[R] Efficiency Question - Nested lapply or nested for loop

[Gabor Grothendieck]

On Fri, Oct 8, 2010 at 11:35 AM, epowell <EPowell1 at med.miami.edu> wrote:
>
> My data looks like this:
>
>> data
>  name G_hat_0_0 G_hat_1_0 G_hat_2_0 G_0 G_hat_0_1 G_hat_1_1 G_hat_2_1 G_1
> 1  rs0  0.488000  0.448625  0.063375   1  0.480875  0.454500  0.064625   1
> 2  rs1  0.002375  0.955375  0.042250   1  0.000000  0.062875  0.937125   2
> 3  rs2  0.050375  0.835875  0.113750   1  0.877250  0.115875  0.006875   0
> 4  rs3  0.000000  0.074750  0.925250   2  0.897750  0.102000  0.000250   0
> 5  rs4  0.000125  0.052375  0.947500   2  0.261500  0.724125  0.014375   1
> 6  rs5  0.003750  0.092125  0.904125   2  0.023000  0.738125  0.238875   1
>
> And my task is:
> For each individual (X) on each row, to find the index corresponding to the
> max of G_hat_X_0, G_hat_X_1, G_hat_X_2 and then increment the cell of the
> confusion matrix with the row corresponding to that index and the column
> corresponding to G_X.
>
> For example, in the first row and the first individual, the index with the
> max value (0.488000) is 0 and the G_0 value is 1, so I would increment
> matrix index of the first row and second column. (Note that the ranges
> between rows and columns are one off.  That is accounted for in the code.)
>
> In reality the data will be much bigger, containing 10000 rows and a
> variable number of columns (inds) between 10 and 500.
>
> The correct result is:
>
>> cmat
>        tru_rr tru_rv tru_vv
> call_rr      2      2      0
> call_rv      0      4      0
> call_vv      0      0      4
>

If we reform data into a 3d array, arr, it can be vectorized like this
where the two args of table correspond to Gmax and Gtru:

arr <- array(t(data[-1]), c(4, 2, 6))
table(apply(arr[-4,,], 2:3, which.max), arr[4,,] + 1)

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