Hi Jeff,
Indeed the data.table package does provide a much cleaner way to achieve
the same functionality, and a lot of other functionality as bonus.
Thanks for letting me know about it.
On Tue, 7 Apr 2015 at 15:41 Jeff Newmiller wrote:
> You might find the data.table package helpful. It uses a
You might find the data.table package helpful. It uses an index sorted with a
radix sort and minimizes moving the data around in memory.
---
Jeff NewmillerThe . . Go Live...
DCN:
Hi all,
Thanks for the responses.
Herve's example is a good small size example of what I wanted.
> y <- c(16, -3, -2, 15, 15, 0, 8, 15, -2)
> someCoolFunc(-2, y)
[1] 3 9
> someCoolFunc(15, y)
[1] 4 5 8
The requirement is that I want someCoolFunc() to run in O(number of
matches) time, instead of
Hi Keshav,
findMatches() in the S4Vectors/IRanges packages (Bioconductor) I think
does what you want:
library(IRanges)
y <- c(16L, -3L, -2L, 15L, 15L, 0L, 8L, 15L, -2L)
x <- c(unique(y), 999L)
hits <- findMatches(x, y)
Then:
> hits
Hits object with 9 hits and 0 metadata columns:
On Mon, 06 Apr 2015, Keshav Dhandhania writes:
> Hi,
>
> I know that one can find all occurrences of x in a vector v by doing
>> which(x == v).
>
> However, if I need to do this again and again, where v is remaining the
> same, then this is quite inefficient. In my particular case, I need to do
>
split() might help, but you should give a more complete
explanation of your problem.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Apr 6, 2015 at 1:56 PM, Keshav Dhandhania
wrote:
> Hi,
>
> I know that one can find all occurrences of x in a vector v by doing
> > which(x == v).
>
> Howeve
On Apr 6, 2015, at 1:56 PM, Keshav Dhandhania wrote:
> Hi,
>
> I know that one can find all occurrences of x in a vector v by doing
>> which(x == v).
>
> However, if I need to do this again and again, where v is remaining the
> same, then this is quite inefficient. In my particular case, I need
Hi,
I know that one can find all occurrences of x in a vector v by doing
> which(x == v).
However, if I need to do this again and again, where v is remaining the
same, then this is quite inefficient. In my particular case, I need to do
this millions of times, and length(v) = 100 million.
Does an
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