Grepl Over Every Column in Dataframe-CodePudding

I have this dataframe in R:

col1 = c("abc", "cbe", "ddd")
col2 = c("hbc", "lbc","uhr")
id = c(1,2,3)

example = data.frame(id, col1, col2)

I want to select all rows that contain "bc" in any column. This would look like this:

 id col1 col2
1  1  abc  hbc
2  2  cbe  lbc

I know how to do this for a single column:

# almost works
select_col_1 = example[grepl("bc", example$col1, ignore.case = TRUE)]

But is there a way to repeat this for every column in the data frame?

I tried to do this with the "lapply" statement:

lapply(example, function(x){x[grepl('bc', example, ignore.case = TRUE)])

But I don't think I am doing this correctly.

Can someone please show me how to do this?

Thank you!

CodePudding user response：

You can do this using the dplyr package which supplies filter() to select rows, everything() to select all columns and if_any() for a result in any column.

library(dplyr)
example %>% 
  filter(if_any(everything(), ~ grepl("bc", .)))

Result:

  id col1 col2
1  1  abc  hbc
2  2  cbe  lbc

CodePudding user response：

If you want to use lapply, you can use the following code:

col1 = c("abc", "cbe", "ddd")
col2 = c("hbc", "lbc","uhr")
id = c(1,2,3)

example = data.frame(id, col1, col2)

example[Reduce(`|`, lapply(example[-1], grepl, pattern="bc")),]
#>   id col1 col2
#> 1  1  abc  hbc
#> 2  2  cbe  lbc

^{Created on 2022-07-15 by the reprex package (v2.0.1)}