CODEWITHSUNDEEP
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GSandro_Strongs
GSandro_Strongs

Reputation: 911

The function dplyr mutate does not to apply a function stored in one of the dataframe/data.table object

I have a data.table object which contains 3 columns (spec_fun, param_1, param_2). I wanted to use mutate to create a new column based on the result of the 3 mentioned columns but I got the error "Error in mutate() caused by error in do.call() ! 'what' must be a function or character string". I tried to create a separated function but it didn't work. This is my data.table object:

library(mixtools)
library(dplyr)
library(data.table)
my_data <- data.table(spec_fun = c("weibull", "rgamma", "rgamma"),
                        param_1 = c(0.7767294, 0.7362431, 0.8118086),
                        param_2 = c(6.8048834,0.1088231,0.1357522))

Here the code which generates an error:

final <- my_data %>%
    mutate(my_med = median(do.call(spec_fun,
                                   args = lapply(list(1000, param_1, param_2), 
                                                 function(x) if(!is.na(x)) x))))

Thanks in advance.

Upvotes: 6

Views: 119

Answers (6)

ThomasIsCoding
ThomasIsCoding

Reputation: 102529

You can run

set.seed(0)
my_data %>%
    mutate(my_med = map_dbl(str_glue(
        "median({spec_fun}(1000, {param_1}, {param_2}))"
    ), ~ eval(str2lang(.x))))

or

library(rlang)

set.seed(0)
my_data %>%
    mutate(my_med = mapply(
        \(f, ...) median(exec(f, ...)),
        spec_fun, list(1000), param_1, param_2
    ))

which gives

   spec_fun   param_1   param_2   my_med
     <char>     <num>     <num>    <num>
1: rweibull 0.7767294 6.8048834 4.492247
2:   rgamma 0.7362431 0.1088231 4.047691
3:   rgamma 0.8118086 0.1357522 3.734777

Upvotes: 1

Edo
Edo

Reputation: 7858

Since you're using two different working framework (tidyverse and data.table), I'll suggest you one working way for each.

library(purrr) # new library to use pmap_dbl !
library(data.table)
  
my_data <- data.table(spec_fun = c("rweibull", "rgamma", "rgamma"),
                      param_1 = c(0.7767294, 0.7362431, 0.8118086),
                      param_2 = c(6.8048834, 0.1088231, 0.1357522))

Tidyverse

set.seed(1)
my_data$my_med <- pmap_dbl(my_data, ~median(do.call(..1, args = list(1000, ..2, ..3))))

data.table

set.seed(1)
my_data[, my_med := median(do.call(spec_fun, args = list(1000, param_1, param_2))), by = .I][]

Output for both

#>    spec_fun   param_1   param_2   my_med
#>      <char>     <num>     <num>    <num>
#> 1: rweibull 0.7767294 6.8048834 4.515552
#> 2:   rgamma 0.7362431 0.1088231 4.047691
#> 3:   rgamma 0.8118086 0.1357522 3.734777
  • pmap allows you to elaborate rowwise, _dbl transforms the output from a list to a numeric vector
  • by = .I allows the data.table to work by row

Notes:

  • added r to rweibull
  • removed library(mixtools) because it served no purpose here
  • removed lapply(..., function(x) if(!is.na(x)) x))) because it made no difference

Upvotes: 3

Friede
Friede

Reputation: 7979

Base R, similar but different.

my_data |> 
  transform(m = mapply(\(f, a, b) getFunction(f)(n = 1e3, a, b),
            spec_fun, param_1, param_2) |> median())

Or transpose and operate column-wise (length 3 again).

my_data = setNames(as.data.frame(t(my_data[-1])), my_data$spec_fun)
sapply(names(my_data), 
       \(x) getFunction(x)(n = 1e3, my_data[[x]][1], my_data[[x]][2]) |> meadin())

Typo-corrected Data

my_data = data.frame(spec_fun = c("rweibull", "rgamma", "rgamma"), # corrected
           param_1 = c(0.7767294, 0.7362431, 0.8118086),
           param_2 = c(6.8048834,0.1088231,0.1357522))

Upvotes: 3

Roland
Roland

Reputation: 132969

You should not use mutate with a data.table. Also, it is unclear why you are simulating the median instead of calculating it.

library(data.table)
my_data <- data.table(spec_fun = c("rweibull", "rgamma", "rgamma"),
                      param_1 = c(0.7767294, 0.7362431, 0.8118086),
                      param_2 = c(6.8048834,0.1088231,0.1357522))
#simulate medians
set.seed(42)
my_data[, sim_med := mapply(\(f, p1, p2) median(getFunction(f)(n = 1e3, p1, p2)), 
                            spec_fun, param_1, param_2)]

#use quantile function
my_data[, q50 := getFunction(sub("^r", "q", spec_fun))(p = 0.5, param_1, param_2), by = spec_fun]
#   spec_fun   param_1   param_2  sim_med      q50
#     <char>     <num>     <num>    <num>    <num>
#1: rweibull 0.7767294 6.8048834 4.564007 4.245135
#2:   rgamma 0.7362431 0.1088231 4.214413 4.054822
#3:   rgamma 0.8118086 0.1357522 3.805266 3.775631

Upvotes: 4

jay.sf
jay.sf

Reputation: 73562

You might use mapply to avoid row-wise operations.

> set.seed(42)
> my_data  |> 
+   transform(
+     my_med=mapply(\(f, x, y) {
+       do.call(f, 
+               lapply(list(1000, x, y), \(x) {
+                 if (!is.na(x)) x
+               })) |> median()
+     }, spec_fun, param_1, param_2))
  spec_fun   param_1   param_2   my_med
1 rweibull 0.7767294 6.8048834 4.564007
2   rgamma 0.7362431 0.1088231 4.214413
3   rgamma 0.8118086 0.1357522 3.805266

Upvotes: 2

Edward
Edward

Reputation: 19339

Add a rowwise()

my_data %>%
  rowwise() %>%
  mutate(my_med = median(do.call(spec_fun,
                                 args = lapply(list(1000, param_1, param_2), 
                                               function(x) if(!is.na(x)) x))))

# A tibble: 3 × 4
# Rowwise: 
  spec_fun param_1 param_2 my_med
  <chr>      <dbl>   <dbl>  <dbl>
1 rweibull   0.777   6.80    4.13
2 rgamma     0.736   0.109   3.86
3 rgamma     0.812   0.136   3.77

Note "rweibull" typo:

my_data <- data.table(spec_fun = c("rweibull", "rgamma", "rgamma"),
                        param_1 = c(0.7767294, 0.7362431, 0.8118086),
                        param_2 = c(6.8048834,0.1088231,0.1357522))

Upvotes: 4

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