# Re: [R] Bilateral matrix

```> On May 17, 2018, at 6:40 AM, Miluji Sb <miluj...@gmail.com> 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 Dallas from other cities), so
> these two cities are not being included in the output.```
```
William told you to make sure that the two location factors had the same levels
(aka Labels). At the moment they do not. Dallas and OKC are missing from the
Labels in current_location. Bert showed you how to do that.

--
David.

> I have provided 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 = "factor"), current_location. =
> structure(c(8L,
> 3L, 3L, 8L, 4L, 1L, 4L, 5L, 6L, 4L, 7L, 2L), .Label = c("Atlanta",
> "Austin", "Boston", "Cambridge", "Durham", "Lynn", "New Orleans",
> "New York"), class = "factor")), class = "data.frame", row.names = c(NA,
> -12L))
>
> ### ideal output
> structure(list(previous_location. = structure(c(3L, 9L, 8L, 10L,
> 2L, 5L, 1L, 7L, 4L, 6L), .Label = c("Atlanta", "Austin", "Boston",
> "Cambridge", "Dallas", "Durham", "Lynn", "New Orleans", "New York",
> "OKC"), class = "factor"), Boston = c(0L, 1L, 1L, 0L, 0L, 0L,
> 0L, 0L, 0L, 0L), New.York = c(1L, 0L, 0L, 1L, 0L, 0L, 0L, 0L,
> 0L, 0L), New.Orleans = c(0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L,
> 0L), OKC = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), Austin = c(0L,
> 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L), Dallas = c(0L, 0L, 0L, 0L,
> 0L, 0L, 0L, 0L, 0L, 0L), Atlanta = c(0L, 0L, 0L, 0L, 0L, 1L,
> 0L, 0L, 0L, 0L), Lynn = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L,
> 0L), Cambridge = c(0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L), Durham = c(0L,
> 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L)), class = "data.frame", row.names =
> c(NA,
> -10L))
>
> Sincerely,
>
> Milu
>
> On Wed, May 16, 2018 at 5:12 PM, Bert Gunter <bgunter.4...@gmail.com> wrote:
>
>> 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 <miluj...@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.4...@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.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
>>>>> ng-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
>>>>> ng-guide.html
>>>>> and provide commented, minimal, self-contained, reproducible code.
>>>>>
>>>>
>>>>
>>>
>>
>
>       [[alternative HTML version deleted]]
>
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