Encontrar as freqüências mais altas em várias colunas - r, data.table, sapply

Eu tenho um data_frame contendo 10 colunas e 2000 linhas. Meus dados de amostra seriam semelhantes a:

rs_id   Code    Combination_Ag  A.Ag    Combination_Bg  B.Ag    Combination_Cg  C.Ag
rs_1    0   1:01/1:01   1   13:02/13:02 1   03:04/03:04 6   1:01/1:01   1
rs_1    0   1:01/11:01  2   13:02/49:01 2   03:04/15:02 1   1:01/15:01  1
rs_1    1   1:01/2:01   6   13:02/57:01 1   03:04/7:01  2   1:01/3:01   1
rs_1    2   1:01/2:05:  1   13:02/8:01  1   06:02/06:02 3   1:01/4:04   1
rs_1    2   1:01/24:02  3   14:01/14:02 1   06:02/15:02 1   1:01/4:04   3
rs_2    0   1:01/3:01   1   14:01/7:02  1   06:02/2:02: 1   1:01/4:07   1
rs_2    1   1:01/31:01  1   15:01/15:01 1   06:02/3:03  1   1:01/7:01   2
rs_2    1   11:01/2:01  4   15:01/18:01 1   06:02/3:04  1   10:01/14:01 1
rs_2    2   11:01/25:01 1   15:01/44:02 2   06:02/4:01  1   10:01/3:01  5

Eu estou tentando encontrar a combinação mais alta (A.Ag, B.Bg C.Ag) para rs_id = 0, 1 e 2. Como posso conseguir isso? A saída seria

rs_1    0   1:01/11:01   2   13:02/49:01    2   03:04/03:04 6   1:01/1:01   1
rs_1    1   1:01/2:01    6   13:02/57:01    1   03:04/7:01  2   1:01/3:01   1
rs_1    2   1:01/24:02   3   06:02/06:02    3   06:02/15:02 1   1:01/4:04   3
rs_2    0   1:01/3:01    1   14:01/7:02     1   06:02/2:02: 1   1:01/4:07   1
rs_2    1   11:01/2:01   4   15:01/18:01    1   06:02/3:04  1   10:01/14:01 1
rs_2    2   11:01/25:01  1   15:01/44:02    2   06:02/4:01  1   10:01/3:01  5

Respostas:

3 para resposta № 1

Essa abordagem remodela os dados do formato amplo para o longo (fusão dois medir colunas simultaneamente), escolhe a linha com o topo Ag valor para cada combinação única de rs_id, Codee variable. Por fim, o resultado é remodelado novamente do formato longo para o formato grande com a ordem da coluna reorganizada para retornar o resultado esperado:

library(data.table)
cols <- c("Combination", "Ag")
melt(setDT(DF), measure.vars = patterns("Combination", "[A-D][.]Ag"),
value.name = cols)[
, variable := forcats::lvls_revalue(variable, LETTERS[1:4])][
, .SD[which.max(Ag)], by = .(rs_id, Code, variable)][
, dcast(.SD, rs_id + Code ~ variable, value.var = cols)][
, setcolorder(.SD, c(1:2, as.vector(outer(c(0, 4), 3:6, "+"))))]

   rs_id Code Combination_A Ag_A Combination_B Ag_B Combination_C Ag_C Combination_D Ag_D
1:  rs_1    0    1:01/11:01    2   13:02/49:01    2   03:04/03:04    6     1:01/1:01    1
2:  rs_1    1     1:01/2:01    6   13:02/57:01    1    03:04/7:01    2     1:01/3:01    1
3:  rs_1    2    1:01/24:02    3    13:02/8:01    1   06:02/06:02    3     1:01/4:04    3
4:  rs_2    0     1:01/3:01    1    14:01/7:02    1   06:02/2:02:    1     1:01/4:07    1
5:  rs_2    1    11:01/2:01    4   15:01/15:01    1    06:02/3:03    1     1:01/7:01    2
6:  rs_2    2   11:01/25:01    1   15:01/44:02    2    06:02/4:01    1    10:01/3:01    5

Editar

O OP pediu uma explicação sobre o último dos acorrentados data.table expressões setcolorder(.SD, c(1:2, as.vector(outer(c(0, 4), 3:6, "+")))).

Esta expressão ordena as colunas do resultado por referência, ou seja, sem copiar. Ao remodelar vários value.vars as colunas são agrupadas por value.var:

melt(setDT(DF), measure.vars = patterns("Combination", "[A-D][.]Ag"),
value.name = cols)[
, variable := forcats::lvls_revalue(variable, LETTERS[1:4])][
, .SD[which.max(Ag)], by = .(rs_id, Code, variable)][
, dcast(.SD, rs_id + Code ~ variable, value.var = cols)]

   rs_id Code Combination_A Combination_B Combination_C Combination_D Ag_A Ag_B Ag_C Ag_D
1:  rs_1    0    1:01/11:01   13:02/49:01   03:04/03:04     1:01/1:01    2    2    6    1
2:  rs_1    1     1:01/2:01   13:02/57:01    03:04/7:01     1:01/3:01    6    1    2    1
3:  rs_1    2    1:01/24:02    13:02/8:01   06:02/06:02     1:01/4:04    3    1    3    3
4:  rs_2    0     1:01/3:01    14:01/7:02   06:02/2:02:     1:01/4:07    1    1    1    1
5:  rs_2    1    11:01/2:01   15:01/15:01    06:02/3:03     1:01/7:01    4    1    1    2
6:  rs_2    2   11:01/25:01   15:01/44:02    06:02/4:01    10:01/3:01    1    2    1    5

enquanto o OP espera que a saída seja agrupada por variable. Portanto, a ordem da coluna desejada é

c(1, 2, 3, 7, 4, 8, 5, 9, 6, 10).

1 e 2 denotar o id.var colunas. as.vector(outer(c(0, 4), 3:6, "+"))) é apenas uma maneira de economizar digitando 3, 7, 4, 8, 5, 9, 6, 10.

outer(c(0, 4), 3:6, "+")

     [,1] [,2] [,3] [,4]
[1,]    3    4    5    6
[2,]    7    8    9   10

as.vector(outer(c(0, 4), 3:6, "+"))

[1]  3  7  4  8  5  9  6 10

Editar 2

O código pode ser mais simplificado. A chamada para as.vector() não é necessário dentro c() Como c() transforma matrizes em vetores. Então, ao invés de

c(1:2, as.vector(outer(c(0, 4), 3:6, "+")))

nós podemos escrever

c(1:2, outer(c(0, 4), 3:6, "+"))

Dados

Observe que eu completei os cabeçalhos de coluna ausentes para as duas últimas colunas.

library(data.table)
DF <- fread(
"rs_id   Code    Combination_Ag  A.Ag    Combination_Bg  B.Ag    Combination_Cg  C.Ag   Combination_Dg  D.Ag
rs_1    0   1:01/1:01   1   13:02/13:02 1   03:04/03:04 6   1:01/1:01   1
rs_1    0   1:01/11:01  2   13:02/49:01 2   03:04/15:02 1   1:01/15:01  1
rs_1    1   1:01/2:01   6   13:02/57:01 1   03:04/7:01  2   1:01/3:01   1
rs_1    2   1:01/2:05:  1   13:02/8:01  1   06:02/06:02 3   1:01/4:04   1
rs_1    2   1:01/24:02  3   14:01/14:02 1   06:02/15:02 1   1:01/4:04   3
rs_2    0   1:01/3:01   1   14:01/7:02  1   06:02/2:02: 1   1:01/4:07   1
rs_2    1   1:01/31:01  1   15:01/15:01 1   06:02/3:03  1   1:01/7:01   2
rs_2    1   11:01/2:01  4   15:01/18:01 1   06:02/3:04  1   10:01/14:01 1
rs_2    2   11:01/25:01 1   15:01/44:02 2   06:02/4:01  1   10:01/3:01  5"
)

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