similar to: Find missing days

#Hi R-users, #I have an example DF like this: y1 <- rnorm(10) + 6.8 y2 <- rnorm(10) + (1:10*1.7 + 1) y3 <- rnorm(10) + (1:10*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:3,10) f <- gl(2,15, labels=paste("lev", 1:2, sep="")) g <- seq(as.Date("2000/1/1"), by="day", length=30) DF <- data.frame(x=x,y=y, f=f, g=g) DF$wdays <- weekdays(DF$g)

Hi R-users, I have a data.frame like this (modificated from https://stat.ethz.ch/pipermail/r-help/2007-August/138124.html). y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:5,12) f <- gl(3,20, labels=paste("lev", 1:3, sep="")) d <- data.frame(x=x,y=y, f=f) and this is how I can

Hi all, I have a question about transforming the data from summary function. Let's say I have a data frame like this: > x = data.frame(a = c(rep("lev1", 5), rep("lev2", 5)), b = c(rnorm(5)+2, rnorm(5))) > x a b 1 lev1 1.5964765 2 lev1 2.2945609 3 lev1 3.5285787 4 lev1 1.4439838 5 lev1 2.2948826 6 lev2 1.7063506 7 lev2 -0.4042742 8 lev2

#Hi R-users, #Suppose that I have a data.frame like this: y1 <- rnorm(10) + 6.8 y2 <- rnorm(10) + (1:10*1.7 + 1) y3 <- rnorm(10) + (1:10*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:3,10) f <- gl(2,15, labels=paste("lev", 1:2, sep="")) g <- seq(as.Date("2000/1/1"), by="day", length=30) DF <- data.frame(x=x,y=y, f=f, g=g) DF$g[DF$x == 1]

Hi Im pretty new to R and I have run in to a problem. How do I name all the levels in this list. Lev1 <- c("A1","A2") Lev2 <- c("B1","B2") Lev3 <- c("C1","C2") MyList <- lapply(Lev1,function(x){ lapply(Lev2,function(y){ lapply(Lev3,function(z){ paste(unlist(x),unlist(y),unlist(z)) })})}) I would like to name the different

Dear list members, I have problems to interpret the coefficients from a lm model involving the interaction of a numeric and factor variable compared to separate lm models for each level of the factor variable. ## data: y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:20,3) f <- gl(3,20,

Hi folks, I would like to trim the trailing spaces in my factor variables using lapply (described in this post by Marc Schwartz: http://tolstoy.newcastle.edu.au/R/e2/help/07/08/22826.html) but the code is not functioning (in this example there is only one factor with trailing spaces): y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <-

R-users E-mail: r-help@r-project.org Hi! R-users. I am just wondering what the definition of "dffits" in R language is. Let me show you an simple example. function() { library(MASS) xx <- c(1,2,3,4,5) yy <- c(1,3,4,2,4) data1 <- data.frame(x=xx, y=yy) lm.out <- lm(y~., data=data1, x=T) lev1 <- lm.influence(lm.out)$hat sig1 <-

Hi all, I'm having a little bit of trouble with some date conversions and am hoping someone can help me out. Thanks in advance. OK, I have two sources of data that provide date info in a csv file differently. I've attached a small zipped file with two text files that illustrate both. (Is it ok to send attachments to this list? Not sure. It's very small.) I need to be able to

Hi everybody, I am beginer in R and I need your precious help. I want to create a small function in R as in sas to retrieve date. I have a file with data that import in R. DATE PAYS nb_pays.ILI. 1 24/04/2009 usa 0 2 24/04/2009 usa 0 3 24/04/2009 Mexique 0 4 24/04/2009

Hi R-users, What is the best way to achieve a table which contains all days and months between years 2007-2020? I would like to calculate number of days in each month within those years (to data frame). Regards, Lauri [[alternative HTML version deleted]]

I am trying to calculate the median of each row of a data frame where the data frame consist of columns of class Date. Below are my test data and best attempt at using apply. I didn't see a solution via Google or the Baron search site. I'd be grateful for any suggestions or solutions. I'm using R 2.0.0 on Mac OS X. Thank you, Stephen Weigand ### Test data date1 <- c(1000,

Hi All, I have an script in R which accepts user inputs for certain parameters, particularly dates, which the user inputs as character strings. eg: > date1 <- "2009-01-21" The script later parses the input via the as.Date function: > as.Date(date1) However, as.Date encounters an error when the string does not represent an actual date. eg: > date1 <-

Hi, Try this: el<- read.csv("el.csv",header=TRUE,sep="\t",stringsAsFactors=FALSE) ?elsplit<- split(el,el$st) ? datetrial<-data.frame(date1=seq.Date(as.Date("1930.1.1",format="%Y.%m.%d"),as.Date("2010.12.31",format="%Y.%m.%d"),by="day")) elsplit1<- lapply(elsplit,function(x)

I am currently puzzled by a seach path behavior. I have a library of a dozen routines getlabs(), getssn(), getecg(), ... that interface to local repositories and pull back patient information. All have a the first 6 arguments in common, and immediately call a second routine to do initial processing of these 6. The functions "joe" and "fred" below capture the relevant

Full_Name: Alexander L. Belikoff Version: 2.8.1 OS: Ubuntu 9.04 (x86_64) Submission from: (NULL) (67.244.71.200) I'm trying to reshape the following data frame: ID DATE1 DATE2 VALUE_TYPE VALUE 'abcd1233' 2009-11-12 2009-12-23 'TYPE1' 123.45 ... VALUE_TYPE is a string and is a factor with only 2 values

I have time series observation on daily frequencies : library(zoo) SD=1 date1 = seq(as.Date("01/01/01", format = "%m/%d/%y"), as.Date("12/31/02", format = "%m/%d/%y"), by = 1) len1 = length(date1); data1 = zoo(matrix(rnorm(len1, mean=0, sd=SD*0.5), nrow = len1),  date1) plot(data1) Now I want to calculate 1. Quarterly statistics like mean, variance etc

Hi, I have a query about sqldf, and dates in general. I couldnt find much on the net or on the forums, hence I am here. Here is the issue: I want to write a function that accepts 3 arguments: date1, date2 and a dataframe, say 'df'. Within the function, I want to populate a temp dataframe which essentially contains the output of the query "select * from df where DATE between date1

I Have been trying to convert a vector of dates into julian dates using the following commands: as.date & as.numeric. I can convert a date no problem by doing the following: as.numeric (as.date ("9/21/2004")) but as soon as I try to do an entire vector I am given the following: as.numeric (as.date(date1)) Error in as.date(date1) : Cannot coerce to date format I have tried

i have a data frame with 2 columns of dates. with str(dataframe) i have ensured myself that they were indeed formatted as dates. one column has NA's in it. the aim is now to make a third column that chooses date1 if it is a date, and choose date2 if it is a NA. i am trying df$date3=ifelse(is.na(df$date1), df$date2, df$date1). this leads to unexpected behaviour: the resulting column is