similar to: Plot matrix

Suppose that I have a matrix like: m <- rbind(c(1,2,3,4), c(2,3,2,1)) Is there any way to efficiently plot the _lines_ as if I was doing: plot(m[1,], type="l") points(m[2,], type="l", col="red") (of course, in the "real world" there much more than just 2 lines and 4 columns...) Alberto Monteiro

Dear list, I have the following matrix. How can I make the following plot? 1. The x-axis has index 1:7, and the first column is plotted against index 1, second against 2, and so on. 2. I want the points from the left upper conner including the antidiagonal to be plotted with col=2, and the rest with col=3 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,] 0.589 0.857 0.923 0.944 0.954 0.963

Hi all, I am not a skillful R programmer and has I am handling with large dataframes (about 30000 x 300) I am in need of an efficient function. I have 4 dataframes with the same dimension. I need to generate other dataframe with the some dimension than the others where in each position it has the median value of the 4 values in the same position coming from the 4 dataframes. Grateful by your

I want to create a list based on the information from a data.frame, Model. So I tried the following: MyList <- list(colnames(Model)[2] = levels(Model$(colnames(Model)[2]))) but it failed with an error: Error: unexpected '=' in "list(colnames(Model)[2] =" I have the following problems with this command line: (1) I wanted to use colnames(Model)[2] as a tag for the list:

Dear I try to add three different lines (solid, short dash, long dash) in to current barplot. I saw there are types of lines ("p","b", ect). However, they are not what I request. Has nayone have this experience? Many Thanks! Xin [[alternative HTML version deleted]]

Hi, the R code: a <- matrix(c(1,5,4,3,7,10,34,4,3,8,6,5,12,17,45,3,2,45,46,47,3,4,22,12,21), nrow=5) rownames(a) <- c("a","b","c","d","e") a a[which(a[,3] < 8), ] a[which(a[,3] < 6), ] produces the following output: > a [,1] [,2] [,3] [,4] [,5] a 1 10 6 3 3 b 5 34 5 2 4 c 4 4 12 45 22 d

Hi, Is there are function similar to excel vlookup in R. Please let me know. Thanks, Sachin ____________________________________________________________________________________ [[alternative HTML version deleted]]

Hi list I loaded the content of a file dureetasks.txt to variable a. This file contains an interger per line. when I print a vector, it displays correctly. however, when I try to print the histogram, I get this error message > a=read.table("dureetasks.txt") > hist(a) Error in hist.default(a) : 'x' must be numeric Can you help please? regards -- PhD candidate in

Hi, I have a table of x rows and y columns. The table is huge and so i'd like to create a subset of the data containing rows where any of the y values are below a threshold, say 1e-4. Is there a simple way of doing this in R? thanks Rich

Hi, I have two dataframes one with 144 rows and 160 columns (SDF1) and one with 12 rows and 160 columns (SDF2). Now I'm trying to divide rows 1:12 with SDF2, rows 13:24 with SDF2, rows 25:36 with SDF 2, . In S-Plus the following code works fine: DFS = SDF1[1:144,1:60] / as.vector(SDF2[1:12,1:160]) but in R when I try to implement the formula I get the following error: "/

Dear mailing list members, I am a new R user, I would like to plot the follewing data var1 <- c(1,2,1,1,2,1,2,1,2,2) var2 <- round(rgamma(10,2,1)/0.1)*0.1 var3 <- c(0,1,0,1,0,0,0,0,1,0) var4 <- c(1,2,2,2,1,1,1,1,1,1) Var <- data.frame(var1,var2,var3,var4) Var <- Var[sort(Var$var1),] tt <- Var$var1+((runif(length(Var$var1))/6)-(0.5/6)) labelname <- c("time 1",

Hi, I am trying to plot two distribution graph on one plot. But I dont know how. I set them to the same x, y limit, even same x, y labels. Code: x1=rnorm(25, mean=0, sd=1) y1=dnorm(x1, mean=0, sd=1) x2=rnorm(25, mean=0, sd=1) y2=dnorm(x2, mean=0, sd=1) plot(x1, y1, type='p', xlim=range(x1,x2), ylim=range(y1, y2), xlab='x', ylab='y') plot(x2, y2, type='p',

Hello R list, I have two matrices of identical dimensions, and I want to fit a straight line for each pair of columns and plot the resulting lines. I got it to work with a for loop, but there must be a better way, > n<-5 > N<-10 > > data.x<-matrix(1:(n*N),ncol=n) > data.y<-matrix(1:(n*N) + rnorm(n*N,sd=1),ncol=n) > >

Hi everybody, I have the following problem. I have a vector containing character elements, such as: list = c("aa","bb","cc","dd","ee") I want to create an index which identifies the elements that are different from, e.g. "aa" and "bb". When I do the following: jj = list!="aa" & list!="bb" > jj

Hi, Is there a way in R to get the string which matches the expression, where the expression is a substring of the parent string. Lets say, I have $i <- "transcript:ENST0000112334 pid:ENSP000012345" What I need is the string "pid:ENSP000012345" from $i using the query "ENSP". Appreciate your comments. Praveen Surendran School of Medicine and

I'm importing a lot of text tables of data (from Latent Gold) that includes hashes in some of the column names ("Cluster#1", "Cluster#2", etc.). Is there an easy way to strip the offending hashes out before pushing the text into a table or data frame? I thought I'd use gsub, e.g., but can't figure out how to read in a text file without reading it into a table or

hello, If I wanna export data.frame from R to html have you got some ideas to do this? ___________________________________________________________________________ [[alternative HTML version deleted]]

Hi, I like to know the simple way to list the R package names in our linux system. Thanks Ying [[alternative HTML version deleted]]

Hi All, is there a function in R that allows me to work with fractions without transforming them to floats (or whatever) in between? Something that would calculate something like: (1/2 + 1/8) * 1/2 = 5/16 without ever transforming to 0.5 and 0.125? Best, Federico -- Federico C. F. Calboli Department of Epidemiology and Public Health Imperial College, St Mary's Campus Norfolk Place,

Dear group, Here is my data frame: > dput(u) structure(list(DESCRIPTION = structure(c(2L, 5L, 6L, 7L, 9L, 11L, 12L, 15L, 14L, 16L, 1L, 10L, 3L, 4L, 13L, 8L, 17L), .Label = c("COFFEE C Jul/10", "COPPER May/10", "CORN Jul/10", "CORN May/10", "COTTON NO.2 Jul/10", "CRUDE OIL miNY May/10", "GOLD Jun/10", "HENRY HUB