[R] "unfurling" rankings into a matrix of preferences

[Dimitri Liakhovitski]

With NAs it'd be:

rk <- c(2,NA,4,3,1, NA)
outer(rk, rk, "<") + 0

Wow, I still can't believe it - just one line!


On Mon, Jul 27, 2015 at 5:31 PM, Dimitri Liakhovitski
<dimitri.liakhovitski at gmail.com> wrote:
> Wow!
>
> On Mon, Jul 27, 2015 at 5:28 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote:
>> ## I leave it to you to add the NA edges
>>
>>> rk <- c(2,4,3,1)
>>
>>> outer(rk,rk,"<")+0
>>
>>      [,1] [,2] [,3] [,4]
>> [1,]    0    1    1    0
>> [2,]    0    0    0    0
>> [3,]    0    1    0    0
>> [4,]    1    1    1    0
>>
>>
>>
>> Cheers,
>> Bert
>>
>> Bert Gunter
>>
>> "Data is not information. Information is not knowledge. And knowledge
>> is certainly not wisdom."
>>    -- Clifford Stoll
>>
>>
>> On Mon, Jul 27, 2015 at 12:38 PM, Dimitri Liakhovitski
>> <dimitri.liakhovitski at gmail.com> wrote:
>>> I have 5 items in total (1:5), but I show a person only 4 items (1:4)
>>> and ask this person to rank items 1:4 in terms of preferences (1 is
>>> best, 2 is second best, 4 is worst), and I get a vector of ranks:
>>> ranks <- c(2,4,3,1)
>>>
>>> # That means that this person liked item 4 best and item 2 worst.
>>>
>>> I would like to "unfirl" this vector of ranks into a matrix of
>>> preferences where if the row item prefers the column item, then it's a
>>> 1. Otherwise, it's a zero. So, the output should be a 5 by 5 matrix
>>> (because overall we have 5 items, not 4, but item 5 did not
>>> participate in rankings), and it would always have zeros in a
>>> diagonal.:
>>>
>>> 0    1    1    0 NA
>>> 0    0    0    0 NA
>>> 0    1    0    0 NA
>>> 1    1    1    0 NA
>>> NA NA NA NA 0
>>>
>>> I can loop through all possible pairs the person saw and fill the
>>> matrix accordingly, but it seems like a lot of looping. Could one do
>>> it in a more elegant way?
>>>
>>> Thank you very much!
>>>
>>>
>>> --
>>> Dimitri Liakhovitski
>>>
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>
>
>
> --
> Dimitri Liakhovitski



-- 
Dimitri Liakhovitski