similar to: Matrix/dataframe indexing

You rang sir? library(tidyverse) xx = 1:10 yr1 = yr2 = yr3 = rnorm(10) dat1 <- data.frame(xx , yr1, yr2, y3) dat1 %>% select(!starts_with("yr")) or for something a bit more exotic as I have been trying to learn a bit about the "data.table package library(data.table) xx = 1:10 yr1 = yr2 = yr3 = rnorm(10) dat2 <- data.table(xx , yr1, yr2, yr3) dat2[, !names(dat2)

Steven, Just want to add a few things to what people wrote. In base R, the methods mentioned will let you make a copy of your original DF that is missing the items you are selecting that match your pattern. That is fine. For some purposes, you want to keep the original data.frame and remove a column within it. You can do that in several ways but the simplest is something where you sat the

I have a data frame containing variables "yr3",...,"yr28". How do I remove them with a wild card----something similar to "del yr*" in Windows/doc? Thank you. > colnames(mydata) ? [1] "year"?????? "weight"???? "confeduc"?? "confothr" "college" ? [6] ... ?[41] "yr3"??????? "yr4"???????

I'm trying to simulate trend data over a five year period. I want different trend profiles...the simplest being a linear trend. I've been using the following code: patBdta1 <- NULL for(i in 1:100) patBdta1 <- rbind(patBdta1,c(yr1= mean(rbinom(50,1,.50)), yr2 =mean(rbinom(50,1,.51)), yr3 =mean(rbinom(50,1,.52)),

Thanks to all. Very helpful. Steven from iPhone > On Jan 14, 2023, at 3:08 PM, Andrew Simmons <akwsimmo at gmail.com> wrote: > > ?You'll want to use grep() or grepl(). By default, grep() uses extended > regular expressions to find matches, but you can also use perl regular > expressions and globbing (after converting to a regular expression). > For example: >

mydata[, -grep("^yr",colnames(mydata))] On Sat, Jan 14, 2023 at 8:57 AM Steven T. Yen <styen at ntu.edu.tw> wrote: > I have a data frame containing variables "yr3",...,"yr28". > > How do I remove them with a wild card----something similar to "del yr*" > in Windows/doc? Thank you. > > > colnames(mydata) > [1]

You'll want to use grep() or grepl(). By default, grep() uses extended regular expressions to find matches, but you can also use perl regular expressions and globbing (after converting to a regular expression). For example: grepl("^yr", colnames(mydata)) will tell you which 'colnames' start with "yr". If you'd rather you use globbing:

x["V2"] is more efficient than using drop=FALSE, and perfectly normal syntax (data frames are lists of columns). I would ignore the naysayers, or put a comment in if you want to accelerate their uptake. As I understand it, one of the main reasons tibbles exist is because of drop=TRUE. List-slice (single-dimension) indexing works equally well with both standard and tibble types of data

Hello Avi, while something like d$something <- ... may seem like you're directly modifying the data it does not actually do so. Most R objects try to be immutable, that is, the object may not change after creation. This guarantees that if you have a binding for same object the object won't change sneakily. There is a data structure that is in fact mutable which are environments. For

I am new to this thread. At the risk of presenting something that has been shown before, below I demonstrate how a column in a data frame can be dropped using a wild card, i.e. a column whose name starts with "th" using nothing more than base r functions and base R syntax. While additions to R such as tidyverse can be very helpful, many things that they do can be accomplished simply

Valentin, You are correct that R does many things largely behind the scenes that make some operations fairly efficient. >From a programming point of view, though, many people might make a data.frame and not think of it as a list of vectors of the same length that are kept that way. So if they made a copy of the original data with fewer columns, they might be tempted to think the original

Hi, I'm Roslina, PhD student of University of South Australia, Australia from school Maths and Stats. I use S-Plus before and now has started using R-package. I used to analyse rainfall data using julian date. Is there any similar function that you can suggest to me to be used in R-package? Thank you so much for your attention and help [[alternative HTML version deleted]]

i have converted my data into date format like below: > day=as.Date(originaldate,"%m/%d/%Y") > day[1:5] [1] "2008-04-12" "2011-07-02" "2011-09-02" "2008-04-12" "2008-04-12" I wish to select only those observations from 2007 to 2009, how can I select from this list? [[alternative HTML version deleted]]

Dear WizaRds, having met some of you in person in Vienna, I think even more fondly of this community and hope to continue on this route. It was great talking with you and learning from you. Thank you. I am trying to work through an artificial example in post stratification. This is my dataset: library(survey) age <- data.frame(id=1:8, stratum=rep(

Hi All; I have just successfully installed Microsoft: Age of Empires II, Age of Empires II-Conquerors expansion, Age of Empires III on Wine 1.3.7. No dll overides, no pissing about nothing, just dropped in the CD, mapped it to using winecfg and installed. Empires II installed directly from disk and would not run after the install until I applied the no_Cd_patch and not are fine. AoEIII has 3

Dear list I'm using the reshape command and want to reshape a wide data set to a long one e.g. I have the variables y1,y2,y3,age1,age2,age3,sex,ethnic I want my new long data set to consist of the variables y (which has been created from y1,y2,y3), age (which has been created from age1,age2,age3), sex and ethnic I have tried to use the command:

Hi, I am an R newbie and one thing I am having trouble with binding variables that I have created within one data frame into a new data frame when using na.omit(). To illustrate this problem I will give the example I am working on and the approah I have been using:- data.frame1<-filepath.... attach(data.frame1) #create a new variable using a function new.variable<-rep(1,length(weight3))

Hello all, First I have done extensive searches on this forum and others and nothing seems to work. So I decided to post thinking someone could point me to the write post or give me some help. I have drawn a 100 samples from a fictitious population (N=1000), and then randomly selected 25% of the 100 samples. I would like to now merge the data.frame from the 100 samples with the data.frame for

Hi all, I am running regressions with many covariates, most of which remain the same each time (control variables). Instead of writing 30 demographic variables every regression, is there a way I could call them all at once using a variable called, perhaps "demog"? I have tried: > demog <- list(age1, age2, age3) but I get an error when I try to call a list in a regression. I also

Hi all, The last e-mails about beginners gave me the courage to post a question; from a beginner's perspective, there are a lot of questions that I'm tempted to ask. But I'm trying to find the answers either in the documentation, either in the about 15 free books I have, either in the help archives (I often found many similar questions posted in the past). Being an (still actual)