similar to: matrix

I have a matrix, called data. I used the code below to rearrange the data such that the first column remains the same, but the y value falls under either columns 2, 3 or 4, depending on the value of z. If z=1 for example, then the value of y will fall under column 2, if z=2, the value of y falls under column 3, and so on. data x y z [1,] 50 13 1 [2,] 14 8 2 [3,] 3 7 3 [4,] 4 16 1 [5,] 6

I am trying to generate a line graph with quarterly time buckets (with nice labels) on the x-axis. The first block of code below will generate the graph with nicely formatted x-axis labels, but the "type=" and "col=" options are not recognized when factors are used for the x-axis. The second block, where the quarter values are mapped into dates, will plot the line nicely but

Hi Bill, I put stringsAsFactors = FALSE still did not work. tdat <- read.table(textConnection("A B C Y A12 B03 C04 0.70 A23 B05 C06 0.05 A14 B06 C07 1.20 A25 A23 A12 3.51 A16 A25 A14 2,16"),header = TRUE ,stringsAsFactors = FALSE) tdat$D <- 0 tdat$E <- 0 tdat$D <- (ifelse(tdat$B %in% tdat$A, tdat$A[tdat$B], 0)) tdat$E <- (ifelse(tdat$B %in% tdat$A, tdat$A[tdat$C], 0))

Dear R-users, I am using R version 2.10.1 under windows. In a barplot, I want to mark one of the bars with a special border color. For example: barplot(c(3, 7, 11), border = c(NA, "red", NA)) But how to do this when the bars are stacked? for example: barplot(matrix(1:6, ncol=3)) # border of second bar (i.e. the one with total height = 7) should be red again, I try: barplot(matrix(1:6,

Thank you Rui, I did not get the desired result. Here is the output from your script A B C Y D E 1 A12 B03 C04 0.70 0 0 2 A23 B05 C06 0.05 0 0 3 A14 B06 C07 1.20 0 0 4 A25 A23 A12 3.51 1 1 5 A16 A25 A14 2,16 4 4 On Wed, Dec 13, 2017 at 4:36 PM, Rui Barradas <ruipbarradas at sapo.pt> wrote: > Hello, > > Here is one way. > > tdat$D <- ifelse(tdat$B %in% tdat$A,

Hello All, I am trying to create a column of weights based off of factor levels from another column. I am using the weights to calculate L scores. Here is an example where the first column are scores, the second is my "factor" and the third I want to be a column of weights. I can do what I want with an ifelse statement (see below), but I am wondering if anyone knows of a cleaner way

I'm trying to categorize a continuous variable (yes, I know that's horrible, but I'm trying to reproduce some exercises from a textbook) and don't really know an efficient way to do this. I have a data frame that looks like: surv_time relapse sex log_WBC rx 1 35 0 1 1.45 0 2 34 0 1 1.47 0 3 32 0 1 2.20 0 4 32

Hi, Is there any equivalent for ifelse (except if (cond) expr1 else expr2) which takes an atomic element as argument but returns vector since ifelse returns an object of the same length as its argument? x = c(1,2,3) y = c(4,5,6,7) z = 3 ifelse(z <= 3,x,y) would return x and not 1 thanks

Hi, I can use the following code to generate a matrix, each column of which is 'x'. But I have to specify '5' twice in the second command. I am wondering if there is a better way to do it. > x=1:10 > matrix(rep(x,5),nc=5) > t(matrix(rep(x,5),nc=5)) Regards, Peng

I'm new to R (previously used SAS primarily) and I have a genetics data frame consisting of genotypes for each of 300+ subjects (ID1, ID2, ID3, ...) at 3000+ genetic locations (SNP1, SNP2, SNP3...). A small subset of the data is shown below: SNP_ID SNP1 SNP2 SNP3 SNP4 Maj_Allele C G C A Min_Allele T A T G ID1 CC GG CT AA ID2 CC GG CC AA ID3 CC GG nc AA

R users, My intention is to take factors out of DF, create list of tables and export these tables separately using write.table or sink function. write.table writes tables out as DF:s, should I use sink instead? Here is my example: a <- data.frame( indx = 1:20, var1 = rep(c("I20", "I40", "A50", "B60"), each=5),

Thank you Jeff and All, Within a given time period (say 700 days, from the start day), I am expecting measurements taken at each time interval;. In this case "0" means measurement taken, "1" not taken (stopped or opted out and " -1" don't consider that time period for that individual. This will be compared with the actual measurements taken (Observed-

Use the stringsAsFactors=FALSE argument to read.table when making your data.frame - factors are getting in your way here. Bill Dunlap TIBCO Software wdunlap tibco.com On Wed, Dec 13, 2017 at 3:02 PM, Val <valkremk at gmail.com> wrote: > Thank you Rui, > I did not get the desired result. Here is the output from your script > > A B C Y D E > 1 A12 B03 C04 0.70 0 0

# read.table is NOT part of the data.table package #library(data.table) DFM <- read.table( text= 'obs start end 1 2/1/2015 1/1/2017 2 4/11/2010 1/1/2011 3 1/4/2006 5/3/2007 4 10/1/2007 1/1/2008 5 6/1/2011 1/1/2012 6 10/5/2004 12/1/2004 ',header = TRUE, stringsAsFactors = FALSE) # cleaner way to compute D DFM$start <- as.Date( DFM$start, format="%m/%d/%Y" ) DFM$end

Hello, does anyone knows this predict is not resulting? # lm predict dfTestes3sitesCriptic$Velocity_corrected <- ifelse (dfTestes3sitesCriptic$Season == "A" & dfTestes3sitesCriptic$BeforeAfter == "Before", (dfTestes3sitesCriptic$Velocity * mVel3ABefAfter), (ifelse (dfTestes3sitesCriptic$Season == "Sp" & dfTestes3sitesCriptic$BeforeAfter ==

Hi all, Assume the following function that generate a random number. z1<-function (n, eta) { wv <- runif(n) wz <- (-1/eta) * log(wv) wz } y <- z1(100,4) mean(y) I want to run this function say 1000 times and I want to count if the mean(y) outside the following range 0.20 to 0.30. How do I do it in R? Thanks in advance

Dear All! I'm looking for equivalent of Matematica function "Which" which works as follows: z = Which[x<10,0.3, 10<=x<20,0.5, 20<=x<100,1] where x is a vector I can replace it with custom function with set of ifelse but I'm looking for simpler and faster (much faster) solution best wishes Jarek

Dear R users, I have a problem since I try to plot my datas with different colors. plot(tvar, var, xlab="zeit [s]",ylab="Variation [%]", col = ifelse(var <= varstability, 'green','red')) this works well! But since I add a type="l" to my plot, it will color all the plot with green!!! Is there any solution? I avoid to use teachingDemos. Thanks. --

Since the number of choices is small (6), how about this? Starting with Jeff's initial DFM: DFM <- structure(list(obs = 1:6, start = structure(c(16467, 14710, 13152, 13787, 15126, 12696), class = "Date"), end = structure(c(17167, 14975, 13636, 13879, 15340, 12753), class = "Date"), D = c(700, 265, 484, 92, 214, 57), bin = structure(c(6L, 3L, 5L, 1L, 3L, 1L), .Label

Hi everyone, I am at my wits end with what I believe would be considered simple by a more experienced R user. I want to know how to add a variable to a dataframe whose values?are conditional?on the values of an existing?variable.?I can't seem to make an ifelse statement work?for my situation.?The existing variable?in my dataframe is?a character variable named DOW which contains abbreviated