Re: [R] Compact Indicator Matrices

From: Douglas Bates <bates_at_stat.wisc.edu>
Date: Mon, 12 May 2008 11:55:19 -0500

On Mon, May 12, 2008 at 11:27 AM, amarkos <amarkos_at_gmail.com> wrote: > Thanks, it works!

> Could you please provide the direct method you mentioned for the > multivariate case?

I'm not sure what you mean. I looked at what I wrote and I don't see anything that would fit that description.

May I suggest that you continue to cc: the R-help list on the discussion. I can't always respond rapidly to requests and there are many who read the list that can.

> On May 12, 4:30 pm, "Douglas Bates" <ba..._at_stat.wisc.edu> wrote:

>> On Sun, May 11, 2008 at 9:49 AM, amarkos <amar...@gmail.com> wrote:
>> > On May 11, 4:47 pm, "Douglas Bates" <ba..._at_stat.wisc.edu> wrote:
>>
>> >> Do you mean that you want to collapse similar rows into a single row
>> >> and perhaps a count of the number of times that this row occurs?
>>
>> > Let me rephrase the problem by providing an example.
>>
>> > Input:
>>
>> > A =
>> > [,1] [,2]
>> > [1,] 1 1
>> > [2,] 1 3
>> > [3,] 2 1
>> > [4,] 1 2
>> > [5,] 2 1
>> > [6,] 1 2
>> > [7,] 1 1
>> > [8,] 1 2
>> > [9,] 1 3
>> > [10,] 2 1
>>
>> An important question here is do you start with two or more variables
>> like the columns of your matrix A? If so, there is a more direct
>> method of getting the answers that you want. The natural way to store
>> such variables in R is as factors. I prefer to use letters instead of
>> numbers to represent the levels of a factor (that way I don't confuse
>> a factor with a numeric variable when I look at rows) so I would
>> create a data frame with two factors instead of a matrix.
>>
>> > V1 <- factor(c(1,1,2,1,2,1,1,1,1,2), labels = LETTERS[1:2])
>> > V2 <- factor(c(1,3,1,2,1,2,1,2,3,1), labels = letters[1:3])
>> > df <- data.frame(f1 = V1, f2 = V2)
>> > df
>>
>> f1 f2
>> 1 A a
>> 2 A c
>> 3 B a
>> 4 A b
>> 5 B a
>> 6 A b
>> 7 A a
>> 8 A b
>> 9 A c
>> 10 B a
>>
>> You could produce the indicator matrix and check for unique rows, etc.
>> - I will show that below - but all you need is the interaction of the
>> two factors
>>
>> > df$f12 <- with(df, f1:f2)[drop = TRUE]
>> > df
>>
>> f1 f2 f12
>> 1 A a A:a
>> 2 A c A:c
>> 3 B a B:a
>> 4 A b A:b
>> 5 B a B:a
>> 6 A b A:b
>> 7 A a A:a
>> 8 A b A:b
>> 9 A c A:c
>> 10 B a B:a> str(df)
>>
>> 'data.frame': 10 obs. of 3 variables:
>> $ f1 : Factor w/ 2 levels "A","B": 1 1 2 1 2 1 1 1 1 2
>> $ f2 : Factor w/ 3 levels "a","b","c": 1 3 1 2 1 2 1 2 3 1
>> $ f12: Factor w/ 4 levels "A:a","A:b","A:c",..: 1 3 4 2 4 2 1 2 3 4
>>
>> > table(df$f12)
>>
>> A:a A:b A:c B:a
>> 2 3 2 3> as.numeric(df$f12)
>>
>> [1] 1 3 4 2 4 2 1 2 3 4
>>
>> Notice that this shows you that there are four distinct combinations
>> that occur 2, 3, 2 and 3 times respectively; the first combination
>> occurs in rows 1 and 7, it consists of the first level of f1 and the
>> first level of f2, etc.
>>
>> If you really do want the indicator matrix you could generate it as
>>
>> > (ind <- cbind(model.matrix(~ 0 + f1, df), model.matrix(~ 0 + f2, df)))
>>
>> f1A f1B f2a f2b f2c
>> 1 1 0 1 0 0
>> 2 1 0 0 0 1
>> 3 0 1 1 0 0
>> 4 1 0 0 1 0
>> 5 0 1 1 0 0
>> 6 1 0 0 1 0
>> 7 1 0 1 0 0
>> 8 1 0 0 1 0
>> 9 1 0 0 0 1
>> 10 0 1 1 0 0> unique(ind)
>>
>> f1A f1B f2a f2b f2c
>> 1 1 0 1 0 0
>> 2 1 0 0 0 1
>> 3 0 1 1 0 0
>> 4 1 0 0 1 0
>>
>> but working with the factors is generally much simpler than working
>> with the indicators.
>>
>>
>>
>> > # Indicator matrix
>> > A <- data.frame(lapply(data.frame(obj), as.factor))
>>
>> > nocases <- dim(obj)[1]
>> > novars <- dim(obj)[2]
>>
>> > # variable levels
>> > levels.n <- sapply(obj, nlevels)
>> > n <- cumsum(levels.n)
>>
>> > # Indicator matrix calculations
>> > Z <- matrix(0, nrow = nocases, ncol = n[length(n)])
>> > newdat <- lapply(obj, as.numeric)
>> > offset <- (c(0, n[-length(n)]))
>> > for (i in 1:novars)
>> > Z[1:nocases + (nocases * (offset[i] + newdat[[i]] - 1))] <- 1
>>
>> > #######
>>
>> > Output:
>>
>> > Z =
>>
>> > [,1] [,2] [,3] [,4] [,5]
>> > [1,] 1 0 1 0 0
>> > [2,] 1 0 0 0 1
>> > [3,] 0 1 1 0 0
>> > [4,] 1 0 0 1 0
>> > [5,] 0 1 1 0 0
>> > [6,] 1 0 0 1 0
>> > [7,] 1 0 1 0 0
>> > [8,] 1 0 0 1 0
>> > [9,] 1 0 0 0 1
>> > [10,] 0 1 1 0 0
>>
>> > Z is an indicator matrix in the Multiple Correspondence Analysis
>> > framework.
>> > My problem is to collapse identical rows (e.g. 2 and 9) into a single
>> > row and
>> > store the row ids.
>>
>> > ______________________________________________
>> > R-h..._at_r-project.org mailing list
>> >https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>>
>> ______________________________________________
>> R-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> Angelos Markos
> Dr. of Applied Informatics,
> University of Macedonia, Greece
>

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