similar to: Computing row differences in new columns

Hi all, I have a data set with time interval and depending on the interval I want to create 5 more variables . Sample data below 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 6,10/15/2004,12/1/2004 First, I want get interval between the start date and end dates (End-start). obs, Start , end, datediff

Ii is difficult to provide useful help, because you have failed to read and follow the posting guide. In particular: 1. Plain text, not HTML. 2. Use dput() or provide code to create your example. Text printouts such as that which you gave require some work to wrangle into into an example that we can test. Specifically: 3. Have you gone through any R tutorials?-- it sure doesn't look like

Thank you all for the useful suggestion. I did some of my homework. library(data.table) DFM <- read.table(header=TRUE, 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',stringsAsFactors = FALSE) DFM DFM$D =as.numeric(difftime(as.Date(DFM$end,format="%m/%d/%Y"),

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-

# 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

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

Hello-- There must be a better way to do this. I have a class method in my model that finds averages and does a few calculations using find_by_sql. The problem I¹m encountering is that all computed columns from MySQL come back as type string. E.g., def self.find_averages(domain_id) if @@domain_average return @@domain_average else @@domain_average =

Hi all, I have the following problem: I have a csv-file consisting of timestamp values (no dates), e.g.: Timestamp1;Timestamp2; 05:24:43;05:25:05; 15:47:02;15:47:22; 18:36:05;18:36:24; 15:21:24;15:22:04; I need a vector with the difference of the two timestamps, so I read the data with the read.csv-function: myObj <- read.csv("file.csv",header=TRUE,sep=";"). I have then

Hi, I have a dataset with 6 categorical variables. I have used this following code to make the variables u1-u6 ordered factors and this works well. cat1cat2 cat3 cat4 cat5 cat6 ? 0 ? ?? 1 ? ? 1????? 0 ??? 0? ?? 1 ? 1 ? ?? 1 ? ? 0 ? ?? 0 ? ? 0 ? ? 0 ....... .... ############ data<-read,table("example.txt") data <- as.data.frame(lapply(data, ordered)) ############ Now,

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Hello, I'd like to use DBI to store lm objects in a database. I've to analyze many of linear models and I cannot store them in a single R-session (not enough memory). Also it'd be nice to have them persistent. Maybe it's possible to create a compact binary representation of the object (the kind of format created created by "save"), so that one doesn't need to write

hi guys I need to interpolate values for the zero coupon yield curve. Following data is given date days rate 1996 01

I'm new with R. I want to plot 3 lines in one graph. This is my data: print(x) V1 V2 V3 V41 -4800 25195.73 7415.219 7264.282 -2800 15195.73 5415.219 7264.28 I tried using matplot, but I cannot get exactly what I want. This is what I get, and this is my code: matplot(x[,1],x[,-1],type='b', xlab = "epsilon_h", ylab = "Value2", xlim=

I would like to subset data from a larger dataset and generate a smaller dataset. However, I don't want to use sample() because it does it randomly. I would like to take non-random subsamples, for example, every 2nd number, or every 3rd number. Is there a procedure that does this? Thanks, Nate -- View this message in context:

A couple of days ago I set up my printer on my FreeBSD server and set it up so I could print from my FreeBSD client. This all works very well, printing from botht the server and the FBSD client. However, I also have samba set up for file serving with an XP box. I have been trying for quite some time now to set samba up to enable me to print from the XP client. My situation at the moment is that

Hi I am trying to process tabular data as follows: Data in the input file is of the form genome1 genome2 tree-dist log10escore Genome1 and genome2 are alphabetic. Tree-dist and log10escore are numeric. I wish to extract only those rows from this table where the log10escore is less than -3. data <-read.table(filename); data$log10escore = data$log10escore[ data$log10escore < -3]; I

When I try writing the current device to a jpeg file I keep getting a gray background for the jpeg images using dev.print. I have tried using the option bg="white" and bg="#FFFFFF" but neither seem to have any effect. If instead of writing to a screen device I open a jpeg device I can get any background color I want. This works, but I prefer to write to the screen and then

There seems to be a minor problem with reproducing numbers from rnorm with Box-Muller. The pattern suggests it might have something to do with the value that gets dropped when an odd number of numbers is requested. (Details below.) Also, could "user-supplied" be added as an option for normal.kind in RNGkind. I'm sure the Box-Muller in R is better than my own attempt, but I would

Hi all, I'd appreciate if anyone can help me with this... I have a data frame that looks like this: 1 + name1 1 2 3 2 + name2 5 9 10 2 - name3 56 74 93 1 - name4 65 75 98 I need to rearrange this in a way so that the rows with "1" in the first column, and "-" in the second column; then columns 4 and 6 should switch places. That is, column 6 would be now column 4 and

Hello, A collegue of mine has compared the runtime of a linear model + anova in SAS and S+. He got the same results, but SAS took a bit more than a minute whereas S+ took 17 minutes. I've tried it in R (1.9.0) and it took 15 min. Neither machine run out of memory, and I assume that all machines have similar hardware, but the S+ and SAS machines are on windows whereas the R machine is Redhat