similar to: Simulation help

Hi all, I am hoping someone can help me out with this: If I have dataframe of years and ages and the first column and first row are filled with leading values: Df<- age1 age2 age3 Yr1 1 0.4 0.16 Yr2 1.5 0 0 Yr3 0.9 0 0 Yr4 1 0 0 Yr5 1.2 0 0 Yr6 1.4 0 0 Yr7 0.8 0 0 Yr8 0.6 0 0 Yr9 1.1 0 0 Now the rest of the cells need to be filled according to the previous year and age

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"???????

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

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]]

I'm trying to create a function which will allow me to subset a data set based on values of various specified variables. I also want to then apply some other function(s) (e.g., summary). This is what I've tried so far.... > test.fx <- function(dta, expvar, expval) { + newdta <- subset(dta, eval(expvar)>expval) + summary(newdta$eval(expvar)) + } > >

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(

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

Is there a way to associate text descriptions with variables in a data.frame? For example... Let's say that in my data.frame I have a variable named var1. var1 represents the responses to the question "When was the last time you saw your physician?" When I tabulate the variable var1 I'd like the output to be a bit more descriptive and contain the more descriptive

Is there any such thing as a program file in R? That is, is it possible to compile and save code for performing data management and analytic tasks for a given project into a single/multiple file(s) (i.e., like a script file (S-Plus), a do file (STATA), or a *.sas file (SAS))? Any words of wisdom or a point in the direction of some documentation would be helpful. If it is not possible to work

svymean w/ svyby is working for me... > svyby(~visitcnt, ~agegrp3.f, svymean, design=svydes) agegrp3.f visitcnt se.visitcnt 18-44 18-44 8.755552 0.4953235 45-64 45-64 10.131555 0.5347806 65+ 65+ 9.588802 0.4323629 svyquantile is working for me... > svyquantile(~visitcnt, quantiles=c(.25, .5, .75), ties="rounded", design=svydes) 0.25

Is there any existing function for creating contingency tables that will display counts, row, column, and cell percentages in the same tables....anything similar to crosstabs in S? Marc W. Zodet, MS Health Statistician Center for Cost and Financing Studies Division of Statistical Research and Methods 2101 East Jefferson Street, Suite 500 Rockville, Maryland 20852 Phone: 301-594-7072 Fax:

I need some guidance from someone who is familiar/has some experience with the survey package. The data that I am using is from the Medical Expenditure Panel Survey (www.meps.ahrq.gov <http://www.meps.ahrq.gov/> ). The STRATA and PSU variables are varstr01 and varpsu01 respectively. When I try to specify them with the svydesign function I get an error message. An excerpt of my session