Removing variables from data frame with a wile card
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) %like% "yr", with=FALSE ]
On Sat, 14 Jan 2023 at 12:28, <avi.e.gross at gmail.com> wrote:
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 column to NULL as in: mydata$NAME <- NULL using the mydata["NAME"] notation can do that for you by using a loop of unctional programming method that does that with all components of your grep. R does have optimizations that make this less useful as a partial copy of a data.frame retains common parts till things change. For those who like to use the tidyverse, it comes with lots of tools that let you select columns that start with or end with or contain some pattern and I find that way easier. -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of Steven Yen Sent: Saturday, January 14, 2023 7:49 AM To: Andrew Simmons <akwsimmo at gmail.com> Cc: R-help Mailing List <r-help at r-project.org> Subject: Re: [R] Removing variables from data frame with a wile card 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:
grepl("^yr", colnames(mydata))
will tell you which 'colnames' start with "yr". If you'd rather you
use globbing:
grepl(glob2rx("yr*"), colnames(mydata))
Then you might write something like this to remove the columns starting
with yr:
mydata <- mydata[, !grepl("^yr", colnames(mydata)), drop = FALSE]
On Sat, Jan 14, 2023 at 1:56 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] "year" "weight" "confeduc" "confothr" "college" [6] ... [41] "yr3" "yr4" "yr5" "yr6" "yr7" [46] "yr8" "yr9" "yr10" "yr11" "yr12" [51] "yr13" "yr14" "yr15" "yr16" "yr17" [56] "yr18" "yr19" "yr20" "yr21" "yr22" [61] "yr23" "yr24" "yr25" "yr26" "yr27" [66] "yr28"...
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______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
John Kane Kingston ON Canada [[alternative HTML version deleted]]