Subsetting a dataframe within a function

Question

I am struggling to reference columns with a function.

Take the data:

  dat = data.frame(height = c(20, 20, 40, 50, 60, 10), weight = c(100, 200, 300, 200, 140, 240),
               age = c(19, 20, 20, 19, 10, 11))
  Age_list <- c(19)

Take the example:

toy_func <- function(df,list,column){

  for (i in list){
  toy_output <- df[,column == i]
  }

  Return(toy_output)
}

And run:

tst <- toy_func(dat,Age_list,"age")

The output is a dataframe with no variables. I would like to generate an output where the initial dataframe dat has been filtered by ages that equal 19.

The loop is necessary as I plan to iterate through each unique item in the age column. Essentially i am writting a function to partition a dataframe by the unique values in one of its columns.

Thanks in advance, John

partitioning a data.frame by it's unique values can be done using the split function. See what happens with split(mtcars, mtcars$gear) — phiver
– phiver, Commented Aug 31, 2018 at 13:27
Don't think this is possible because i want to write a general function and so we can't use $ — JDraper
– JDraper, Commented Aug 31, 2018 at 13:31
There is a function to do what you mentioned "...partition a dataframe by the unique values in one of its columns" and it's called split as @phiver mentioned. In your case you can check it with split(dat, dat$age). Is this the output you wanted? — AntoniosK
– AntoniosK, Commented Aug 31, 2018 at 13:58

s__ · Accepted Answer · 2018-08-31 14:30:30Z

2

Using the advice of @phiver, you can try this:

dat = data.frame(height = c(20, 20, 40, 50, 60, 10),
                 weight = c(100, 200, 300, 200, 140, 240),
                 age = c(19, 20, 20, 19, 10, 11))
Age_list <- c(19,20)

for (i in Age_list){
  toy_output <- split(dat,dat$age==i)$`TRUE`
  print(toy_output)
  }

Result:

  height weight age
1     20    100  19
4     50    200  19
  height weight age
2     20    200  20
3     40    300  20

EDIT:

You can do a simple workaround to put it in a function:

toy_func <- function(df,list,x){

for (i in list){
  toy_output <- split(df,df[,x]==i)$`TRUE`
  print(toy_output)
              }
           }

toy_func(dat,Age_list ,3)

  height weight age
1     20    100  19
4     50    200  19
  height weight age
2     20    200  20
3     40    300  20

EDIT2:

And this is going to give you a data.frame with the result, with the i column that tells you which iteration generates each row:

toy_func <- function(df,list,x){
   datalist = list()
        for (i in list){
            toy_output <- data.frame(split(df,df[,3]==i)$`TRUE`)
            toy_output$i <- i
            datalist[[i]] <- toy_output
                       }
   print(do.call(rbind, datalist))
  }

toy_func(dat,Age_list ,3)

  height weight age  i
5     60    140  10 10
2     20    200  20 20
3     40    300  20 20

edited Aug 31, 2018 at 14:30

answered Aug 31, 2018 at 13:33

s__

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3 Comments

JDraper Over a year ago

Thanks @s_t - that works on my toy example. Could you explain the $TRUE`` part. I assume you used that to get around the issues i had with referencing columns names in a function

s__ Over a year ago

Because putting an == is going to give the false and the true result of the equation, and you only need the ==TRUE result, so the $ choose the true. You can have a proof doing this split(dat, dat$age==20), that gives you a list as class. Hope this comment and the answer help you.

JDraper Over a year ago

Hi, sorry i just realized this won't work within the function because of the use of $. The idea was to write a generalized function.

Roman · Accepted Answer · 2018-08-31 14:12:20Z

0

You can try

foo <- function(x, y, z) x[ x[[y]] %in% z, ]
foo(dat, "age", c(19, 20))
  height weight age
1     20    100  19
2     20    200  20
3     40    300  20
4     50    200  19

If you need the output as a sorted list you can write

lapply(c(19,20), foo, x=dat, y="age")
[[1]]
  height weight age
1     20    100  19
4     50    200  19

[[2]]
  height weight age
2     20    200  20
3     40    300  20

answered Aug 31, 2018 at 14:12

Roman

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