R Programming - pmax with data frame - r

Question

I was trying to use pmax function in my program. I had a data frame of numbers and I was trying to compare it with a single number. Output had NA's.
I figured out that data frame does not work with pmax so I changed data frame to matrix. And it worked. I was curious as to why data frame was returning NA. Is it something to do with recycling?
code:-
mat <- matrix(runif(500), nrow = 20, ncol = 5)
df <- as.data.frame(mat)
pmax(mat, .5) # No NA's
pmax(df, .5) # Many NA's

Answer 1

It is one of the problems where the replication of the value on the second argument is not recycled fully, i.e. it depends on the number of columns. This could be the reason
rep(0.5, ncol(df))[df < 0.5]
#[1] 0.5 0.5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
#[41] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
Note that for the first 2 values, the 0.5 is correctly being changed as the logical matrix is TRUE for those elements df < 0.5 while it is not the case because 0.5 is replicated only based on the number of columns.
Suppose, we look at pmax, the line
mmm[change] <- each[change]
is problematic. We can check the output by printing the output of 'each' and 'each[change]. If we modify the function to include the print statement
pmax2 <- function (..., na.rm = FALSE)
{
elts <- list(...)
if (length(elts) == 0L)
stop("no arguments")
if (all(vapply(elts, function(x) is.atomic(x) && !is.object(x),
NA))) {
mmm <- .Internal(pmax(na.rm, ...))
mostattributes(mmm) <- attributes(elts[[1L]])
}
else {
mmm <- elts[[1L]]
has.na <- FALSE
as <- methods::as
asL <- function(x) if (isS4(x))
as(x, "logical")
else x
for (each in elts[-1L]) {
l1 <- length(each)
l2 <- length(mmm)
if (l2 && (l2 < l1 || !l1)) {
if (l1%%l2)
warning("an argument will be fractionally recycled")
mmm <- rep(mmm, length.out = l1)
}
else if (l1 && (l1 < l2 || !l2)) {
if (l2%%l1)
warning("an argument will be fractionally recycled")
each <- rep(each, length.out = l2)
}
na.m <- is.na(mmm)
na.e <- is.na(each)
if (has.na || (has.na <- any(na.m) || any(na.e))) {
if (any(na.m <- asL(na.m)))
mmm[na.m] <- each[na.m]
if (any(na.e <- asL(na.e)))
each[na.e] <- mmm[na.e]
}
nS4 <- !isS4(mmm)
if (isS4(change <- mmm < each) && (nS4 || !isS4(each)))
change <- as(change, "logical")
change <- change & !is.na(change)
print(change)
mmm[change] <- each[change]
print(each)
print(each[change])
if (has.na && !na.rm)
mmm[na.m | na.e] <- NA
if (nS4)
mostattributes(mmm) <- attributes(elts[[1L]])
}
}
mmm
}
Now, we check the print output based on applying pmax2 on 'df'
invisible(pmax2(df, 0.5))
# V1 V2 V3 V4 V5
# [1,] TRUE TRUE TRUE TRUE FALSE
# [2,] TRUE FALSE TRUE TRUE TRUE
# [3,] FALSE FALSE TRUE TRUE FALSE
# [4,] FALSE TRUE TRUE TRUE TRUE
# [5,] FALSE TRUE TRUE FALSE TRUE
# [6,] FALSE FALSE TRUE TRUE TRUE
# [7,] TRUE TRUE TRUE FALSE TRUE
# [8,] FALSE FALSE TRUE FALSE FALSE
# [9,] FALSE FALSE TRUE FALSE TRUE
#[10,] TRUE TRUE TRUE TRUE FALSE
#[11,] FALSE TRUE TRUE TRUE TRUE
#[12,] TRUE TRUE FALSE TRUE FALSE
#[13,] FALSE TRUE TRUE TRUE FALSE
#[14,] FALSE TRUE FALSE FALSE TRUE
#[15,] TRUE FALSE FALSE FALSE TRUE
#[16,] FALSE TRUE FALSE TRUE FALSE
#[17,] TRUE FALSE TRUE FALSE FALSE
#[18,] TRUE FALSE TRUE FALSE TRUE
#[19,] FALSE FALSE TRUE TRUE TRUE
#[20,] TRUE FALSE TRUE FALSE TRUE
#[1] 0.5 0.5 0.5 0.5 0.5
# [1] 0.5 0.5 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
#[41] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
Note that this is exactly the same output we got with the rep mentioned earlier.
However, on a matrix this is not executed because of the if/else statements
invisible(pmax2(mat, 0.5))
nothing is printed
It is better to apply pmax on a matrix when compared with a single element than on a data.frame and that element. Otherwise, we can unlist the data.frame or convert it to matrix
all.equal(c(pmax(mat, .5)), pmax(unlist(df), .5), check.attributes = FALSE)
#[1] TRUE
data
set.seed(24)
mat <- matrix(runif(500), nrow = 20, ncol = 5)
df <- as.data.frame(mat)