How to replace string for every row in specfic column using dplyr and stringr

Question

I have the following tibble:

library(tidyverse)

df <- tibble::tribble(
  ~sample, ~colB, ~colC,
  "foo",   1,  2,
  "bar_x",   2,  3,
  "qux.6hr.ID",   3,  4,
  "dog",   1,  1
)


df
#> # A tibble: 4 x 3
#>       sample  colB  colC
#>        <chr> <dbl> <dbl>
#> 1        foo     1     2
#> 2      bar_x     2     3
#> 3 qux.6hr.ID     3     4
#> 4        dog     1     1

df <- factor(final_df$samples, levels=c("bar_x","foo","qux.6hr.ID","dog"))

    df
#> [1] foo        bar_x      qux.6hr.ID dog       
#> Levels: bar_x foo qux.6hr.ID dog

What I want to do is for every row in sample column remove these substrings: _x and .6hr if exist. The final table looks like this:

     sample  colB  colC
        foo     1     2
        bar     2     3
     qux.ID     3     4
        dog     1     1

How can I achieve that?

df %>% mutate(sample = gsub('_x|\\.6hr', '', sample)) or equivalently with stringr, df %>% mutate(sample = str_replace_all(sample, '_x|\\.6hr', '')) — alistaire
– alistaire, Commented Jun 3, 2017 at 5:08
@alistaire Actually my df contain factor. See my update. Sorry. How can I modify your code? — pdubois
– pdubois, Commented Jun 3, 2017 at 5:17
gsub still works, though it coerces to character. You could make a call to levels<-, but it's a little awkward in dplyr syntax. The forcats package supplies an alternative: df %>% mutate(sample = factor(sample), sample = forcats::fct_relabel(sample, function(x){str_replace_all(x, '_x|\\.6hr', '')})) though you have to structure the second parameter as a function à la lapply. — alistaire
– alistaire, Commented Jun 3, 2017 at 5:33

akrun · Accepted Answer · 2017-06-03 05:22:40Z

10

We can use

df %>% 
     mutate(sample = gsub("_x|\\.\\d+[A-Za-z]+", "", sample))
# A tibble: 4 x 3 
#   sample  colB  colC
#    <chr> <dbl> <dbl>
#1    foo     1     2
#2    bar     2     3
#3 qux.ID     3     4
#4    dog     1     1

If the 'sample' column is factor class either we can wrap with factor on the output of gsub or do this on the levels of sample

levels(df$sample) <- gsub("_x|\\.\\d+[A-Za-z]+", "", levels(df$sample))
df$sample
#[1] foo    bar    qux.ID dog   
#Levels: bar foo qux.ID dog

edited Jun 3, 2017 at 5:22

answered Jun 3, 2017 at 5:12

akrun

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2 Comments

pdubois Over a year ago

Actually my df contains factor. See my update. Sorry. How can I modify your code?

akrun Over a year ago

@pdubois gsub will take factor as well. If you retain as factor, then wrap the output with factor i.e. mutate(sample = factor(gsub(..

GGAnderson · Accepted Answer · 2021-09-09 21:14:22Z

2

And here's a solution using the purrr:map function, which has the added benefit of returning the same result whether "sample" is chr or factor.

df %>%
   mutate(sample = map_chr(sample, ~str_replace(.x, 
                                         pattern = "_x|\\.\\d+[A-Za-z]+", 
                                         replacement = "")))
# A tibble: 4 x 3
#  sample  colB  colC
#  <chr>  <dbl> <dbl>
#1 foo        1     2
#2 bar        2     3
#3 qux.ID     3     4
#4 dog        1     1

answered Sep 9, 2021 at 21:14

GGAnderson

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