Using Anonymous Functions to return both original and modified variables-CodePudding

Apologies in advance, I'm struggling with how to phrase this question and I hope my repex is sufficiently explanatory.

I'm trying to use a combination of tidyverse's mutate_ and anonymous functions to apply a transformation to a variable with the goal of returning the original variable and its transformed counter part to the data frame. I'm struggling on how to rename the output the function returns, while keeping the original. So far it only returns the modified copy of the original.

So far, my code takes the original inputs, and returns their transformed versions to the data such as in the current_output object. I tried to use assign() nested in return() to allow for more flexible naming and to return a new variable all together like in the desired_output object but no such luck.

library(tidyverse)

data <- data.frame(X1 = c(2, 2, 2, 2),
                   X1a = c(3, 3, 3, 3),
                   X2 = c(3, 3, 3, 3))

data%>%
  mutate_at(vars(X1,
                 X1a),
            function(x){ 
              X2 = get("X2", parent.frame())
              X1a = x * X2
              return(assign(paste(x, "X2", "_Product", sep = "_"), X1a))
            }) -> data2

desired_output <- data.frame(X1 = c(2, 2, 2, 2),
                           X1a = c(3, ,3 , 3),
                           X2 = c(3, 3, 3, 3),
                           X1_X2_Product = c(6, 6, 6, 6),
                           X1a_X2_Product = c(9, 9, 9, 9))

current_output <- data.frame(X1 = c(6, 6, 6, 6),
                             X1a = c(9, 9, 9, 9),
                             X2 = c(3, 3, 3, 3))

CodePudding user response：

We could use across and its .names argument:

library(dplyr)
data %>% 
  mutate(across(c(X1, X1a), ~ .x* X2, .names = "{.col}_X2_Product"))

  X1 X1a X2 X1_X2_Product X1a_X2_Product
1  2   3  3             6              9
2  2   3  3             6              9
3  2   3  3             6              9
4  2   3  3             6              9

CodePudding user response：

This answer does not respect all the constraints required by the question, but is here for comedic effect:

library(base)  % I know...

data$X1_X2_Product  <- data$X1  * data$X2
data$X1a_X2_Product <- data$X1a * data$X2