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 )
On Tue, May 8, 2018 at 9:21 AM, Huzefa Khalil <huzefa.kha...@umich.edu>
wrote:
> 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",
> "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", row.names = c(NA,
> -15L))
>
> dcast(temp1, previous_location ~ current_location)
>
> On Tue, May 8, 2018 at 12:10 PM, Miluji Sb <miluj...@gmail.com> wrote:
> > I have data on current and previous location of individuals. I would like
> > to have a matrix with bilateral movement between locations. I would like
> > the final output to look like the second table below.
> >
> > I have tried using crosstab() from the ecodist but I do not have another
> > variable to measure the flow. Ultimately I would like to compute the
> > probability of movement between cities (movement to city_i/total movement
> > from city_j).
> >
> > Is it possible to aggregate the data in this way? Any guidance would be
> > highly appreciated. Thank you!
> >
> > # Original data
> > 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",
> > "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", row.names = c(NA,
> > -15L))
> >
> > # Expected output
> > structure(list(X = structure(c(3L, 1L, 2L), .Label = c("Austin",
> > "Houston", "OKC"), class = "factor"), Boston = c(2L, NA, NA),
> > New.York = c(NA, 2L, 2L), Cambridge = c(2L, NA, NA)), class =
> > "data.frame", row.names = c(NA,
> > -3L))
> >
> > Sincerely,
> >
> > Milu
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@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@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]]
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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