Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE))) x R1 R2 1 C B 2 B B 3 C E 4 E C 5 E B 6 D E 7 E E 8 D F 9 C D 10 A E Notice that the factor levels between the two factors, R1 and R2, slide by one level; that is, factor R1 does not have level F while factor R2 does not have level A. I want to swap the factor levels based on the combined levels of the two factors as shown below: tl <- unique(c(levels(tmp$R1), levels(tmp$R2))) for(ii in 1:dim(tmp)[1]) { kk <- which(tl %in% tmp[ii,'R2'], arr.ind = TRUE) - which(tl %in% tmp[ii,'R1'], arr.ind = TRUE) if(kk%%2!=0) { # swap the their levels between the two factors qq <- tmp[ii,]$R1 tmp[ii,]$R1 <- tmp[ii,]$R2 tmp[ii,]$R2 <- qq } } How to go about this case? Thanks! On Thu, Jul 6, 2017 at 5:16 PM, Ista Zahn <istazahn at gmail.com> wrote:> How about > > foo <- with(list(r1 = tmp$R1, > r2 = tmp$R2, > swapme = (as.numeric(tmp$R1) - as.numeric(tmp$R2)) %% 2 != 0), > { > tmp[swapme, "R1"] <- r2[swapme] > tmp[swapme, "R2"] <- r1[swapme] > tmp > }) > > Best, > Ista > > On Thu, Jul 6, 2017 at 4:06 PM, Gang Chen <gangchen6 at gmail.com> wrote: >> Suppose that we have the following dataframe: >> >> set.seed(1) >> (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace >> TRUE), R2 = sample(LETTERS[1:5], 10, replace = TRUE))) >> >> x R1 R2 >> 1 1 B B >> 2 2 B A >> 3 3 C D >> 4 4 E B >> 5 5 B D >> 6 6 E C >> 7 7 E D >> 8 8 D E >> 9 9 D B >> 10 10 A D >> >> I want to do the following: if the difference between the level index >> of factor R1 and that of factor R2 is an odd number, the levels of the >> two factors need to be switched between them, which can be performed >> through the following code: >> >> for(ii in 1:dim(tmp)[1]) { >> kk <- which(levels(tmp$R2) %in% tmp[ii,'R2'], arr.ind = TRUE) - >> which(levels(tmp$R1) %in% tmp[ii,'R1'], arr.ind = TRUE) >> if(kk%%2!=0) { # swap the their levels between the two factors >> qq <- tmp[ii,]$R1 >> tmp[ii,]$R1 <- tmp[ii,]$R2 >> tmp[ii,]$R2 <- qq >> } >> } >> >> More concise and efficient way to do this? >> >> Thanks, >> Gang >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code.

Untested, but I expect that setting the levels to be the same across the two factors levels(tmp$R1) <- levels(tmp$R2) <- LETTERS[1:6] and proceeding as before should be fine. Best, Ista On Jul 6, 2017 6:54 PM, "Gang Chen" <gangchen6 at gmail.com> wrote: Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE))) x R1 R2 1 C B 2 B B 3 C E 4 E C 5 E B 6 D E 7 E E 8 D F 9 C D 10 A E Notice that the factor levels between the two factors, R1 and R2, slide by one level; that is, factor R1 does not have level F while factor R2 does not have level A. I want to swap the factor levels based on the combined levels of the two factors as shown below: tl <- unique(c(levels(tmp$R1), levels(tmp$R2))) for(ii in 1:dim(tmp)[1]) { kk <- which(tl %in% tmp[ii,'R2'], arr.ind = TRUE) - which(tl %in% tmp[ii,'R1'], arr.ind = TRUE) if(kk%%2!=0) { # swap the their levels between the two factors qq <- tmp[ii,]$R1 tmp[ii,]$R1 <- tmp[ii,]$R2 tmp[ii,]$R2 <- qq } } How to go about this case? Thanks! On Thu, Jul 6, 2017 at 5:16 PM, Ista Zahn <istazahn at gmail.com> wrote:> How about > > foo <- with(list(r1 = tmp$R1, > r2 = tmp$R2, > swapme = (as.numeric(tmp$R1) - as.numeric(tmp$R2)) %% 2!= 0),> { > tmp[swapme, "R1"] <- r2[swapme] > tmp[swapme, "R2"] <- r1[swapme] > tmp > }) > > Best, > Ista > > On Thu, Jul 6, 2017 at 4:06 PM, Gang Chen <gangchen6 at gmail.com> wrote: >> Suppose that we have the following dataframe: >> >> set.seed(1) >> (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace >> TRUE), R2 = sample(LETTERS[1:5], 10, replace = TRUE))) >> >> x R1 R2 >> 1 1 B B >> 2 2 B A >> 3 3 C D >> 4 4 E B >> 5 5 B D >> 6 6 E C >> 7 7 E D >> 8 8 D E >> 9 9 D B >> 10 10 A D >> >> I want to do the following: if the difference between the level index >> of factor R1 and that of factor R2 is an odd number, the levels of the >> two factors need to be switched between them, which can be performed >> through the following code: >> >> for(ii in 1:dim(tmp)[1]) { >> kk <- which(levels(tmp$R2) %in% tmp[ii,'R2'], arr.ind = TRUE) - >> which(levels(tmp$R1) %in% tmp[ii,'R1'], arr.ind = TRUE) >> if(kk%%2!=0) { # swap the their levels between the two factors >> qq <- tmp[ii,]$R1 >> tmp[ii,]$R1 <- tmp[ii,]$R2 >> tmp[ii,]$R2 <- qq >> } >> } >> >> More concise and efficient way to do this? >> >> Thanks, >> Gang >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html>> and provide commented, minimal, self-contained, reproducible code.[[alternative HTML version deleted]]

how about gdata functions? set.seed(1) (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace = TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE))) tmp.orig=tmp library(gdata) bigMap=mapLevels(x=list(factor(tmp[,"R1"]),factor(tmp[,"R2"])),combine = T,codes=FALSE) mapLevels(tmp[,"R1"]) = bigMap mapLevels(tmp[,"R2"]) = bigMap dd=as.integer(tmp[,"R2"])-as.integer(tmp[,"R1"]) odd=dd%%2!=0 swap=tmp[odd,"R2"] tmp[odd,"R2"]=tmp[odd,"R1"] tmp[odd,"R1"]=swap cbind(tmp,odd,tmp.orig) cheers Peter> On 7. Jul 2017, at 00:54, Gang Chen <gangchen6 at gmail.com> wrote: > > Thanks a lot, Ista! I really appreciate it. > > How about a slightly different case as the following: > > set.seed(1) > (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace > TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE))) > > x R1 R2 > 1 C B > 2 B B > 3 C E > 4 E C > 5 E B > 6 D E > 7 E E > 8 D F > 9 C D > 10 A E > > Notice that the factor levels between the two factors, R1 and R2, > slide by one level; that is, factor R1 does not have level F while > factor R2 does not have level A. I want to swap the factor levels > based on the combined levels of the two factors as shown below: > > tl <- unique(c(levels(tmp$R1), levels(tmp$R2))) > for(ii in 1:dim(tmp)[1]) { > kk <- which(tl %in% tmp[ii,'R2'], arr.ind = TRUE) - which(tl %in% > tmp[ii,'R1'], arr.ind = TRUE) > if(kk%%2!=0) { # swap the their levels between the two factors > qq <- tmp[ii,]$R1 > tmp[ii,]$R1 <- tmp[ii,]$R2 > tmp[ii,]$R2 <- qq > } > } > > How to go about this case? Thanks! > > > On Thu, Jul 6, 2017 at 5:16 PM, Ista Zahn <istazahn at gmail.com> wrote: >> How about >> >> foo <- with(list(r1 = tmp$R1, >> r2 = tmp$R2, >> swapme = (as.numeric(tmp$R1) - as.numeric(tmp$R2)) %% 2 != 0), >> { >> tmp[swapme, "R1"] <- r2[swapme] >> tmp[swapme, "R2"] <- r1[swapme] >> tmp >> }) >> >> Best, >> Ista >> >> On Thu, Jul 6, 2017 at 4:06 PM, Gang Chen <gangchen6 at gmail.com> wrote: >>> Suppose that we have the following dataframe: >>> >>> set.seed(1) >>> (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace >>> TRUE), R2 = sample(LETTERS[1:5], 10, replace = TRUE))) >>> >>> x R1 R2 >>> 1 1 B B >>> 2 2 B A >>> 3 3 C D >>> 4 4 E B >>> 5 5 B D >>> 6 6 E C >>> 7 7 E D >>> 8 8 D E >>> 9 9 D B >>> 10 10 A D >>> >>> I want to do the following: if the difference between the level index >>> of factor R1 and that of factor R2 is an odd number, the levels of the >>> two factors need to be switched between them, which can be performed >>> through the following code: >>> >>> for(ii in 1:dim(tmp)[1]) { >>> kk <- which(levels(tmp$R2) %in% tmp[ii,'R2'], arr.ind = TRUE) - >>> which(levels(tmp$R1) %in% tmp[ii,'R1'], arr.ind = TRUE) >>> if(kk%%2!=0) { # swap the their levels between the two factors >>> qq <- tmp[ii,]$R1 >>> tmp[ii,]$R1 <- tmp[ii,]$R2 >>> tmp[ii,]$R2 <- qq >>> } >>> } >>> >>> More concise and efficient way to do this? >>> >>> Thanks, >>> Gang >>> >>> ______________________________________________ >>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >>> https://stat.ethz.ch/mailman/listinfo/r-help >>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >>> and provide commented, minimal, self-contained, reproducible code. > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.