Creating a Function in dplyr that Operates on Columns through Variable/String Manipulation

Question

I am working with a dataset that contains many columns that are similarly named (ex thing_1 , thing_2, blargh_1, blargh_2, fizz_1, fizz_2), and I've been trying to write a function that takes in a string (such as fizz) and performs some operation on all superstrings of the column (such as fizz_1 + fizz_2).

So far, I have structured my code into something that looks something like:

newData <- data %>%
    mutate(fizz = f("fizz"))

f <- function(name) {
name_1 + name_2
}

where f as written obviously doesn't work. I've toyed around with assign, but not been terribly successful. I'm also open to other ways to tackle the problem (maybe a function that takes in a dataset and the string). Thanks!

Community · Accepted Answer · 2020-06-20 09:12:55Z

4

If we are creating a function then make use of the select_helpers which can take starts_with or ends_with or match as arguments

library(dplyr)
library(purrr)
f1 <- function(data, name){
          data %>%
              mutate(!! name := select(., starts_with(name)) %>% reduce(`+`))
    }

f1(df1, "fizz")
f1(df1, "blargh")
f1(df1, "thing")
#   thing_1 thing_2 thing_3 fizz_1 fizz_2 blargh_1 blargh_2 thing
#1       1       6      11      2      3        4        5    18
#2       2       7      12      3      4        5        6    21
#3       3       8      13      4      5        6        7    24
#4       4       9      14      5      6        7        8    27
#5       5      10      15      6      7        8        9    30

Or specify select(., matches(str_c("^", name, "_\\d+$")))

data

df1 <- data.frame(thing_1 = 1:5, thing_2 = 6:10, thing_3 = 11:15, 
    fizz_1 = 2:6, fizz_2 = 3:7, blargh_1 = 4:8, blargh_2 = 5:9)

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Dec 11, 2019 at 22:17

akrun

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4 Comments

Zhiqiang Wang Over a year ago

Interesting! Is is possible to explain !! name := ? thanks

Reliably Sourced Over a year ago

Thanks! That's a very neat and compact solution

akrun Over a year ago

@ZhiqiangWang That notation is specific to tidyverse for assignment (:=) with lhs being a object. here, we evaluate (!!) to get the value in name instead of creating name as the column name

Zhiqiang Wang Over a year ago

Thanks a lot. This is handy and I will definitely try it with my work!

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