[R] Need help to split a given matrix is a "sequential" way

[Ivan Calandra]

Hi,
Not sure exactly how, but I think using a combination of unique() and 
split() could do what you're looking for.
I hope it will help you
Ivan

Le 3/30/2010 09:20, Megh a écrit :
> I need to split a given matrix in a sequential order. Let my matrix is :
>
>    
>> dat<- cbind(sample(c(100,200), 10, T), sample(c(50,100, 150, 180), 10,
>> T), sample(seq(20, 200, by=20), 10, T)); dat
>>      
>        [,1] [,2] [,3]
>   [1,]  200  100   80
>   [2,]  100  180   80
>   [3,]  200  150  180
>   [4,]  200   50  140
>   [5,]  100  150   60
>   [6,]  100   50   60
>   [7,]  100  100  100
>   [8,]  200  150  100
>   [9,]  100   50  120
> [10,]  200   50  180
>
> Now I need to split above matrix according to unique numbers in the 2nd
> column. Therefore I have following :
>
>    
>> dat1<- dat[which(dat[,1] == unique(dat[,1])[1]),]
>> dat2<- dat[-which(dat[,1] == unique(dat[,1])[1]),]; dat1; dat2
>>      
>       [,1] [,2] [,3]
> [1,]  200  100   80
> [2,]  200  150  180
> [3,]  200   50  140
> [4,]  200  150  100
> [5,]  200   50  180
>       [,1] [,2] [,3]
> [1,]  100  180   80
> [2,]  100  150   60
> [3,]  100   50   60
> [4,]  100  100  100
> [5,]  100   50  120
>
> Now each of dat1 and dat2 needs to be splited according to the it's 2nd
> column i.e.
>
>    
>> dat11<- dat1[which(dat1[,2] == unique(dat1[,2])[1]),]
>> dat12<- dat1[which(dat1[,2] == unique(dat1[,2])[2]),]
>> dat13<- dat1[which(dat1[,2] == unique(dat1[,2])[3]),]; dat11; dat12;
>> dat13
>>      
> [1] 200 100  80
>       [,1] [,2] [,3]
> [1,]  200  150  180
> [2,]  200  150  100
>       [,1] [,2] [,3]
> [1,]  200   50  140
> [2,]  200   50  180
>
> similarly for dat2..............
>
> This kind of sequential spliting would continue for
> (no_of_cols_of_ogirinal_matrix -1) times. It would be greate if again I can
> put all those matrices within a "list" object for further calculations.
>
> Therefore you see if the original matrix is of small_size then that can be
> handled manually. However for a moderately large matrix that task would be
> very clumbersome. Therefore I am looking for some mechanized way to do that
> for an arbitrary matrix.
>
> Can anyone here help me on this regard?
>
> Thank you so much for your kind attention.
>
>    

-- 
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calandra at uni-hamburg.de

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