nalluri pratap schreef:
Hi All,
I have two variables X, Y. The question is if the value of X is equal to
one, then the values in Y have to be reversed other wise it should not perfom
any action. I think this should be done using lapply function?
Example
Y values : 1 2
Hi All,
I have two variables X, Y. The question is if the value of X is equal to
one, then the values in Y have to be reversed other wise it should not perfom
any action. I think this should be done using lapply function?
Example
Y values : 1 2 3 NA
X Y (ORIGINAL) Y
Am Montag, den 17.07.2006, 19:42 -0400 schrieb jim holtman:
You were down at least 5300 levels in subroutine calls. Here is the
first couple of lines from 'traceback()':
5363: vavb(a, b, n, v, x)
...
5316: vavb(a, b, n, v, x)
The other thing is that there is no 'x' defined in the
I was checking out your function and thinking of trying to make it more
efficient when I noticed your comment above.
:-D
Yes, bioinformatics is full
of this stuff - have a look at Bioconductor.
I'm about it, but it'll take some time. It's a vast repository...
For the meantime, I've
Please look at
http://www.turbulence.org/Works/song/
This is a website by Martin Wattenberg that visually displays
the types of patterns you are looking for. He gave a paper
at the Joint Statistics Meetings in Minneapolis 2005.
__
On Mon, 17 Jul 2006, John Wiedenhoeft wrote:
Hi there,
I'm having myself a hard time writing an algorithm for finding patterns
within a given melody. In a vector I'd like to find ALL sequences that
occur at least twice, without having to check all possible patterns via
pattern matching.
Am Dienstag, den 18.07.2006, 22:09 -0400 schrieb Jim Lemon:
Hi John,
Minor bug - I zeroed the hit counter in the wrong place.
find.replay-function(tunestring,maxlen) {
return(matchlist)
}
Dear Jim,
many, many thanks for your effords :-D!!! Your program is great and very
elegant
Hi there,
I'm having myself a hard time writing an algorithm for finding patterns
within a given melody. In a vector I'd like to find ALL sequences that
occur at least twice, without having to check all possible patterns via
pattern matching.
I finally found a solution in a style that I'm used
Thanks for your response!
Am Montag, den 17.07.2006, 17:36 -0400 schrieb jim holtman:
It would help if you could provide the calling script and the data
that you are using.
The code I sent is included in .Rprofile. One of my data vectors would
be v = c(1, 1, 1, 5, 6, 1, 1, 1, 1, 6, 1, 1, 6,
