CODEWITHSUNDEEP
rdplyrtidyverse
Nick
Nick

Reputation: 3374

Vectorised column operations in dplyr

I am looking for a tidy way of incorporating vectorised operations on columns using dplyr.

Basically, having a simple df as follows:

library(dplyr)
df <- data.frame("X" = runif(1:10), 
             "Y" = runif(1:10), "Z" = runif(1:10)) %>% 
 tbl_df()

I am now looking to apply the following vectorised formula:

Formula <- "X / Y + lag(Z)"

Of course the following won't work as it is looking for a column 'X / Y + lag(Z)':

df %>% mutate(Result := !!sym(Formula))

Can anyone suggest a simple way of applying a formula from a vector directly in a pipe on columns to achieve:

df %>% mutate(Result = X/Y+lag(Z))

Upvotes: 2

Views: 133

Answers (2)

akrun
akrun

Reputation: 886938

With tidyverse, parse_expr can be used

library(dplyr)    
df <- df %>% 
         mutate(Calc_Col = !! rlang::parse_expr(Formula))

and if we need to pass the column name as variable, use the := (as @Nick mentioned in the comments)

Name <- "Calc_Col" 
df <- df %>% 
        mutate(!!Name := !!rlang::parse_expr(Formula))

Upvotes: 1

Nautica
Nautica

Reputation: 2018

Is this what you're looking for?

set.seed(1)
df <- data.frame("X" = runif(1:10), 
                 "Y" = runif(1:10), "Z" = runif(1:10)) %>% 
    tbl_df()

Formula <- "X / Y + lag(Z)"

df <- df %>% mutate(Result = eval(parse(text = Formula)))

        X      Y      Z Result
    <dbl>  <dbl>  <dbl>  <dbl>
 1 0.153  0.0158 0.527  NA    
 2 0.322  0.231  0.327   1.93 
 3 0.479  0.0958 0.365   5.33 
 4 0.764  0.537  0.105   1.79 
 5 0.180  0.223  0.0243  0.913
 6 0.178  0.538  0.975   0.355
 7 0.869  0.820  0.845   2.03 
 8 0.356  0.263  0.0628  2.20 
 9 0.0399 0.710  0.968   0.119
10 0.863  0.422  0.825   3.02

parse an unevaluated expression, then evaluate it.

Upvotes: 1

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