R - using map to apply a list function to dataframe column and create new columns with elements of the list

Question

I have a dataframe with and id column and an eats column, and a separate food list. I want to process the dataframe so that a column is added for each food in the food list which is populated with 1 if the food is present in eats and 0 otherwise.

txt <- tibble(id = c(1, 2, 3),
          eats = c("apple, oats, banana, milk, sugar",
                   "oats, banana, sugar",
                   "chocolate, milk, sugar"))

food_list <- c("apple", "oats", "chocolate")

for (i in food_list){
  print(i)
  txt <- txt %>% 
    mutate(!!i := if_else(stringr::str_detect(eats, i), 1, 0))
}

I could do this using a for loop but struggling to do it without a loop. I Will be very grateful if someone can point me to how this can be done without using for loops and instead using the purrr library map functions.

Thanks!

akrun · Accepted Answer · 2021-07-29 00:34:24Z

5

We could use map as

library(purrr)
library(dplyr)
library(stringr)
txt <- map_dfc(food_list, ~ txt %>%
      transmute(!! .x := +(stringr::str_detect(eats, .x)))) %>% 
    bind_cols(txt, .)

-output

txt
# A tibble: 3 x 5
     id eats                             apple  oats chocolate
  <dbl> <chr>                            <int> <int>     <int>
1     1 apple, oats, banana, milk, sugar     1     1         0
2     2 oats, banana, sugar                  0     1         0
3     3 chocolate, milk, sugar               0     0         1

In base R, this can be done in on-liner

txt[food_list] <- +(sapply(food_list, grepl, x = txt$eats))

edited Jul 29, 2021 at 0:34

answered Jul 29, 2021 at 0:02

akrun

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1 Comment

AnilGoyal Over a year ago

Ohh I just noticed that my answer is nearly similar to yours, but believe me I didn't copy the idea. +1 already :)

Anoushiravan R · Accepted Answer · 2021-07-29 00:19:30Z

4

You can use cbind and str_detect , with map_df:

library(dplyr)
library(purrr)
library(stringr)

cbind(txt, map_dfc(food_list, ~+str_detect(txt$eats, .x))%>%set_names(food_list))

  id                             eats apple oats chocolate
1  1 apple, oats, banana, milk, sugar     1    1         0
2  2              oats, banana, sugar     0    1         0
3  3           chocolate, milk, sugar     0    0         1

edited Jul 29, 2021 at 0:19

Anoushiravan R

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answered Jul 29, 2021 at 0:16

GuedesBF

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Comments

Anoushiravan R · Accepted Answer · 2021-07-29 00:18:41Z

3

Here is an alternative solution:

library(dplyr)
library(tidyr)

txt %>%
  separate_rows(eats, sep = ", ") %>%
  rowwise() %>%
  mutate(ext = match(eats, food_list)) %>%
  drop_na() %>%
  pivot_wider(names_from = eats, values_from = ext, values_fn = length, values_fill = 0) %>%
  right_join(txt, by = "id") %>%
  relocate(id, eats)

# A tibble: 3 x 5
     id eats                             apple  oats chocolate
  <dbl> <chr>                            <int> <int>     <int>
1     1 apple, oats, banana, milk, sugar     1     1         0
2     2 oats, banana, sugar                  0     1         0
3     3 chocolate, milk, sugar               0     0         1

answered Jul 29, 2021 at 0:18

Anoushiravan R

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2 Comments

GuedesBF Over a year ago

Quite an adventure, going through pretty much the whole tidyverse to get the answer!

Anoushiravan R Over a year ago

Yes it's more tidyr, dplyr and a little bit of base R :)

AnilGoyal · Accepted Answer · 2021-07-29 15:53:59Z

You may use base R's Reduce like this

Reduce(function(a, b) {
  a[[b]] <- +(grepl(b, a[["eats"]]))
  a
}, init = txt, food_list)

# A tibble: 3 x 5
     id eats                             apple  oats chocolate
  <dbl> <chr>                            <int> <int>     <int>
1     1 apple, oats, banana, milk, sugar     1     1         0
2     2 oats, banana, sugar                  0     1         0
3     3 chocolate, milk, sugar               0     0         1

You may also use purrr::reduce similarly, where you can use (i) walrus operator and (ii) bang bang operators, instead of subsetting

library(tidyverse)
txt <- tibble(id = c(1, 2, 3),
              eats = c("apple, oats, banana, milk, sugar",
                       "oats, banana, sugar",
                       "chocolate, milk, sugar"))

food_list <- c("apple", "oats", "chocolate")

reduce(food_list, .init = txt, ~ .x %>% 
         mutate(!!.y := +str_detect(eats, .y))
         )
#> # A tibble: 3 x 5
#>      id eats                             apple  oats chocolate
#>   <dbl> <chr>                            <int> <int>     <int>
#> 1     1 apple, oats, banana, milk, sugar     1     1         0
#> 2     2 oats, banana, sugar                  0     1         0
#> 3     3 chocolate, milk, sugar               0     0         1

^{Created on 2021-07-29 by the reprex package (v2.0.0)}

Ronak Shah · Accepted Answer · 2021-07-29 06:13:07Z

Add word boundaries (\\b) to the values in food_list so that words are matched completely.

For example, see the difference in outputs in the following case -

library(stringr)
x <- c('apple', 'pineapple')

str_detect(x, 'apple')
#[1] TRUE TRUE

str_detect(x, '\\bapple\\b')
#[1]  TRUE FALSE

The same goes for grepl in base R -

food_list <- c("apple", "oats", "chocolate")
food_pat <- sprintf('\\b%s\\b', food_list)
txt[food_list] <- lapply(food_pat, function(x) as.integer(grepl(x, txt$eats)))
txt

# A tibble: 3 x 5
#     id eats                             apple  oats chocolate
#  <dbl> <chr>                            <int> <int>     <int>
#1     1 apple, oats, banana, milk, sugar     1     1         0
#2     2 oats, banana, sugar                  0     1         0
#3     3 chocolate, milk, sugar               0     0         1

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