> ll<-list(c("1","2","3"),c("2","3","4"))
> ll
[[1]]
[1] "1" "2" "3"
[[2]]
[1] "2" "3" "4"
> dd<-data.frame(matrix(nrow=10,ncol=3))
> dd
X1 X2 X3
1 NA NA NA
2 NA NA NA
3 NA NA NA
4 NA NA NA
5 NA NA NA
6 NA NA NA
7 NA NA NA
8 NA NA NA
9 NA NA NA
10 NA NA NA
> dd[1,]<-ll[[1]]
> dd
http://cran.r-project.org/web/packages/diagram/vignettes/diagram.pdf
It might be difficult to include very large amounts of data, but names and
counts should be fine.
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Thanks Chuck,
I was trying to implement something more complicated than what I had to and
after finding the reduce() function in bioconductor, everything went
smoothly.
Thanks again
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Try to do this
> list <- seq(1,5,1)
> list<-list[-3]
> list
[1] 1 2 4 5
list[-x] will remove the xth value in your list. Also you can do something
like
> list[-c(1:4)]
[1] 5
To remove values at indexes 1-4
> list[-c(1,4)]
[1] 2 3 5
To remove values at indexes 1 and 4
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So I have a bunch of c(start,end) points and want to consolidate them into as
few c(start,end) as possible.
For example:
sample startend
A 5 10
B 7 18
C 14
D 16 20
I'd want the function to return the two distinct
> list<-seq(2,10,2)
> list
[1] 2 4 6 8 10
> list[-which(2==list)]
[1] 4 6 8 10
using the which() will let you remove things from a list that have a
specified value... I usually use the
blah<- blah[-which(TRUE==is.na(blah)) ]
which will remove all NA values in your list
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I too think I worded it incorrectly...
so the second two columns of the matrix are the start and end of an interval
however, because some of the intervals overlap, I want to limit the number
of intervals I have to deal with.
So therefore,
(5 10)should merge with(7 18) making
Awesome, that works great, and it cuts my runtime down by a lot.
thanks baptiste
I totally forgot that I could just check to see if the number was inbetween
via less than and greater than rather than having to construct the vector.
findInterval also helps with another problem I was having somew
So basically I want to do this -
4 %in% 1:10
should return true
Would there be another way of doing this without having to do the 1:10 part?
I am using a very large data set and trying to do
459124 %in% 103000:983000
multiple times for many values, and it is taking quite a long time
Also, I
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