Removing variables from data frame with a wile card

?s 16:54 de 15/01/2023, Sorkin, John escreveu:
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 using base R.

# Create data frame with three columns
one <- rep(1,10)
one
two <- rep(2,10)
two
three <- rep(3,10)
three
mydata <- data.frame(one=one, two=two, three=three)
cat("Data frame with three columns\n")
mydata

# Drop the column whose name starts with th, i.e. column three
# Find the location of the column
ColumToDelete <- grep("th",colnames((mydata)))
cat("The colomumn to be dropped is the column called three, which is column",ColumToDelete,"\n")
ColumToDelete

# Drop the column whose name starts with "th"
newdata2 <- mydata[,-ColumnToDelete]
cat("Data frame after droping column whose name is three\n")
newdata2

I hope this helps.
John

________________________________________
From: R-help <r-help-bounces at r-project.org> on behalf of Valentin Petzel <valentin at petzel.at>
Sent: Saturday, January 14, 2023 1:21 PM
To: avi.e.gross at gmail.com
Cc: 'R-help Mailing List'
Subject: Re: [R] Removing variables from data frame with a wile card

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 example compare

L <- list()
local({L$a <- 3})
L$a

with

E <- new.env()
local({E$a <- 3})
E$a

The latter will in fact work, as the same Environment is modified, while in the first one a modified copy of the list is made.

Under the hood we have a parser trick: If R sees something like

f(a) <- ...

it will look for a function f<- and call

a <- f<-(a, ...)

(this also happens for example when you do names(x) <- ...)

So in fact in our case this is equivalent to creating a copy with removed columns and rebind the symbol in the current environment to the result.

The data.table package breaks with this convention and uses C based routines that allow changing of data without copying the object. Doing

d[, (cols_to_remove) := NULL]

will actually change the data.

Regards,
Valentin

14.01.2023 18:28:33 avi.e.gross at gmail.com:

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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Hello,

Actually, Bill had addressed this in his post yesterday [1].
With your example,

one <- rep(1,10)
two <- rep(2,10)
three <- rep(3,10)
mydata <- data.frame(one=one, two=two, three=three)

ColumToDelete <- grep("fo",colnames((mydata)))
ColumToDelete
#> integer(0)
ColumToDeleteLogical <- grepl("fo",colnames((mydata)))
ColumToDeleteLogical
#> [1] FALSE FALSE FALSE

# Drop the column whose name starts with "fo"
# empty data.frame
mydata[, -ColumToDelete]
#> data frame with 0 columns and 10 rows

# nothing is deleted
mydata[, !ColumToDeleteLogical]
#>    one two three
#> 1    1   2     3
#> 2    1   2     3
#> 3    1   2     3
#> 4    1   2     3
#> 5    1   2     3
#> 6    1   2     3
#> 7    1   2     3
#> 8    1   2     3
#> 9    1   2     3
#> 10   1   2     3

[1] https://stat.ethz.ch/pipermail/r-help/2023-January/476682.html

Hope this helps,

Rui Barradas

Removing variables from data frame with a wile card

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