Cfrgtkky

Firstly it is my first question on StackOverflow, I hope I will write it the good way. If not, don't hesitate to tell me... And sorry for my approximative english!

I would like to use the mutate function from dplyr to change the type of the columns of a data.frame, but without knowing in advance the new type. Thus I would like to create dynamically the function name (for example "as.numeric", "as.factor"), taking the new type from another data.frame.

Here is a concrete example (what I want to do is for data.frames with more than 100 variables, so you will understand I don't want to do this manually!):

library(tidyverse)



df <- data.frame(Name = c("Roger", "Steve"), Age = c("40", "32"), stringsAsFactors = FALSE)

glimpse(df)



Observations: 2

Variables: 2

$ Name <chr> "Roger", "Steve"

$ Age  <chr> "40", "32"



types <- data.frame(Field = c("Name", "Age"), OldType = c("character", "character"), NewType = c("factor", "integer"), stringsAsFactors = FALSE)

glimpse(types)



Observations: 2

Variables: 3

$ Field   <chr> "Name", "Age"

$ OldType <chr> "character", "character"

$ NewType <chr> "factor", "integer"

I searched during a long time and found a lot of documentation on quasiquotation, and I tried a few things, but without never getting the expected result. Here are two attempts I made:

# First attempt

for(i in 1:nrow(types)){

  field <- types$Field[i]

  field_quo <- enquo(field)

  new_type <- paste0("as.", types$NewType[i], "(", field, ")")

  new_type_quo <- enquo(new_type)

  df <- df %>% mutate(!!field_quo := !!new_type_quo)

}

glimpse(df)



Observations: 2

Variables: 2

$ Name <chr> "as.factor(Name)", "as.factor(Name)"

$ Age  <chr> "as.integer(Age)", "as.integer(Age)"

=> the function calls are considered as strings, and the value of the columns are replaced instead of their types.

# Second attempt

for(i in 1:nrow(types)){

  field <- types$Field[i]

  field_quo <- ensym(field)

  new_type <- paste0("as.", types$NewType[i], "(", field, ")")

  new_type_quo <- ensym(new_type)

  df <- df %>% mutate(!!field_quo := !!new_type_quo)

}

Here I get an error:

Error in mutate_impl(.data, dots) : Binding not found: as.factor(Name).

I guess the mutate function considers what is in parenthesis as a whole variable name?

I have tried other things but without success. I must admit that I am not a R expert and I have difficulties to fully understand this concept of quasiquotation, despite the quality of the documentation. So I know that I am doing the things wrong, but Idon't know why nor how to do it right... Can someone help?

Thanks!

interesting question. I think I found a solution using map2 from purrr package in tidyverse.

# Data

df <- data.frame(Name = c("Roger", "Steve"), Age = c("40", "32"), stringsAsFactors = FALSE)

types <- data.frame(Field = c("Name", "Age"), OldType = c("character", "character"), NewType = c("factor", "integer"), stringsAsFactors = FALSE)



library(tidyverse)



# Create a column with function names that is needed. I.e. adding as.

types <- types %>% 

  mutate(newType2 = paste0("as.", NewType))



# Then loop over column names and functions

df2 <- map2_dfc(types$Field, 

         types$newType2, 

         ~df %>% 

           select_(.x) %>% 

           mutate_all(.y)

  ) %>% as_tibble()

gives you

> df2

# A tibble: 2 x 2

  Name    Age

  <fct> <int>

1 Roger    40

2 Steve    32

But for easy data type conversion try. which gives you a suitable data type for each column. However, it never suggests factors. But then you can use convert if you want.

library(hablar)



df %>% 

    retype() %>%

    convert(fct(Name))

I think what you want is get. This allows you to retrieve an object by passing in its name as a character, so get('as.factor') will return the as.factor function. Fitting this into your previous attempts:

for(i in 1:nrow(types)) {

    field <- sym(types$Field[i])

    typeFun <- get(paste0('as.', types$NewType[i]))

    df <- df %>% 

        mutate(!!field := typeFun(!!field))

}

glimpse(df)



Observations: 2

Variables: 2

$ Name <fct> Roger, Steve

$ Age  <int> 40, 32