name <- c("Jon", "Bill", "Maria")
agenn <- c(23, 41, 32)
agelk <- c(23, 41, 32)
agepm <- c(23, 41, 32)
df <- data.frame(name, age,agelk,agepm)
print (df)
I would like to drop columns with column names that contain c("epm","enn","jkk").
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Does this answer your question? How to drop columns by name pattern in R?TarJae– TarJae2021-05-01 14:08:06 +00:00Commented May 1, 2021 at 14:08
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4 Answers
Using dplyr:
library(dplyr)
df %>%
select(-contains(c("epm", "enn", "jkk")))
#> name agelk
#> 1 Jon 23
#> 2 Bill 41
#> 3 Maria 32
Comments
Using data.table and %like%
df[,!colnames(df) %like% paste0(c("epm","enn","jkkk"),collapse="|")]
name agelk
1 Jon 23
2 Bill 41
3 Maria 32
1 Comment
NelsonGon
Edited to ensure we drop as per OP and also to use
%like% once for more efficiency. Revert edit if you prefer.Using matches
library(dplyr)
df %>%
select(-matches('epm|enn|jkk'))
# name agelk
#1 Jon 23
#2 Bill 41
#3 Maria 32
Comments
Here a base R approach:
First of all your code ;)
name <- c("Jon", "Bill", "Maria")
agenn <- c(23, 41, 32)
agelk <- c(23, 41, 32)
agepm <- c(23, 41, 32)
df <- data.frame(name, agenn,agelk,agepm)
Create your values do drop:
drop_val <- c("epm","enn","jkk")
You can check with grepl if a string exists in another. So let's loop for every string you want to remove and compare if it exists in colnames(df). Well and if none of your strings fits, you keep them -> otherwise remove.
df[!grepl(paste0(drop_val,collapse="|" ),names(df))]
Output:
name agelk 1 Jon 23 2 Bill 41 3 Maria 32
1 Comment
NelsonGon
Edited to remove for loops, and avoid explicit
==FALSE checks. Revert back if you please.