[R] Bilateral matrix

[Bert Gunter]

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

> lvls <- sort(union(x,y))
> x <- factor(x, levels = lvls)
> y <- factor(y, levels = lvls)

> xtabs( ~ x + y)
   y
x   a b c d e
  a 0 0 1 0 0
  b 0 0 0 0 1
  c 0 0 1 0 0
  d 0 0 1 1 1
  e 0 0 1 1 0

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 Wed, May 16, 2018 at 7:49 AM, Miluji Sb <milujisb using gmail.com> wrote:

> 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]]
>>> >
>>> > ______________________________________________
>>> > 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.
>>>
>>> ______________________________________________
>>> 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.
>>>
>>
>>
>

	[[alternative HTML version deleted]]