John,
I am very familiar with the evolving tidyverse and some messages a while back
included people who wanted this forum to mainly stick to base R, so I leave out
examples.
Indeed, the tidyverse is designed to make it easy to select columns with all
kinds of conditions including using r
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 th
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
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
The -grep(pattern,colnames) as a subscript is a bit dangerous. If no
colname matches the pattern then all columns will be omitted (because -0 is
the same as 0, which means no column). !grepl(pattern,colnames) avoids this
problem.
> mydata <- data.frame(A=1:3,B=11:13)
> mydata[, -grep("^yr", colna
Thanks to all. Very helpful.
Steven from iPhone
> On Jan 14, 2023, at 3:08 PM, Andrew Simmons 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
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