Behavior of function "identical" with factors - r

Question

R stores factors as integers. Therefore, when using the function identical, it cannot find when two factors are of the same name if they have different levels.
Here's an MWE:
y <- structure(list(portfolio_date = structure(c(1L, 1L, 1L, 2L, 2L,
2L), .Label = c("2000-10-31", "2001-04-30"), class = "factor"),
security = structure(c(2L, 2L, 1L, 3L, 2L, 4L), .Label = c("Currency Australia (Fwd)",
"Currency Euro (Fwd)", "Currency Japan (Fwd)", "Currency United Kingdom (Fwd)"
), class = "factor")), .Names = c("portfolio_date", "security"
), row.names = c(10414L, 10417L, 10424L, 21770L, 21771L, 21774L
), class = "data.frame")
x <- structure(list(portfolio_date = structure(1L, .Label = "2000-10-31", class = "factor"),
security = structure(1L, .Label = "Currency Euro (Fwd)", class = "factor")),
.Names = c("portfolio_date", "security"), row.names = 10414L, class = "data.frame")
identical(y[1,], x)
Returns FALSE
But if we look at the objects, they appear identical to the user
y[1,]
portfolio_date security
10414 2000-10-31 Currency Euro (Fwd)
x
portfolio_date security
10414 2000-10-31 Currency Euro (Fwd)
Ultimately I want to be able to do something like the following:
apply(y, 1, identical, x)
10414 10417 10424 21770 21771 21774
TRUE TRUE FALSE FALSE FALSE FALSE
which(apply(y, 1, identical, x))
1 2
Any suggestions as to how to achieve this? Thanks.

Answer 1

One option is to use the rowwise from dplyr to check row-by-row; If you need to compare the row.names at the same time then you need to create an id column for both, otherwise, it will return TRUE for the first two rows.
library(dplyr)
x$id <- row.names(x)
y$id <- row.names(y)
rowwise(y) %>% do(check = isTRUE(all.equal(., x, check.attributes = F))) %>% data.frame
check
1 TRUE
2 FALSE
3 FALSE
4 FALSE
5 FALSE
6 FALSE

Answer 2

In order to perform the comparison, the factors need to be converted into character objects.
By using base R alone here is a solution:
apply(apply(y, 2, as.character), 1, identical, apply(x, 2, as.character))
The inner apply loops convert each column in the source and target data frames to character objects and the outer apply loops through the rows.
If the x data frame has more than one row, the actual behavior may not be as expected.

Answer 3

Use the package 'compare'.
library(compare)
result <- NULL
for (i in 1:NROW(y)){
one <- compare(y[i,], x, dropLevels=T)
two <- one$detailedResult[1]==T & one$detailedResult[2]==T
result <- c(result, two)
}
as.character(result)#TRUE TRUE FALSE FALSE FALSE FALSE

Answer 4

Solution for data posted in OP
The example posted in the OP can be easily treated by using droplevels().
Let us first look at why the comparison identical(y[1,], x) returns FALSE:
str(y[1,])
#'data.frame': 1 obs. of 2 variables:
#$ portfolio_date: Factor w/ 2 levels "2000-10-31","2001-04-30": 1
#$ security : Factor w/ 4 levels "Currency Australia (Fwd)",..: 2
whereas
str(x)
#'data.frame': 1 obs. of 2 variables:
#$ portfolio_date: Factor w/ 1 level "2000-10-31": 1
#$ security : Factor w/ 1 level "Currency Euro (Fwd)": 1
So the difference lies in the factors, even though both objects are displayed in the same way, as shown in the OP's question.
This is where the function droplevels() is useful: it removes unused factors. By applying droplevels() to y[1,] with its redundant factors, we obtain:
identical(droplevels(y[1,]), x)
#[1] TRUE
If x also contains unused factors, it will be necessary to wrap it into droplevels(), too. In any case, it won't do any harm:
identical(droplevels(y[1,]), droplevels(x))
#[1] TRUE
General solution
Using droplevels() may not work if the real data is more complex than the data posted in the "MWE" of the OP. Such situations may include, e.g., equivalent entries in x and y[1,] that are stored as different factor levels. An example where droplevels() fails is given in the data section at the end of this answer.
The following solution represents an efficient possibility to treat such general situations. It works for the data posted in the OP as well as for the more complicated case of the data posted below.
First, two auxiliary vectors are created that contain only the characters of each row. By using paste() we can concatenate each row to a single character string:
temp_x <- apply(x, 1, paste, collapse=",")
temp_y <- apply(y, 1, paste, collapse=",")
With these vectors it becomes easily possible to compare rows of the original data.frames, even if the entries were originally stored as factors with different levels and numbering.
To identify which rows are identical, we can use the %in% operator, which is more appropriate than the function identical() in this case, as the former checks for equality of all possible row combinations, and not just individual pairs.
With these simple modifications the desired output can be obtained quickly and without further loops:
setNames(temp_y %in% temp_x, names(temp_y))
#10414 10417 10424 21770 21771 21774
# TRUE TRUE FALSE FALSE FALSE FALSE
which(temp_y %in% temp_x)
#[1] 1 2
y[temp_y %in% temp_x,]
# portfolio_date security
#10414 2000-10-31 Currency Euro (Fwd)
#10417 2000-10-31 Currency Euro (Fwd)
data
x <- structure(list(portfolio_date = structure(1:2, .Label = c("2000-05-15",
"2000-10-31"), class = "factor"), security = structure(c(2L, 1L),
.Label = c("Currency Euro (Fwd)", "Currency USD (Fwd)"),
class = "factor")), .Names = c("portfolio_date", "security"),
class = "data.frame", row.names = c("10234", "10414"))
y <- structure(list(portfolio_date = structure(c(1L, 1L, 1L, 2L, 2L, 2L),
.Label = c("2000-10-31", "2001-04-30"), class = "factor"),
security = structure(c(2L, 2L, 1L, 3L, 2L, 4L),
.Label = c("Currency Australia (Fwd)", "Currency Euro (Fwd)",
"Currency Japan (Fwd)", "Currency United Kingdom (Fwd)"),
class = "factor")), .Names = c("portfolio_date", "security"),
row.names = c(10414L, 10417L, 10424L, 21770L, 21771L, 21774L),
class = "data.frame")