similar to: sum part of a vector

Hello Dear R community,  This is my problem.  I have a data set (dataframe) called "mydat". It consist of 3 numerical variable.  They are Centrecode, FSUSN and Round. I want to create unique ID by combining these 3 variables. Follwing commands gives me what I need. mydat1 <- paste(mydat$Centrecode, mydat$FSUSN,mydat$Round,sep="") newds <- data.frame(mydat1)    For a

Dear all, I have got two vectors coding for a stimulus presented in the current trial (mydat$Stimulus) and a prediction in the same trial (mydat$Prediciton), respectively. By applying an if-conditional I want to create a new vector that indicates if there is a match between both vectors in the same trial. That is, if the prediction equals the stimulus. When I pick out some trials randomly, I get

Thank you Jim, I read the data as you suggested but I could not find K1 in col1. rbind(preval,mydat) Col1 Col2 col3 1 <NA> <NA> <NA> 2 X1 <NA> <NA> 3 Y1 <NA> <NA> 4 K2 <NA> <NA> 5 W1 <NA> <NA> 6 Z1 K1 K2 7 Z2 <NA> <NA> 8 Z3 X1 <NA> 9 Z4 Y1 W1 On Sat, Feb 24, 2018 at 6:18 PM, Jim

Hi Val, My fault - I assumed that the NA would be first in the result produced by "unique": mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) val23<-unique(unlist(mydat[,c("Col2","col3")])) napos<-which(is.na(val23)) preval<-data.frame(Col1=val23[-napos],

Dear List, I am trying to plot date vs. time, but am having problems getting my y-axis labels how I want them.? When left on its own R plots time at 6 hour intervals from 03:00 to 23:00.? I am wanting 6 hour intervals from 2:00 to 22:00.? I realize yaxp doesn't work in plot(), so I am trying to get it to work in par().? However, now I get the ticks where I want them but the time is output

HI Jim and all, I want to put one more condition. Include col2 and col3 if they are not in col1. Here is the data mydat <- read.table(textConnection("Col1 Col2 col3 K2 X1 NA Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) The desired out put would be Col1 Col2 col3 1 X1 0 0 2 K1 0 0 3 Y1 0 0 4 W1 0 0 6 K2 X1

Hi, Is there a way to build up a vector, item by item. In perl, we can "push" an item onto an array. How can we can do this in R? I have a loop that generates values as it goes. I want to end up with a vector of all the loop results. In perl it woud be: for(item in list){ result <- 2*item^2 (Or whatever formula, this is just a pseudo example) Push(@result_list,

Dear R forum, The example below is just an indicative one and I have constructed it. My real life data and conditions are different. I have a data.frame as given below mydat = data.frame(A = c(19, 20, 19, 19, 19, 18, 16, 18, 19, 20), B = c(19, 20, 20, 19, 20, 18, 19, 18, 17, 16)) if (length(mydat$A) > 10) { stop("A has length more than 10") }else if (max(mydat$B) > 18) {

Sorry , I hit the send key accidentally here is my complete message. Thank you Jim and all, I got it. I have one more question on the original question What does this "[-1] " do? preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1], Col2=NA,col3=NA) mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2

Dear 'HTH' R friends I have a small dataframe as given below. I need to sort this database based on date in a decending order. I am not sure whether I have defined the date column in a proper format. mydat<-data.frame(date = (c("1/31/2010", "2/28/2010", "3/31/2010", "4/30/2010", "5/31/2010", "6/30/2010",

Dear all R-users, (My apologies if this subject is wrong) I have dataset: mydat <- as.data.frame( matrix(c(14,0,1,0,1,1, 25,1,1,0,1,1, 5,0,0,1,1,0, 31,1,1,1,1,1, 10,0,0,0,0,1), nrow=5,ncol=6,byrow=TRUE)) dimnames(mydat)[[2]]

Jim has been exceedingly patient (and may well continue to be so), but this smells like "failure to launch". At what point will you start showing your (failed) attempts at solving your own problems so we can help you work on your specific weaknesses and become self-sufficient? -- Sent from my phone. Please excuse my brevity. On February 25, 2018 7:55:55 AM PST, Val <valkremk at

Dear R helpers Suppose I have a data frame as given below mydat = data.frame(x = c(1,1,1, 2, 2, 2, 2, 2, 5, 5, 6), y = c(10, 10, 10, 8, 8, 8, 7, 7, 2, 2, 4)) mydat         x     y 1      1     10 2      1     10 3      1     10 4      2       8 5      2       8 6      2       8 7      2       7 8      2       7 9      5       2 10    5       2 11    6       4 unique(mydat$x) will give me 1,

hi Val, Your problem seems to be that the data are read in as a factor. The simplest way I can think of to get around this is: mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1], Col2=NA,col3=NA) rbind(preval,mydat)

On 24/02/2018 1:53 PM, William Dunlap via R-help wrote: > x1 = rbind(unique(preval),mydat) > x2 <- x1[is.na(x1)] <- 0 > x2 # gives 0 > > Why introduce the 'x2'? x1[...] <- 0 alters x1 in place and I think that > altered x1 is what you want. > > You asked why x2 was zero. The value of the expression > f(a) <- b > and assignments

Hi I am attempting to use the lattice bwplot function to generate boxplots of numerous parameters (1-panel/parameter) by site (x-axis). The parameters have quite different ranges of values, so it would be best to have a separate y-axis range for each panel. Below is a basic example of what I am trying to do. As is seen, the y-axes need to be scaled individually to make this useful. Any

Dear all, Suppose I have a dataset like below, then I take for example, 100 random sample "class" variable where contains "yes" and "no" respectively, 70% and 30%. I need a new 100 random sample from mydat dataset, but I can't get the result. Thanks you very much for any helps. Best, Muhammad Subianto mydat <- data.frame(size=c(30,12,15,10,12,12,25,30,20,14),

Hello everyone, A recent (in 2.5 I suspect) change in R is giving me trouble. I want to apply a function (tolower) to all the columns of a data.frame and get a data.frame in return. Currently, on a data.frame, both apply (for arrays) and lapply (for lists) work, but each returns its native class (resp. matrix and list): apply(mydat,2,tolower) # gives a matrix lapply(mydat,tolower) # gives

When I import a simple dataset, run LDA, and then try to use the model to forecast out of sample data, I get a forecast for the training set not the out of sample set. Others have posted this question, but I do not see the answers to their posts. Here is some sample data: Date Names v1 v2 v3 c1 1/31/2009 Name1 0.714472361 0.902552278 0.783353694 a 1/31/2009 Name2 0.512158919 0.770451596

Thank you Jim and all, I got it. I have one more question on the original question What does this "[-1] " do? preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1], Col2=NA,col3=NA) mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE)