This is better by a factor of 4:
len = 10
d = replicate(len, list(pH = 3,marker = TRUE,position = "A"),FALSE)
system.time(
{
pHAll = data.frame(
pH = unlist(lapply(d,"[[",1)),
pH = unlist(lapply(d,"[[",2)),
pH = unlist(lapply(d,"[[",3)))
}
)
Dieter
--
View this messag
Greg Hirson wrote:
>
> I'd approach this by first making a matrix, then converting to a data
> frame with appropriate types.
Well, I knew that matrixes are faster for numerics, but I also knew that the
required conversion to character would be a show-stopper. My second wisdom
was bogus. Your
The as.numeric(d.df$pH) should have been an
as.numeric(as.character(d.df$pH)).
Sorry for the confusion.
Greg
On 1/5/10 12:33 AM, Greg Hirson wrote:
> Dieter,
>
> I'd approach this by first making a matrix, then converting to a data
> frame with appropriate types. I'm sure there is a way to do
Dieter,
I'd approach this by first making a matrix, then converting to a data
frame with appropriate types. I'm sure there is a way to do it with
structure in one step. Operations on matrices are usually faster than on
dataframes.
len <- 10
d <- replicate(len, list(pH = 3, marker = TRUE
I have very large data sets given in a format similar to d below. Converting
these to a data frame is a bottleneck in my application. My fastest version
is given below, but it look clumsy to me.
Any ideas?
Dieter
# ---
len = 10
d = replicate(len, list(pH = 3,marker = TRU
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