Map values in dataframe row to new values

Question

I have a dataframe, say:

df <- tibble(question = 1:3, a = c('chicken', 'apple', 'beer'), b = c('chicken', 'banana', 'beer'), c = c('beef', 'apple', 'wine'))

question a       b       c    
>      <int> <chr>   <chr>   <chr>
1        1 chicken chicken beef 
2        2 apple   banana  apple
3        3 beer    beer    wine

And I would like to replace the the values in row 2 with some mapping:

apple -> 1
banana -> 2

So that the resulting output is:

question a       b       c    
>      <int> <chr>   <chr>   <chr>
1        1 chicken chicken beef 
2        2       1       2        1
3        3 beer    beer    wine

I've looked at case_match, but it appears to want a vector and my data is rows rather than columns. I think I can use across() to just get this to apply to all columns in a particular row, but not sure how to fit these pieces together.

EDIT: I'm not interested in replacing the value at a specific index of column and row. I'm interested in recoding all the values in a row by a particular mapping.

Does this answer your question? [How to replace certain values in a specific rows and columns with NA in R?](https://stackoverflow.com/questions/54615462/how-to-replace-certain-values-in-a-specific-rows-and-columns-with-na-in-r) — SamR, Feb 08 '23 at 17:24

Jon Spring · Accepted Answer · 2023-02-08T17:37:13.873

5

"If the row isn't 2, leave the values in a:c alone, otherwise swap in these new values":

df |>
  mutate(across(a:c, ~if_else(row_number() != 2, .,
                        case_match(., "apple" ~ "1", "banana" ~ "2"))))

Result

  question       a       b    c
1        1 chicken chicken beef
2        2       1       2    1
3        3    beer    beer wine

df <- data.frame(question = 1:3,
           a = c("chicken", "apple", "beer"),
           b = c("chicken", "banana", "beer"),
           c = c("beef", "apple", "wine"))

edited Feb 08 '23 at 17:37

answered Feb 08 '23 at 17:32

Jon Spring

55,165
4
35
53

I wonder if this is simply because I'm tired and missing something, but this doesn't actually replace for me. I removed the filtering limitations in across and if_else to see if that would help, but I still get the original dataframe back. data %>% mutate(across(everything(), case_match(., 'apple' ~ '1', 'banana' ~ '2'))) – Anthop Feb 08 '23 at 18:21
Are you saying the output of running the code above with the sample data above is different for you, or that `df` has not been changed in so doing? (We'd need to add `df <- [the code above]` if you want `df` to change.) If you are getting different output, what output are you getting and what version of dplyr are you running? `case_match` was added in dplyr 1.1.0, just recently added to CRAN. – Jon Spring Feb 08 '23 at 19:49
Two things about your code in the comment above: 1) You probably can't run your code `across(everything()...` without more modification if some columns are numeric and others are character. 2) You're missing a `~` before `case_match`. I presume you could run `data %>% mutate(across(a:c, ~case_match(., 'apple' ~ "1", 'banana' ~ "2", . ~ .)))` – Jon Spring Feb 08 '23 at 19:51
I edited my original question to include the line of code defining the df. I also did have the `~` before `case_match` in the code, but screwed up pasting in the comment. Basically I'm running `data <- tibble(question = 1:3, a = c('chicken', 'apple', 'beer'), b = c('chicken', 'banana', 'beer'), c = c('beef', 'apple', 'wine'))` and then `data %>% mutate(across(a:c, ~case_match(., 'apple' ~ "1", 'banana' ~ "2", . ~ .)))` and the output still has apple and banana in it. – Anthop Feb 08 '23 at 20:57
Odd. I get the expected output with that data and code. What do you get from `packageVersion("dplyr")`? – Jon Spring Feb 08 '23 at 21:17
`packageVersion("dplyr")` returns `[1] ‘1.1.0’`. – Anthop Feb 08 '23 at 22:04
Let us [continue this discussion in chat](https://chat.stackoverflow.com/rooms/251738/discussion-between-jon-spring-and-anthop). – Jon Spring Feb 08 '23 at 22:19

score 4 · Answer 2 · answered Feb 08 '23 at 17:29

You could use across with case_when and condition with row_number to apply on only the second row like this:

library(dplyr)
df %>%
  mutate(across(a:c, ~ case_when(row_number() == 2 & . == "apple" ~ '1',
                                 row_number() == 2 & . == "banana" ~ '2',
                                 TRUE ~ .)))
#>   question       a       b    c
#> 1        1 chicken chicken beef
#> 2        2       1       2    1
#> 3        3    beer    beer wine

^{Created on 2023-02-08 with reprex v2.0.2}

S-SHAAF · Answer 3 · 2023-02-08T17:34:36.843

2

You can create a conditional function using ifelse and include the row index with all columns but the first one to apply the function:

df[2, 2:4] <- ifelse(df[2, 2:4] == "apple", 1,
                    ifelse(df[2, 2:4] == "banana", 2, df[2, 2:4]))

edited Feb 08 '23 at 17:34

answered Feb 08 '23 at 17:29

S-SHAAF

1,863
2
5
14

score 2 · Answer 4 · answered Feb 08 '23 at 17:39

2

Using a named vector

 df1[2, 2:4] <- setNames(c(1, 2), c("apple", "banana"))[unlist(df1[2,-1])]

-output

> df1
  question       a       b    c
1        1 chicken chicken beef
2        2       1       2    1
3        3    beer    beer wine

answered Feb 08 '23 at 17:39

akrun

874,273
37
540
662

score 1 · Answer 5 · answered Feb 08 '23 at 17:30

Here's a way with match:

df[2, ] <- c(1, 2, df[2, ])[match(df[2, ], c("apple", "banana", df[2, ]))]

#  question       a       b    c
#1        1 chicken chicken beef
#2        2       1       2    1
#3        3    beer    beer wine

Might be easier to use in a function:

replace_row <- function(data, row, new, old){
  data[row, ] <- c(new, data[row, ])[match(data[row, ], c(old, data[row, ]))]
  data
}
replace_row(df, 2, c(1, 2), c("apple", "banana"))
#  question       a       b    c
#1        1 chicken chicken beef
#2        2       1       2    1
#3        3    beer    beer wine

Map values in dataframe row to new values

5 Answers5