Thank you very much, though I still don't quite undertdand the
explanation :)
Nevertheless, I just found a seemingly simple (at least quiker to type)
solution after try-and-error:
eval(mapply(function(x) {x; function() x}, c("a", "b")))
Wish it may help future readers.
On Thu, May 03, 2012 at
As the title says, I want to apply a function (which itself returns
a function) to a list (or vector), and get a list (or vector) of
generated functions as the return value, but get unexpected result.
Anyone with an idea about the reason of this phenomenon and a correct
way to implement the requir
Thanks, I'll post to that list if the encountered problem is just about
ggplot2.
And from the date of the issue on GitHub it seems that I might need to
manually work around the problem some more times :)
On Fri, Dec 23, 2011 at 08:10:05AM -0600, Hadley Wickham wrote:
> See https://github.com/hadle
For example, prepare like this
> df.0 <- data.frame(x = 0, y = 0, note = "1")
> df.1 <- subset(df.0, note == "1")
> df.2 <- subset(df.0, note == "2")
Then a call to
> ggplot() + aes(x = x, y = y) +
> geom_point(data = df.1) + geom_point(data = df.2)
produces the error
> Error in eval(expr, envir
Hmm, that's my fault when composing this mail, but the problem was
really encountered at that time.
Nevertheless, neither can I reproduce the problem now, perhaps I just
made another mistake at that time.
Thanks all the same, and sorry for the disturbance anyway :|
On Tue, Oct 04, 2011 at 10:10:56
Ah, now I see...
Thanks very much :)
On Sat, Oct 01, 2011 at 09:27:34AM -0400, Gabor Grothendieck wrote:
> On Sat, Oct 1, 2011 at 5:28 AM, Casper Ti. Vector
> wrote:
> Its linear given c so calculate the residual sum of squares using lm
> (or lm.fit which is faster) given c and opt
Example:
> f <- function(x) { 1 + 2 * log(1 + 3 * x) + rnorm(1, sd = 0.5) }
> y <- f(x <- c(1 : 10)); y
[1] 4.503841 5.623073 6.336423 6.861151 7.276430 7.620131 7.913338 8.169004
[9] 8.395662 8.599227
> nls(x ~ a + b * log(1 + c * x), start = list(a = 1, b = 2, c = 3), trace =
> TRUE)
37.22954
Is there any way to use expression() in legend labels with ggplot2?
It seems that things like
> scale_shape_manual(value = c(
> x = expression(italic(x)),
> y = expression(italic(y))
> ))
don't work.
Thanks very much :)
--
Using GPG/PGP? Please get my current public key (ID: 0xAEF6A134,
Hello list, is there any R function for generating spline for implicit
functions, for example a spline for an two-column data frame where for
one value for one variable there maybe several correspondent values for
another? Many thanks.
Currently I'm using this home-made one:
> spline2d <- function
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