How to drop columns by name pattern in R?

I have this dataframe:

state county city  region  mmatrix  X1 X2 X3    A1     A2     A3      B1     B2     B3      C1      C2      C3

  1      1     1      1     111010   1  0  0     2     20    200       Push      8     12      NA      NA      NA
  1      2     1      1     111010   1  0  0     4     NA    400       Shove      9     NA 

Now I want to exclude columns whose names end with a certain string, say "1" (i.e. A1 and B1). I wrote this code:

df_redacted <- df[, -grep("\\1$", colnames(df))]

However, this seems to delete every column. How can I modify the code so that it only deletes the columns that matches the pattern (i.e. ends with "3" or any other string)?

The solution has to be able to handle a dataframe with has both numerical and categorical values.

I found a simple answer using dplyr/tidyverse. If your colnames contain "This", then all variables containing "This" will be dropped.

library(dplyr) 
df_new <- df %>% select(-contains("This"))

Your code works like a charm if I apply it to a minimal example and just search for the string "A":

df <- data.frame(ID = 1:10,
                 A1 = rnorm(10),
                 A2 = rnorm(10),
                 B1 = letters[1:10],
                 B2 = letters[11:20])
df[, -grep("A", colnames(df))]

So your problem is more a regular expression problem, not how to drop columns. If I run your code, I get an error:

df[, -grep("\\3$", colnames(df))]
Error in grep("\\3$", colnames(df)) : 
  invalid regular expression '\3$', reason 'Invalid back reference'

Update: Why don't you just use this following expression?

df[, -grep("1$", colnames(df))]
   ID         A2 B2
1   1  2.0957940  k
2   2 -1.7177042  l
3   3 -0.0448357  m
4   4  1.2899925  n
5   5  0.7569659  o
6   6 -0.5048024  p
7   7  0.6929080  q
8   8 -0.5116399  r
9   9 -1.2621066  s
10 10  0.7664955  t

Just as an additional answer, since I stumbled across this, when looking for the data.table solution to this problem.

library(data.table)
dt <- data.table(df)
drop.cols <- grep("1$", colnames(dt))
dt[, (drop.cols) := NULL]

For excluding any string you can use...

 # Search string to exclude
 strng <- "1"
 df <- data.frame(matrix(runif(25,max=10),nrow=5))
 colnames(df) <- paste( "EX" , 1:5 )
 df_red <- df[, -( grep(paste0( strng , "$" ) , colnames(df),perl = TRUE) ) ]

	df
#		  EX 1     EX 2        EX 3     EX 4     EX 5
#	1 7.332913 4.972780 1.175947853 6.428073 8.625763
#	2 2.730271 3.734072 6.031157537 1.305951 8.012606
#	3 9.450122 3.259247 2.856123205 5.067294 7.027795
#	4 9.682430 5.295177 0.002015966 9.322912 7.424568
#	5 1.225359 1.577659 4.013616377 5.092042 5.130887

	df_red
#		  EX 2        EX 3     EX 4     EX 5
#	1 4.972780 1.175947853 6.428073 8.625763
#	2 3.734072 6.031157537 1.305951 8.012606
#	3 3.259247 2.856123205 5.067294 7.027795
#	4 5.295177 0.002015966 9.322912 7.424568
#	5 1.577659 4.013616377 5.092042 5.130887

If you are specifically looking for a pattern that appears at the end of the column name, to drop those columns, you can use the following command:

library(dplyr) 
df_new <- df %>% select(-ends_with("linear"))

All the columns that end with the string linear will be dropped.

You can expand it further using regex for a broader pattern search. I have a data frame that has a bunch of columns with "name", "upper_name"and"lower_name"` as they represent confidence intervals for a bunch of series, but I don't need them all. So, using regex, you can do the following:

pattern = "(upper_[a-z]*)|(lower_[a-z]*)"
policyData <- policyData[, -grep(pattern = pattern, colnames(policyData))]

The "|" allows me to include an or statement in the regex so I can do it once with a single patter rather than look for each pattern.