Re: [R] Bilateral matrix

[David Winsemius]

> On May 17, 2018, at 6:40 AM, Miluji Sb  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.

Please read all the replies for meaning.

-- 
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  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  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 
>>> 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 
 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", 

Re: [R] Bilateral matrix

[Miluji Sb]

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.

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  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  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 
>> 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 
>>> 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  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 

Re: [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  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 
> 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 
>> 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  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/posti
>>> 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
>>> 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]]

__
R-help@r-project.org mailing 

Re: [R] Bilateral matrix

[William Dunlap via R-help]

Make current_location and previous_location factors with the same set of
levels.  The levels could be the union of the values in the two columns or
a predetermined list.  E.g.,

> x <- data.frame(previous_location=c("Mount Vernon","Burlington"),
current_location=c("Sedro Woolley","Burlington"))
> allCities <- levels(factor(unlist(x))) # union of observed values
> allCities
[1] "Burlington""Mount Vernon"  "Sedro Woolley"
> x[] <- lapply(x, factor, levels=allCities)
> xtabs(~previous_location + current_location,data=x)
 current_location
previous_location Burlington Mount Vernon Sedro Woolley
Burlington 10 0
Mount Vernon   00 1
Sedro Woolley  00 0

or, using an externally determined set of cities

> allCities <- c("Anacortes","Burlington","Concrete","Mount Vernon","Sedro
Woolley")
> x[] <- lapply(x, factor, levels=allCities)
> xtabs(~previous_location + current_location,data=x)
 current_location
previous_location Anacortes Burlington Concrete Mount Vernon Sedro Woolley
Anacortes 0  000 0
Burlington0  100 0
Concrete  0  000 0
Mount Vernon  0  000 1
Sedro Woolley 0  000 0


Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Wed, May 16, 2018 at 7:49 AM, Miluji Sb  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 
> 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 
> > 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  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", 

Re: [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  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 
> 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  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/posti
>> 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
>> 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]]

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Re: [R] Bilateral matrix

[Bert Gunter]

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 
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  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]]

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Re: [R] Bilateral matrix

[Huzefa Khalil]

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  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
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