CODEWITHSUNDEEP
rcovariancelapply
Jon
Jon

Reputation: 763

Weighted variance-covariance matrices and lapply

I have a list prob with 50 elements. Each element is a 601x3 matrix of probabilities, each row of which represents a complete sample space (i.e., each row of each matrix sums to 1). For instance, here are the first five rows of the first element of prob:

> prob[[1]][1:5,]

           [,1]      [,2]       [,3]
 [1,] 0.6027004 0.3655563 0.03174335
 [2,] 0.6013667 0.3665756 0.03205767
 [3,] 0.6000306 0.3675946 0.03237481
 [4,] 0.5986921 0.3686131 0.03269480
 [5,] 0.5973513 0.3696311 0.03301765

Now, what I want to do is to create the following matrix for each row of each matrix/element in the list prob. Taking the first row, let a = .603, b = .366, and c = .032 (rounding to three decimal places). Then,

> w
         [,1]       [,2]       [,3]
 [1,] a*(1-a)       -a*b       -a*c
 [2,]    -b*a    b*(1-b)       -b*c
 [3,]    -c*a       -c*b    c*(1-c)

Such that:

> w
           [,1]       [,2]       [,3]
 [1,]  0.239391  -0.220698  -0.019296
 [2,] -0.220698   0.232044  -0.011712
 [3,] -0.019296  -0.011712   0.030976

I want to obtain a similar 3x3 matrix 600 more times (for the rest of the rows of this matrix) and then to repeat this entire process 49 more times for the rest of the elements of prob. The only thing I can think of is to call apply within lapply so that I am accessing each row of each matrix one-at-a-time. I'm sure that is not an elegant way to do this (not to mention I can't get it to work), but I can't think of anything else. Can anyone help me out with this? I'd also love to hear suggestions for using a different structure (e.g., is it bad to use matrices within lists?).

Upvotes: 1

Views: 447

Answers (1)

IRTFM
IRTFM

Reputation: 263301

Running this process with lapply on a list of similarly dimensioned matrices should be very simple. If it represents a challenge, then you should post the dput(.) output for a two element list with similar matrices. The challenge is really to do the processing row by row which is illustrated here with the output being a 3x3xN array:

w <- apply(M, 1, function(rw) diag( rw*(1-rw) ) + 
                    rbind( rw*c(0, -rw[1], -rw[1] ), 
                           rw*c(-rw[2],0, -rw[2] ),
                           rw*c(-rw[3], -rw[3], 0)
         )

 )
 w
             [,1]        [,2]        [,3]        [,4]        [,5]
 [1,]  0.23945263  0.23972479  0.23999388  0.24025987  0.24052272
 [2,] -0.22032093 -0.22044636 -0.22056801 -0.22068575 -0.22079962
 [3,] -0.01913173 -0.01927842 -0.01942588 -0.01957412 -0.01972314
 [4,] -0.22032093 -0.22044636 -0.22056801 -0.22068575 -0.22079962
 [5,]  0.23192489  0.23219793  0.23246881  0.23273748  0.23300395
 [6,] -0.01160398 -0.01175156 -0.01190081 -0.01205173 -0.01220435
 [7,] -0.01913173 -0.01927842 -0.01942588 -0.01957412 -0.01972314
 [8,] -0.01160398 -0.01175156 -0.01190081 -0.01205173 -0.01220435
 [9,]  0.03073571  0.03102998  0.03132668  0.03162585  0.03192748

 w <- array(w, c(3,3,5) )
 w
, , 1

            [,1]        [,2]        [,3]
[1,]  0.23945263 -0.22032093 -0.01913173
[2,] -0.22032093  0.23192489 -0.01160398
[3,] -0.01913173 -0.01160398  0.03073571

, , 2

            [,1]        [,2]        [,3]
[1,]  0.23972479 -0.22044636 -0.01927842
[2,] -0.22044636  0.23219793 -0.01175156
[3,] -0.01927842 -0.01175156  0.03102998

.... snipped remaining output

Upvotes: 1

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