Hi,
Try:
b1<- b
b1[!b1>=7]<- NA
lst1 <- split(b1,cumsum(c(0,abs(diff(b>=7)))))
indx <- as.logical(((seq_along(lst1)-1)%%2))
lst1[indx]<- lapply(seq_along(lst1[indx]),function(i) {lst1[indx][[i]]<-
rep(i,length(lst1[indx][[i]]))})
C2 <- unlist(lst1,use.names=FALSE)
all.equal(c1,C2)
#[1] TRUE
A.K.
----- Original Message -----
From: arun <smartpink...@yahoo.com>
To: Benjamin Gillespie <gy...@leeds.ac.uk>
Cc:
Sent: Wednesday, October 9, 2013 8:19 PM
Subject: Re: [R] Possible loop/ if statement query
Hi,
There should be a simpler way with cumsum(diff()).
b=c((1:10),sort(1:9,decreasing=TRUE),(2:12),sort(6:11,decreasing=TRUE),(7:13))
b1<- b
c1=c(
NA,NA,NA,NA,NA,NA,1,1,1,1,1,1,1,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,2,2,2,2,2,2,2,2,2,2,2,NA,3,3,3,3,3,3,3)
rl1<- rle(b1>=7)
indx1<-(cumsum(rl1$lengths)+1)[!rl1$values]
indx2<-(cumsum(rl1$lengths))[rl1$values]
b1[!b1>=7] <- NA
lst1<- split(sort(c(indx1,indx2)),((seq_along(sort(c(indx1,indx2)))-1)%/%2)+1)
mat1<- sapply(seq_along(lst1),function(i) {x<- lst1[[i]]; b1[seq(x[1],x[2])]<-
i; b1 })
indx2New<- !is.na(mat1[,2]) & mat1[,2]==2
indx3New<- !is.na(mat1[,3]) & mat1[,3]==3
mat1[!is.na(mat1[,1]) & mat1[,1]>3,1] <-
c(mat1[,2][indx2New],mat1[,3][indx3New])
all.equal(c1,mat1[,1])
#[1] TRUE
A.K.
----- Original Message -----
From: Benjamin Gillespie <gy...@leeds.ac.uk>
To: "r-help@R-project.org" <r-help@r-project.org>
Cc:
Sent: Wednesday, October 9, 2013 6:39 PM
Subject: [R] Possible loop/ if statement query
Dear r genii,
I hope you can help.
I have vector 'b':
b=c((1:10),sort(1:9,decreasing=TRUE),(2:12),sort(6:11,decreasing=TRUE),(7:13))
and, from 'b' I wish to create vector 'c':
c=c(
NA,NA,NA,NA,NA,NA,1,1,1,1,1,1,1,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,NA,2,2,2,2,2,2,2,2,2,2,2,NA,3,3,3,3,3,3,3)
The rules I want to use to create 'c' are:
A numeric of equal to, or over 7 in 'b' needs to result in a numeric (i.e. not
NA) in 'c';
A numeric of less than 7 in 'b' needs to result in "NA" in 'c';
Where 'groups' of numerics equal to, or over 7 in 'b' are present (i.e. next to
each other in the list), the numerics produced in 'c' all need to be the same;
Each 'group' of numerics in 'b' must result in a unique numeric in 'c' (and,
ideally, they should run in sequence as in 'c' above (1,2,3...).
If anyone has any idea where to start on this or can crack it I'll be most
grateful!!
Many thanks in advance,
Ben Gillespie, Research Postgraduate
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Tel: +44(0)113 34 33345
Mob: +44(0)770 868 7641
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and provide commented, minimal, self-contained, reproducible code.
