create new variables in a dataframe from existing variables using lists of names

Question

I was wondering if there was a way to construction a function to do the following:

Original dataframe, df:

Obs Col1    Col2
1   Y       NA
2   NA      Y
3   Y       Y

Modified dataframe, df:

Obs Col1 Col2 Col1_YN Col2_YN
1   Y    NA   “Yes”   “No”
2   NA   Y    “No     “Yes”
3   Y    Y    “Yes”   “Yes”

The following code works just fine to create the new variables but I have lots of original columns with this structure and the “Yes” “No” format works better when constructing tables.

df$Col1_YN <- as.factor(ifelse(is.na(df$Col1), “No”, “Yes”))
df$Col2_YN <- as.factor(ifelse(is.na(df$Col2), “No”, “Yes”))

I was thinking along the lines of defining lists of input and output columns to be passed to a function, or using lapply but haven’t figured out how to do this.

akrun · Accepted Answer · 2022-01-25 20:19:39Z

4

We can use across to loop over the columns and create the new columns by modifying the .names

library(dplyr)
df1 <- df %>% 
   mutate(across(-Obs,  
   ~ case_when(. %in% "Y" ~ "Yes", TRUE ~ "No"), .names = "{.col}_YN"))

-output

df1
  Obs Col1 Col2 Col1_YN Col2_YN
1   1    Y <NA>     Yes      No
2   2 <NA>    Y      No     Yes
3   3    Y    Y     Yes     Yes

If we want to use lapply, loop over the columns of interest, apply the ifelse and assign it back to new columns by creating a vector of new names with paste

df[paste0(names(df)[-1], "_YN")] <- 
   lapply(df[-1], \(x) ifelse(is.na(x), "No", "Yes"))

data

df <- structure(list(Obs = 1:3, Col1 = c("Y", NA, "Y"), Col2 = c(NA, 
"Y", "Y")), class = "data.frame", row.names = c(NA, -3L))

answered Jan 25, 2022 at 20:19

akrun

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2 Comments

B. Koch Over a year ago

Is there a way to define a list of input columns for ".names" in your answer. Ex: Initial column names: "Tobacco", "Alcohol", "Fentanyl" etc. with new output columns as "Tobacco_YN", "Alcohol_YN", "Fentanyl_YN" etc.?

akrun Over a year ago

@B.Koch it should be mentioned in across i.e. across(c(Tobacco, Alcohol, Fentanyl), ~ case_when(. %in% "Y" ~ "Yes", TRUE ~ "No"), .names = "{.col}_YN"))

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