[R] Efficiency Question - Nested lapply or nested for loop
Gabor Grothendieck
ggrothendieck at gmail.com
Sat Oct 9 00:28:44 CEST 2010
On Fri, Oct 8, 2010 at 12:47 PM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
> 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)
A couple of further improvements are that we can replace the array,
arr, with a matrix, mat, and also we can add dimension names in the
table() call:
mat <- matrix(t(data[-1]), 4)
table(Gmax = apply(mat[-4,], 2, which.max), Gtru = mat[4,] + 1)
--
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