search for: previous_loc

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 in his "Bloom County" comic strip...

...ourse people have not moved from Boston to all the other 24 cities. I would like all the combinations if possible. Thank you again! Sincerely, Milu On Tue, May 8, 2018 at 6:28 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > 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...

.... I > would like all the combinations if possible. > > Thank you again! > > Sincerely, > > Milu > > On Tue, May 8, 2018 at 6:28 PM, Bert Gunter <bgunter.4567 at gmail.com> > wrote: > >> 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 >>...

...rrent_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", "New York"), class = "factor"), previous_location = structure(c(6L, 2L, 4L, 6L, 7L, 5L, 1L, 3L, 6L, 2L, 6L, 2L, 4L, 6L, 7L), .Label = c("Atlanta", "Austin", "Cleveland", "Houston", "New Orleans", "OKC", "Tulsa" ), class = "factor")), class = "data.frame"...

...rrent_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", "New York"), class = "factor"), previous_location = structure(c(6L, 2L, 4L, 6L, 7L, 5L, 1L, 3L, 6L, 2L, 6L, 2L, 4L, 6L, 7L), .Label = c("Atlanta", "Austin", "Cleveland", "Houston", "New Orleans", "OKC", "Tulsa" ), class = "factor")), class = "data.frame"...

...ided a better sample of the data and the ideal output (wide form - a 10x10 bilateral matrix) but haven't been able to do this. Would it be easier if I create variable for each ID - it would be equal to 1 if the person moved? I am a bit lost - thank you again! ### data structure(list(ID = 1:12, previous_location. = structure(c(3L, 9L, 8L, 10L, 2L, 5L, 1L, 7L, 4L, 6L, 10L, 5L), .Label = c("Atlanta", "Austin", "Boston", "Cambridge", "Dallas", "Durham", "Lynn", "New Orleans", "New York", "OKC"), class = &q...