根据 R 中的向量更改列的类型

Question

I'm dealing with changing formats in R.

I have 2 dataframes:

• The main one df
• Another dataframe tmp which describes columns types of df and the New_format on which columns should be converted

Here is a reproducible example:

df <- data.frame(var1 = c("a", "b", "c"),
                 var2 = c(1,2,3), 
                 var3 = c("d", "e", "f"))

tmp <- data.frame(Variable = c("var1", "var2", "var3"), 
                  Format = c("character", "numeric", "character"),
                  New_format = c("character", "integer", "factor"))

I'd like to convert types of columns where New_format is different from Format . I've struggled a lot by using lapply function but did not manage to do it.

It would be really nice if you have any idea :)

Thanks a lot!

Answer 1

You could set up a named mapping between the New_format values and corresponding as.<value> function, like this:

funcs <- list("character"= as.character,"integer"=as.integer, "factor"=as.factor)

Then, in a loop, call the function

for(i in 1:nrow(tmp)) {
  if(tmp[i,"Format"]!=tmp[i,"New_format"]) {
    df[[tmp[i,"Variable"]]] <-funcs[[tmp[i,"New_format"]]](df[[tmp[i,"Variable"]]])   
  }
}

Answer 2

Use readr::type_convert()

library(tidyverse)

types <- paste(map_chr(tmp$New_format, ~str_sub(., 1,1)), collapse = "")

new_df <- type_convert(df, types, guess_integer = T)

str(new_df)
'data.frame':   3 obs. of  3 variables:
 $ var1: chr  "a" "b" "c"
 $ var2: int  1 2 3
 $ var3: Factor w/ 3 levels "d","e","f": 1 2 3

This function requires that the type specifications are passed in either as a cols() statement, or as a string with the new column type indicated by a single letter (eg "c" for character, "f" for factor, and so on).

So either just rename New_format labels to their single-letter versions ("c", "i", "f"), or you can use str_sub and paste with tmp to get the first letters (which type_convert wants for the type argument).

Note: Make sure to set guess_integer = TRUE , otherwise it will default to type double even if you ask for integer.