从例子来看here https://stackoverflow.com/a/69738109/5784831我试图将总和作为“stepfun”类。我想,as.stepfun
是正确的选择,但我的想法行不通。怎么了?
y1 <- c(0, 1, 2, 0)
x1 <- c(1, 2, 3)
f1 <- stepfun(x = x1, y = y1)
print(class(f1))
# [1] "stepfun" "function" # OK!!!
plot(f1)
y2 <- c(0, 1, 0)
x2 <- c(1.5, 2.5)
f2 <- stepfun(x = x2, y = y2)
plot(f2)
fs <- function(x, f1, f2) {
# y <- f1(x) + f2(x) # OK
# y <- as.stepfun(x = x, y = y, ties = "ordered", right = FALSE) # does not work
# return(y) # does not work
return(f1(x) + f2(x))
}
print(class(fs)) # [1] "function"
# attributes(fs) # no new information...
fm <- function(x, f1, f2) {
return(f1(x) * f2(x))
}
print(class(fm)) # [1] "function"
Example as.
for data.frame
其按预期工作:
z <- c(1, 2)
class(z) # [1] "numeric"
class(as.data.frame(z)) # [1] "data.frame"
关于内部结构stepfun
function (x, y, f = as.numeric(right), ties = "ordered", right = FALSE)
{
if (is.unsorted(x))
stop("stepfun: 'x' must be ordered increasingly")
n <- length(x)
if (n < 1)
stop("'x' must have length >= 1")
n1 <- n + 1L
if (length(y) != n1)
stop("'y' must be one longer than 'x'")
rval <- approxfun(x, y[-if (right)
n1
else 1], method = "constant", yleft = y[1L], yright = y[n1],
f = f, ties = ties)
class(rval) <- c("stepfun", class(rval))
attr(rval, "call") <- sys.call()
rval
}