Join two qenv objects

Usage

join(x, y)

# S4 method for qenv,qenv
join(x, y)

# S4 method for qenv.error,ANY
join(x, y)

# S4 method for qenv,qenv.error
join(x, y)

Arguments

x

(qenv)

y

(qenv)

Details

There are some situations where join() cannot be properly performed, such as these three scenarios:

1. Both qenv objects contain an object of the same name but are not identical.
   
   Example:

   
     x <- new_qenv(
       code = c(mtcars1 = "mtcars1 <- mtcars"),
       env = list2env(list(mtcars1 = mtcars))
     )
     y <- new_qenv(
       code = c(mtcars1 = "mtcars1 <- mtcars['wt']"),
       env = list2env(list(mtcars1 = mtcars['wt']))
     )
     z <- join(x, y)
     # Error message will occur
     

   In this example, mtcars1 object exists in both x and y objects but the content are not identical.
   mtcars1 in the x qenv object has more columns than mtcars1 in the y qenv object (only has one column).

2. join() will look for identical @id values in both qenv objects. The index position of these @ids must be the same to determine the evaluation order. Otherwise, join() will throw an error message.
   
   Example:

   
     common_q <- new_qenv(code = "v <- 1", env = list2env(list(v = 1)))
     x <- eval_code(
       common_q,
       "x <- v"
     )
     y <- eval_code(
       common_q,
       "y <- v"
     )
     z <- eval_code(
       y,
       "z <- v"
     )
     q <- join(x, y)
     join_q <- join(q, z)
     # Error message will occur
   
     # Check the order of evaluation based on the id slot
     shared_ids <- intersect(q@id, z@id)
     match(shared_ids, q@id) # Output: 1 3
     match(shared_ids, z@id) # Output: 1 2
     

   The error occurs because the index position of identical @id between the two objects is not the same.

3. The usage of temporary variable in the code expression could cause join() to fail.
   
   Example:

   
     common_q <- new_qenv()
     x <- eval_code(
       common_q,
       "x <- numeric(0)
        for (i in 1:2) {
          x <- c(x, i)
        }"
     )
     y <- eval_code(
       common_q,
       "y <- numeric(0)
        for (i in 1:3) {
          y <- c(y, i)
        }"
     )
     q <- join(x,y)
     # Error message will occur
   
     # Check the value of temporary variable i in both objects
     x@env$i # Output: 2
     y@env$i # Output: 3
     

   join() fails to provide a proper result because of the temporary variable i exists in both objects but has different value.
   To fix this, we can set i <- NULL in the code expression for both objects.

   
     common_q <- new_qenv()
     x <- eval_code(
       common_q,
       "x <- numeric(0)
        for (i in 1:2) {
          x <- c(x, i)
        }
        # dummy i variable to fix it
        i <- NULL"
     )
     y <- eval_code(
       common_q,
       "y <- numeric(0)
        for (i in 1:3) {
          y <- c(y, i)
        }
        # dummy i variable to fix it
        i <- NULL"
     )
     q <- join(x,y)
     

Examples

q1 <- new_qenv(
  code = c(iris1 = "iris1 <- iris", mtcars1 = "mtcars1 <- mtcars"),
  env = list2env(list(
    iris1 = iris,
    mtcars1 = mtcars
  ))
)
q2 <- q1
q1 <- eval_code(q1, "iris2 <- iris")
q2 <- eval_code(q2, "mtcars2 <- mtcars")
qq <- join(q1, q2)
get_code(qq)
#> [1] "iris1 <- iris"     "mtcars1 <- mtcars" "iris2 <- iris"    
#> [4] "mtcars2 <- mtcars"

common_q <- new_qenv(list2env(list(x = 1)), quote(x <- 1))
y_q <- eval_code(common_q, quote(y <- x * 2))
z_q <- eval_code(common_q, quote(z <- x * 3))
join_q <- join(y_q, z_q)
# get_code only has "x <- 1" occurring once
get_code(join_q)
#> [1] "x <- 1"     "y <- x * 2" "z <- x * 3"