Hello there, I have the following info in MyData: 01 0.40 02 0.40 03 0.40 04 0.35 05 0.34 06 0.33 I want to read them using the following code: ifn <- "MyData" dat <- read.table(ifn) names(dat1)<-c("Code","M") names(dat2)<-c("Code","M") I want to have the first three lines in

Hello All: Can somebody please tell me how I can have the average of 'y' on the graph for different values of 'x in the following code? ofn <- "xy.png" bitmap(ofn, type = "png256", width = 21, height = 30, pointsize = 30, bg = "white",res=50) ifn <- sprintf("xy.txt") data <- read.table(ifn) names(data) <-

Hello All: If I do have: x = (2, 4, 5, 5, 6, 4, 5, 2, 1) y = (9, 11.5, 12.5, 13, 14, 19, 20, 21, 22) I wanted to find a simple function in R which calculates the averages of X and Y for every 4 unit increase of Y. The results should look like: x = (4, 6, 3) #where (2+ 4+ 5+ 5)/ 4 = 4 and ... y = (12, 14, 20.5) #where (9+11.5+12.5+13)/ 4 = 12 and ... Thanks, Mohsen

Dear all, I want to design a histogram and I need to have the frequency at certain points. For example I have the following 2 columns: *X Y* 0.1 25 0.4 22 0.45 11 0.55 21 I want the chart to have 4 columns. First column is from 0.0-0.1 (on X) and frequency is 25. Next colum is wider and form 0.1-0.4 with 22 frequency. Next column is narrow with 11 frequency and the last column is

Hello everyone, I am using the following program to get the p-value of some numbers (column 'LR' of the data.dat file). I want to write the 1st and 2nd column of the output file (data.out) as an integer while the program change them to double. Could anybody please tell me how I can write the code which writes the values of the first two columns as integer? Thanks library ('MASS')

Greetings! I would like to be able to specify a fixed (say) lower limit for plotting, while leaving the upper limit "floating, when plotting. The context is that the maximum in the data to be plotted is unpredictable, being the consequence of a simulation, whereas I know that it cannot be less than (say) 0; and I want to fix the lower limit at 0 in any plot, leaving the upper limit to be

Hello All: I was wondering why many plot options are not working with mhtplot. I could not find much info about this function on the web too. Any comments, any detailed manual ... Thanks, MJK [[alternative HTML version deleted]]

Hello all, I am doing a Likelihood Ratio (LR) test in my simulation and I have a vector LR values (each with 1 degree of freedom) at the end of my simulation. Can anybody tell me how I can write a 'R' code which gives me the p-value for each of those LR values. Thanks [[alternative HTML version deleted]]

Hello All, I am drawing a graph having 18 small graphs inside using par(mfrow = c(6,3)) command. My problem is how to specify the margins of the whole 18 graphs. I used par(mar=c(6.5, 6.5, 1.5, 1.5)) for each graph separately already but it does not left any margins for the 'mtext()' for the margins of the whole 18 graphs. Any comments please. Thanks, MJK [[alternative HTML version

Hello, I have used the following data to draw my barplot: BL LR Q 36.35 1.00 1.92 36.91 4.00 0.00 25.70 6.00 0.00 34.38 3.00 1.92 05.32 0.50 0.00 BL<-c(36.35, 36.91, 25.70, 34.38, 05.32) LR<-c(1.00, 4.00, 6.00, 3.00, 0.50) Q<-<(1.92, 0.00, 0.00, 1.92, 0.00) barplot(dt$LR, main='LR Value', col='orange', border='black',

Hi, I wanted to compute the value of the function ifn at certain values of n. But I am receiving the following error when I was using the following code(given at the end). Error: evaluation nested too deeply: infinite recursion / options(expressions=)? I feel that since the function Grx is recursively related, perhaps making the code too complicated too handle. Can anyone let me know if

Hello everyone, I am using the following program to get the p-value of some numbers (column 'LR' of the data.dat file). I want to write the 1st and 2nd column of the output file (data.out) as an integer while the program change them. Could anybody please tell me how I can write the code which writes the values of the first two columns as integer? For your convenience, I have attached the

Hi, I wanted to compute the value of the function ifn at certain values of n. But I am receiving the following error when I was using the following code(given at the end). Error: evaluation nested too deeply: infinite recursion / options(expressions=)? I feel that since the function Grx is recursively related, perhaps making the code too complicated too handle. Can anyone let me know if

Hi, I want to plot grouped bars to compare 95% confidence interval estimates from two models. Each bar represents a 95% confidence interval estimate of a coefficient from one of the two models. Each group represents confidence interval estimates of the same coefficient from the two models. I think such a bar plot will nicely present whether 95% confidence interval estimates of the same

Hi R fans, As a newbie following the five-hour rule (after hitting my head against the wall for five hours, post to this list), I am appealing for some help understanding geom_rect() in ggplot2. What I want to do is very simple. I want to generate a plot of rectangles. Each one represents a business cycle. The x-values will be pairs representing the start and end of each cycle. The y-values

Hello I have created a graph using the following commands: <<< startBReP3O1T <- diffs$BReP3O1T - diffs$diff_BReP3O1T endBReP3O1T <- diffs$BReP3O1T x <- seq(47,89, length = 10) ymin <- min(min(startBReP3O1T), min(endBReP3O1T)) ymax <- max(max(startBReP3O1T), max(endBReP3O1T)) y <- seq(ymin, ymax, length = 10) plot(x,y, type = 'n', xlab = 'Age', ylab =