On 03/16/2012 12:23 PM, Sarah Goslee wrote:
There's almost certainly a better way, but I'd be more inclined to look for
it if you'd provide a small reproducible example so I could actually try it.
Without knowing the structure of your data, it's very hard to offer
alternatives.
Sarah
On Fri, Mar 16, 2012 at 12:59 PM, Walter Anderson<wandrso...@gmail.com> wrote:
I am working on a simulation where I need to count the number of matches for
an arbitrary pattern in a large sequence of binomial factors. My current
code is
for(indx in 1:(length(bin.05)-3))
if ((bin.05[indx] == test.pattern[1])&& (bin.05[indx+1] ==
test.pattern[2])&& (bin.05[indx+2] == test.pattern[3]))
return.values$count.match.pattern[1] =
return.values$count.match.pattern[1] + 1
Since I am running the above code for each simulation multiple times on
sequences of 10,000,000 factors the code is taking longer than I would like.
Is there a better (more "R" way of achieving the same answer?
Walter Anderson
Thank you for responding. I was attempting to keep my code snippet
small, since my last question which contained what I thought was a small
(but complete example) only received responses that the
provided code was not succinct enough.
the provided sequence of factors (bin.05) consist of a vector of up of
factors like this; [T,T,O,T,O,O,T,T,T,...] in a variety of different
proportions. The search pattern consists of any one of the combinations
of the factors (T & O) in a set of three, ie. [T,T,O]
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