similar to: inserting columns in the middle of a dataframe

I have a data frame somewhat like this: myframe <- data.frame (ID=c(2,3,4,5), Hunger =c(415,452,550,318 )) myframe Now I would like to add a column to the right which summarizes the values for Hunger somewhat to reduce the number of values: If the values for Hunger are between 300-400 I would like to insert the number 350, between 400-500 insert 450 between 500-600 insert 550 Does

Folks: I consider my reply below rather clumsy: One has to keep track of index numbers other than that which is inserted and must separately change column names. Is there as "essentially better" way to do this, either via base R or via an R package. I leave it to you to define "essentially better." Thanks. Cheers, Bert On Mon, Aug 1, 2011 at 10:17 AM, Bert Gunter

Hello, can someone give me an example of how to fit a linear model where the response is a matrix, please? (other than a loop over lm calls) The ?lm points me to model.matrix but it is not clear to me how to use a model matrix in an lm call. fit<-lm(headmatrix ~ tailmatrix) causes problems as soon as I use predict(fit,newframe). The variables in tailmatrix are not found in

Colleages I am running R 2.1.0 on a Mac (same problem occurs in Linux). In some situations, I have mixed text/numeric data that is stored as characters in a matrix. If I convert this matrix to a dataframe, the numeric data becomes factors, not what I intend. TEXT <- paste("Text", 1:4, sep="") NUMBERS <- 10 + 4:1 MATRIX <- cbind(TEXT,

Does anybody have any insight into how to make this faster? I suspect, that the rounding going on may be an issue, as is the stepping through data frame rows using integers ... If you have the patience to teach a noob, he will highly appreciate it ;0) Joh digit <- 4 for (minute in seq(from=25,to=lrange[2])){ # Extract all data associtaed with the current time (minute) frame <-

Hi all .., I have a list of 7000 dataframes with similar column headers and I wanted to add a new column to each dataframe based on a certain condition which is the same for all dataframes. When I extract one dataframe and apply my code it works very well as follows :- First suppose this is my first dataframe in the list > OneDF <- Mylist[[1]] > OneDF ID Pdate

Dear All, I have a data frame with 5 column and 201 row data. I want to add one more column between column 1 and 2 with value of 1. So the new column has to be the second column filled with 1. Any help will be appreciated. Thanks for your time. Thanks & Rg Mohan L

Greetings I'm attempting to insert missing data on the smallest size class of cave cricket instars into a data frame. The data involve censusing photoplots (plots) of roosting cave crickets in which we discern in four instars or size classes. I need to insert data on size class one into a data frame that already has data on size classes two through four. The data can be merged by their

If I create an aggregation like this: aggregate(lastYear[,8:10],list(Stadium=lastYear$STAD),mean) I'll get a new data frame, which I can order if I assign it like this: newFrame <- aggregate(lastYear[,8:10],list(Stadium=lastYear$STAD),mean) newFrame[order(newFrame$TEMP),] But.. if I just want to do this in one statement, what can I do? If this was perl, I'd be using something like

This is a continuing issue with the one on the list a long time ago (I couldn't find a solution to it from the web): -------------------------------------------------------------------------- > Using a formula converted with as.formula with lme leads > to an error message. Same works ok with lm, and with > lme and a fixed formula. > > # demonstrates problems with lme and

Using a formula converted with as.formula with lme leads to an error message. Same works ok with lm, and with lme and a fixed formula. # demonstrates problems with lme and as.formula demo<-data.frame(x=1:20,y=(1:20)+rnorm(20),subj=as.factor(rep(1:2,10))) demo.lm1<-lme(y~x,data=demo,random=~1|subj) print(summary(demo.lm1)) newframe<-data.frame(x=1:5,subj=rep(1,5))

Hello, I tried to construct my very first loop today and completly failed :-( Maybe someone can help me? I have a dataframe somewhat like this one: myframe <- data.frame (Timestamp=c("24.09.2012 09:00", "24.09.2012 10:00", "24.09.2012 11:00", "25.09.2012 09:00", "25.09.2012 10:00", "25.09.2012

> On Apr 15, 2018, at 4:08 AM, Allaisone 1 <Allaisone1 at hotmail.com> wrote: > > > Hi all .., > > > I have a list of 7000 dataframes with similar column headers and I wanted to add a new column to each dataframe based on a certain condition which is the same for all dataframes. > > > When I extract one dataframe and apply my code it works very well as

Hi, apologies if this is too simple but I've been stuck on the following for a while: I have a vector of strings: filenames with a name before the extension and a variety of possible extensions I want to select only those files with: 1) a ".tab" extension AND 2) the character sequence "lad" anywhere in the name of the file before the extension. Surely this

I am recoding some data. Many values that should be 1.5 are recorded as 1-2. Some example data and my solution is below. I am curious about better approaches or any other suggestions. Thanks! # example input data myData <- read.table(textConnection("id, v1, v2, v3 a,1,2,3 b,1-2,,3-4 c,,3,4"),header=TRUE,sep=",") closeAllConnections() # the first column is IDs so remove

Do any of you know any simple programming editors for R scripts which offer basic colour-coding and bracket-matching facilities? Dregging through scripts to find a missing comma or parentheses is something I'd rather do less of... Jon Minton [[alternative HTML version deleted]]

Hi, I have a set of dataframes names AINDSLIM, BINDSLIM, CINDSLIM ... NINDSLIM In each dataframe I want to extract two variables, “pid” and “{w}region”, where {w} means “a”, “b”, “c”, ...”n” At the moment my code looks like: > PidRegion <- data.frame(pid=XWAVEID$pid) > this.region <- AINDSLIM[,c("pid", "aregion")] > PidRegion <- merge(PidRegion,

Full_Name: Steven McKinney Version: 2.9.0 OS: Mac OS X 10.5.6 Submission from: (NULL) (142.103.207.10) A corrupt data frame can be constructed as follows: foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4]

Dear R-Users, I have a dataframe (''forexdata'') of daily returns from the foreign exchange market for three currencies - British Pound (bp), Canadian Dollar(cd), Deustche Mark (dm) vis-a-vis the US Dollar and the Date Of Trade(yymmdd). For some dates the returns are missing (recorded as zero) as there were no trades in that currency for that date. My task is to substitute the

Hi All I have a dataframe: myframe<-data.frame(ID=c("first","second"),x=c(1,2),y=c(3,4)) And I have a function myfun: myfun<-function(x,y) x+y I would like to write a function myfun2 that takes myframe and myfun as parameters and returns a list as below: mylist $first [1] 4 $second [2] 6 Could you please help me with this? Doesn't seem like the