Hi
Feeding R-help back in, in case my suggestions might be of use to
someone else ...
Thanks for the examples - so the main benefit you are looking for is the
labelling on the axes (date labels) ?
If you are just trying to avoid the annoying white lines, it may just be
your PNG/PDF viewer; see ...
http://cran.stat.auckland.ac.nz/doc/FAQ/R-FAQ.html#Why-are-there-unwanted-borders
Even so, making this work for date values for x/y seems like a useful
thing. However, ...
Neither your "fix" nor my "fix" actually works for the example that you
have provided. It just reveals another problem within image.default(),
which is the calculation of "midpoints" ...
if (length(x) > 1 && length(x) == nrow(z)) { # midpoints
dx <- 0.5*diff(x)
x <- c(x[1L] - dx[1L], x[-length(x)] + dx,
x[length(x)] + dx[length(x)-1])
}
This calculation does NOT produce the desired result for "Date"s (the
diff() of the resulting modified 'x' is no longer regular).
So this needs a bit more thought - let me know if you come up with a fix
for that calculation before me :)
Paul
On 15/04/21 4:05 am, cda...@posteo.de wrote:
Hi Paul
Thanks for your answer. I just answer to you, I dont know if this is
correct (also the maillist?)
x <- y <- seq(as.Date("2020-1-1"), as.Date("2020-12-31"), l=12)
z <- array(rnorm(length(x)*length(y)), c(length(x), length(y)))
image(x, y, z, useRaster=FALSE) # ok
image(x, y, z, useRaster=TRUE) # error: ‘useRaster = TRUE’ can only be used
with a regular grid
x <- y <- seq_len(12)
image(x, y, z, useRaster=FALSE) # ok
image(x, y, z, useRaster=TRUE) # ok
The difference is the plot quality: useRaster = TRUE yields better quality as
it omits strange white lines (in both png and pdf). I did not find a pattern
when those white lines appear and when not. Hence, I cannot provide a
reproducable example in which you can see the differences induced by the
useRaster argument. In simple examples like the one above there is no
difference between useRaster true and false plots.
Do you have a suggestion on how to continue?
Thanks a lot,
Chris
Am 14.04.2021 04:37 schrieb Paul Murrell:
Hi
I doubt it is intended (to deliberately exclude "difftime" objects).
Can you please supply a full image() example (with 'x' and/or 'y' as Dates and
a 'z') ? So that I can see what ...
image(x, y, z, useRaster=FALSE)
... looks like, so I can see what you want ...
image(x, y, z, useRaster=TRUE)
... to look like.
I also wonder whether switching to ...
dx[1][rep(1, length(dx))]
... might be better than switching to ...
as.numeric(dx)
It produces the same result for "difftime" objects, and may have a better chance of
working better with other objects (although I confess that not having thought of using a
"difftime" for 'x' I am also failing to think of further possibilities for 'x').
Paul
On 14/04/21 3:11 am,cda...@posteo.de <mailto:cda...@posteo.de> wrote:
Hi The function `check_irregular()` defined within
`graphics::image.default()` checks if the `useRaster` argument for
`graphics::image()` can be true or must be false. According to this
function, the following example vector is irregular: ``` time <-
seq(as.Date("2020-1-1"), as.Date("2020-12-31"), l=12)
check_irregular(time, time) # TRUE ``` In my view, this is not
correct. In this case, the `all.equal`-call does not evaluate to true
due to the special class of `dx` (or `dy`). If I slightly rewrite the
function as ``` my_check_irregular <- function (x, y) { dx <-
as.numeric(diff(x)) dy <- as.numeric(diff(y)) (length(dx) &&
!isTRUE(all.equal(dx, rep(dx[1], length(dx))))) || (length(dy) &&
!isTRUE(all.equal(dy, rep(dy[1], length(dy))))) } ``` the correct
answer is obtained (i.e. that the input vector is not irregular based
on the rational behind `all.equal`): ``` my_check_irregular(time,
time) # FALSE ``` The same applies to POSIX* objects. I was wondering
if this is intended or not? Thanks a lot for any answer, Chris
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