# [R] Possible loop/ if statement query

Benjamin Gillespie gybrg at leeds.ac.uk
Thu Oct 10 23:56:31 CEST 2013

```Fantastic - once again, thanks Arun - your knowledge is very impressive!

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From: arun [smartpink111 at yahoo.com]
Sent: 10 October 2013 03:24
To: Benjamin Gillespie
Cc: R help
Subject: Re: [R] Possible loop/ if statement query

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)
# TRUE

A.K.

----- Original Message -----
From: arun <smartpink111 at yahoo.com>
To: Benjamin Gillespie <gybrg at 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,x)]<- 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])
# TRUE

A.K.

----- Original Message -----
From: Benjamin Gillespie <gybrg at leeds.ac.uk>
To: "r-help at R-project.org" <r-help at 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!!

o-------------------------------------------------------------------o
School of Geography, University of Leeds, Leeds, LS2 9JT
o-------------------------------------------------------------------o
Tel: +44(0)113 34 33345
Mob: +44(0)770 868 7641
o-------------------------------o
http://www.geog.leeds.ac.uk/
o-------------------------------------o
@RiversBenG
o--------------o
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