[R] Bilateral matrix

[Miluji Sb]

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. The full dataset has 25 cities but of
course 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 using 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 Berkeley Breathed in his "Bloom County" comic strip )
>
> On Tue, May 8, 2018 at 9:21 AM, Huzefa Khalil <huzefa.khalil using 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 <milujisb using 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]]
>> >
>> > ______________________________________________
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>> ng-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>>
>> ______________________________________________
>> R-help using 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/posti
>> ng-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>

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