Hi Sarah, Outstanding! Thanks! I did not notice that numbers of each row of A
are different. Actually, I have over 10,000 in B and over 5000 ranges in A.
What if I only need to take out all the B rows that fall into the ranges of A?
- remove the repetitive results. Best,Jiang
> Date: Thu, 3 Jan 2013 17:04:59 -0500
> Subject: Re: [R] Index out SNP position
> From: sarah.gos...@gmail.com
> To: zhyjiang2...@hotmail.com
> CC: r-help@r-project.org
>
> Assuming I understand what you want, which I'm not certain of, here's
> one way; there are more (probably some more elegant).
>
> I'm not sure how you'd put them in a vector, since there are different
> numbers of values for each row of A, so instead I've made a list.
> unlist(SNP) will turn it into a vector.
>
> It's also not consistent which column of A has the higher and lower values.
>
> SNP <- lapply(seq_len(nrow(A)), function(x)B[B >= min(A[x,]) & B <=
> max(A[x,])])
>
> > SNP
> [[1]]
> [1] 36003918 35838399 35838589
>
> [[2]]
> [1] 35838589
>
> [[3]]
> numeric(0)
>
> [[4]]
> [1] 36003918
>
> [[5]]
> numeric(0)
>
> On Thu, Jan 3, 2013 at 4:54 PM, JiangZhengyu <zhyjiang2...@hotmail.com> wrote:
> >
> >
> >
> >
> >
> >
> > Dear R experts,
> >
> > I have 2 matix: A& B. I am trying to index B against A - (1) find out B
> > rows that fall between the col 1 and 2 of A& put them into a new vector
> > SNP.I made code as below, but I cannot think of a right way to do it.
> > Could anyone help me with the code? Thanks,Jiang----
> >
> > A <-
> > matrix(c(35838396,35838674,36003908,36004090,36150188,36151202,35838584,35838674,36003908,36003992),
> > ncol = 2)
> > B <- matrix(c(36003918,35838399,35838589,36262559),ncol = 1) nr=nrow(A)
> > rn=nrow(B) for (i in 1:nr)
> > {
> > for (j in 1:rn){if (B[i,1]<=A[j,1] && B[i,1]>=A[j,2]){SNP[i]=B[i,1]}}
> > }
> >
>
> --
> Sarah Goslee
> http://www.functionaldiversity.org
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