Execute dplyr operation only if column exists

2020-06-09 06:02发布

问题:

Drawing on the discussion on conditional dplyr evaluation I would like conditionally execute a step in pipeline depending on whether the reference column exists in the passed data frame.

Example

The results generated by 1) and 2) should be identical.

Existing column

# 1)
mtcars %>% 
  filter(am == 1) %>%
  filter(cyl == 4)

# 2)
mtcars %>%
  filter(am == 1) %>%
  {
    if("cyl" %in% names(.)) filter(cyl == 4) else .
  }

Unavailable column

# 1)
mtcars %>% 
  filter(am == 1)

# 2)    
mtcars %>%
  filter(am == 1) %>%
  {
    if("absent_column" %in% names(.)) filter(absent_column == 4) else .
  }

Problem

For the available column the passed object does not correspond to the initial data frame. The original code returns the error message:

Error in filter(cyl == 4) : object 'cyl' not found

I have tried alternative syntax (with no luck):

>> mtcars %>%
...   filter(am == 1) %>%
...   {
...     if("cyl" %in% names(.)) filter(.$cyl == 4) else .
...   }
 Show Traceback

 Rerun with Debug
 Error in UseMethod("filter_") : 
  no applicable method for 'filter_' applied to an object of class "logical" 

Follow-up

I wanted to expand this question that would account for the evaluation on the right-hand side of the == in filter call. For instance the syntax below attempts to filter on the first available value. mtcars %>%

filter({
    if ("does_not_ex" %in% names(.))
      does_not_ex
    else
      NULL
  } == {
    if ("does_not_ex" %in% names(.))
      unique(.[['does_not_ex']])
    else
      NULL
  })

Expectedly, the call evaluates to an error message:

Error in filter_impl(.data, quo) : Result must have length 32, not 0

When applied to existing column:

mtcars %>%
  filter({
    if ("mpg" %in% names(.))
      mpg
    else
      NULL
  } == {
    if ("mpg" %in% names(.))
      unique(.[['mpg']])
    else
      NULL
  })

It works with a warning message:

  mpg cyl disp  hp drat   wt  qsec vs am gear carb
1  21   6  160 110  3.9 2.62 16.46  0  1    4    4

Warning message: In { : longer object length is not a multiple of shorter object length

Follow-up question

Is there a neat way of expending the existing syntax in order to get conditional evaluation on the right-hand side of the filter call, ideally staying within dplyr workflow?

回答1:

Because of the way the scopes here work, you cannot access the dataframe from within your if statement. Fortunately, you don't need to.

Try:

mtcars %>%
  filter(am == 1) %>%
  filter({if("cyl" %in% names(.)) cyl else NULL} == 4)

Here you can use the '.' object within the conditional so you can check if the column exists and, if it exists, you can return the column to the filter function.

EDIT: as per docendo discimus' comment on the question, you can access the dataframe but not implicitly - i.e. you have to specifically reference it with .



回答2:

I know I'm late to the party, but here's an answer somewhat more in line with what you were originally thinking:

mtcars %>%
  filter(am == 1) %>%
  {
    if("cyl" %in% names(.)) filter(., cyl == 4) else .
  }

Basically, you were missing the . in filter. Note this is because the pipeline doesn't add . to filter(expr) since it is in an expression surrounded by {}.



回答3:

Avoid this trap:

On a busy day, one might do like the following:

library(dplyr)
df <- data.frame(A = 1:3, B = letters[1:3], stringsAsFactors = F)
> df %>% mutate( C = ifelse("D" %in% colnames(.), D, B)) 
# Notice the values on "C" colum. No error thrown, but the logic and result is wrong
  A B C
1 1 a a
2 2 b a
3 3 c a

Why? Because "D" %in% colnames(.) returns only one value of TRUE or FALSE, and therefore ifelse operates only once. Then the value is broadcasted to the whole column!

Correct way:

> df %>% mutate( C = if("D" %in% colnames(.)) D else B)
  A B C
1 1 a a
2 2 b b
3 3 c c


回答4:

Edit: Unfortunately, this was too good to be true

I might be a bit late to the party. But is

mtcars %>% 
 filter(am == 1) %>%
 try(filter(absent_column== 4))

a solution?



回答5:

This code does the trick and is pretty flexible. The ^ and $ are regex used to perform an exact match.

mtcars %>% 
  set_names(names(.) %>% 
              str_replace("am","1") %>% 
              str_replace("^cyl$","2") %>% 
              str_replace("Doesn't Exist","3")
              )