在多列中查找最高频率

Question

我有一个包含 10 列和 2000 行的 data_frame。我的示例数据如下所示：

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

我正在尝试为 rs_id =0、1 和 2 找到最高组合（A.Ag、B.Bg、C.Ag）。我该如何实现？输出将是

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

Answer 1

此方法将数据从宽格式重塑为长格式（同时融合两个测量列），为每个唯一组合选择具有最高 Ag 值的行rs_id、Code 和 variable。最后，结果再次从长格式重塑为宽格式，列顺序重新排列为 return 预期结果：

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

编辑

OP 要求解释最后一个链式 data.table 表达式 setcolorder(.SD, c(1:2, as.vector(outer(c(0, 4), 3:6, "+")))).

此表达式通过引用对结果的列进行排序，即不进行复制。重塑多个 value.var 时，列按 value.var:

分组

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

而 OP 希望输出按 variable 分组。所以所需的列顺序是

c(1, 2, 3, 7, 4, 8, 5, 9, 6, 10)。

1 和 2 表示 id.var 列。 as.vector(outer(c(0, 4), 3:6, "+"))) 只是一种节省输入 3, 7, 4, 8, 5, 9, 6, 10.

的方法

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

编辑 2

代码可以进一步精简。 c() 中不需要调用 as.vector()，因为 c() 将数组转换为向量。所以，而不是

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

我们可以写

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

数据

请注意，我已经完成了最后两列缺失的列 headers。

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"
)

在多列中查找最高频率

Finding the Highest Frequencies in multiple columns

r

sapply

data.table

编辑

编辑 2

数据