Re: [R] Bilateral matrix

> On May 17, 2018, at 6:40 AM, Miluji Sb wrote: > > Dear William and Ben, > > Thank you for your replies and elegant solutions. I am having trouble with > the fact that two of the previous locations do not appear in current > locations (that is no one moved to OKC and

Re: [R] Bilateral matrix

Dear William and Ben, Thank you for your replies and elegant solutions. I am having trouble with the fact that two of the previous locations do not appear in current locations (that is no one moved to OKC and Dallas from other cities), so these two cities are not being included in the output. I

Re: [R] Bilateral matrix

xtabs does this automatically if your cross classifying variables are factors with levels all the cities (sorted, if you like): > x <- sample(letters[1:5],8, rep=TRUE) > y <- sample(letters[1:5],8,rep=TRUE) > xtabs(~ x + y) y x c d e a 1 0 0 b 0 0 1 c 1 0 0 d 1 1 1 e 1 1 0 >

Re: [R] Bilateral matrix

Make current_location and previous_location factors with the same set of levels. The levels could be the union of the values in the two columns or a predetermined list. E.g., > x <- data.frame(previous_location=c("Mount Vernon","Burlington"), current_location=c("Sedro Woolley","Burlington")) >

Re: [R] Bilateral matrix

Dear Bert and Huzefa, Apologies for the late reply, my account got hacked and I have just managed to recover it. Thank you very much for your replies and the solutions. Both work well. I was wondering if there was any way to ensure (force) that all possible combinations show up in the output.

Re: [R] Bilateral matrix

or in base R : ?xtabs?? as in: xtabs(~previous_location + current_location,data=x) (You can convert the 0s to NA's if you like) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed

Re: [R] Bilateral matrix

Dear Miluji, If I understand correctly, this should get you what you need. temp1 <- structure(list(id = 101:115, current_location = structure(c(2L, 8L, 8L, 3L, 6L, 5L, 1L, 2L, 7L, 4L, 2L, 8L, 8L, 3L, 6L), .Label = c("Austin", "Boston", "Cambridge", "Durham", "Houston", "Lynn", "New Orleans",