Python 计算dataframe中特定列（SUM、AVG、STDEV）的所有嵌套级别聚合

我有一张如下所示的表格（简化）：

我想得到一个包含每个列a级别的所有嵌套组合的表，并计算（比如）子组内的平均值：例如，choose-any-2表看起来像（10个唯一的级别组合）：

选择-any-4看起来像（5个独特的级别组合）：

（优先顺序）R，SQL（postgres，ANSI），Python。； 我在R中的当前解决方案（如下）不能很好地扩展

col\u A

的级别数：

require(tidyverse)
df <- tibble(col_A=c("A", "B","C", "D", "E"), col_B=c(37,28,10,11,99), col_C=c(2,7,5,5,4))

nested_subgroup_agg <- function(choice = 2, mydf = NULL) {
  library(tidyverse)
  dfx <-
    combn(c("A", "B", "C", "D", "E"), choice) %>%
    t() %>%
    as_tibble()
  try(if (choice <= 1) {
    stop("Can't Choose less than 2 levels at a time")
  }
  else{
    if (choice == 2) {
      val <- map_dbl(1:nrow(dfx), function(i) {
        (mydf$col_B[mydf$col_A == dfx$V1[i]] + mydf$col_B[mydf$col_A == dfx$V2[i]]) /
          (mydf$col_C[mydf$col_A == dfx$V1[i]] + mydf$col_C[mydf$col_A == dfx$V2[i]])
      })
    }
    else{
      if (choice == 3) {
        val <- map_dbl(1:nrow(dfx), function(i) {
          (mydf$col_B[mydf$col_A == dfx$V1[i]] + mydf$col_B[mydf$col_A == dfx$V2[i]] + mydf$col_B[mydf$col_A == dfx$V3[i]]) /
            (mydf$col_C[mydf$col_A == dfx$V1[i]] + mydf$col_C[mydf$col_A == dfx$V2[i]] + mydf$col_C[mydf$col_A == dfx$V3[i]])
        })
      }
      else{
        if (choice == 4) {
          val <- map_dbl(1:nrow(dfx), function(i) {
            (mydf$col_B[mydf$col_A == dfx$V1[i]] + mydf$col_B[mydf$col_A == dfx$V2[i]] + mydf$col_B[mydf$col_A == dfx$V3[i]] + mydf$col_B[mydf$col_A == dfx$V4[i]]) /
              (mydf$col_C[mydf$col_A == dfx$V1[i]] + mydf$col_C[mydf$col_A == dfx$V2[i]] + mydf$col_C[mydf$col_A == dfx$V3[i]] + mydf$col_C[mydf$col_A == dfx$V4[i]])
          })
        }
      }
    }
  })
  dfx$val <- val
  dfx
}
## Example
df <-
  tibble(
    col_A = c("A", "B", "C", "D", "E"),
    col_B = c(37, 28, 10, 11, 99),
    col_C = c(2, 7, 5, 5, 4)
  )
nested_subgroup_agg(choice = 4, mydf = df)

require（tidyverse）
df一个想法是使用
combn
获得所有行的组合（考虑到每行有1个字母），然后简单地每2行聚合一次，即

#get a df with all combination of rows
new_d <- dd[c(combn(nrow(dd), 2)),]

#Aggregate
#You can use `aggregate` or `lapply(split())`
lapply(split(new_d, rep(seq((nrow(new_d)) / 2), each = 2)), function(i)sum(i$col_C))

使用data.table的选项：

nested_subgroup_agg <- function(choice=2, mydf) {
    ans <- setDT(mydf)[.(g=rep(seq(choose(.N, choice)), each=choice), col_A=c(combn(col_A, choice))), on=.(col_A)][, 
        .(toString(col_A), sum(col_B) / sum(col_C)), g]
    setnames(ans, names(ans)[-1L], c(paste0("Grp_", choice), "val"))[]
}

nested_subgroup_agg(3, DT)

数据：

库（data.table）
我删除了SQL标记，因为您的问题是关于R中的数据帧。
require(tidyverse)
df <- tibble(col_A=c("A", "B","C", "D", "E"), col_B=c(37,28,10,11,99), col_C=c(2,7,5,5,4))

nested_subgroup_agg <- function(choice = 2, mydf = NULL) {
  library(tidyverse)
  dfx <-
    combn(c("A", "B", "C", "D", "E"), choice) %>%
    t() %>%
    as_tibble()
  try(if (choice <= 1) {
    stop("Can't Choose less than 2 levels at a time")
  }
  else{
    if (choice == 2) {
      val <- map_dbl(1:nrow(dfx), function(i) {
        (mydf$col_B[mydf$col_A == dfx$V1[i]] + mydf$col_B[mydf$col_A == dfx$V2[i]]) /
          (mydf$col_C[mydf$col_A == dfx$V1[i]] + mydf$col_C[mydf$col_A == dfx$V2[i]])
      })
    }
    else{
      if (choice == 3) {
        val <- map_dbl(1:nrow(dfx), function(i) {
          (mydf$col_B[mydf$col_A == dfx$V1[i]] + mydf$col_B[mydf$col_A == dfx$V2[i]] + mydf$col_B[mydf$col_A == dfx$V3[i]]) /
            (mydf$col_C[mydf$col_A == dfx$V1[i]] + mydf$col_C[mydf$col_A == dfx$V2[i]] + mydf$col_C[mydf$col_A == dfx$V3[i]])
        })
      }
      else{
        if (choice == 4) {
          val <- map_dbl(1:nrow(dfx), function(i) {
            (mydf$col_B[mydf$col_A == dfx$V1[i]] + mydf$col_B[mydf$col_A == dfx$V2[i]] + mydf$col_B[mydf$col_A == dfx$V3[i]] + mydf$col_B[mydf$col_A == dfx$V4[i]]) /
              (mydf$col_C[mydf$col_A == dfx$V1[i]] + mydf$col_C[mydf$col_A == dfx$V2[i]] + mydf$col_C[mydf$col_A == dfx$V3[i]] + mydf$col_C[mydf$col_A == dfx$V4[i]])
          })
        }
      }
    }
  })
  dfx$val <- val
  dfx
}
## Example
df <-
  tibble(
    col_A = c("A", "B", "C", "D", "E"),
    col_B = c(37, 28, 10, 11, 99),
    col_C = c(2, 7, 5, 5, 4)
  )
nested_subgroup_agg(choice = 4, mydf = df)

#get a df with all combination of rows
new_d <- dd[c(combn(nrow(dd), 2)),]

#Aggregate
#You can use `aggregate` or `lapply(split())`
lapply(split(new_d, rep(seq((nrow(new_d)) / 2), each = 2)), function(i)sum(i$col_C))

dput(dd)
structure(list(col_A = structure(1:5, .Label = c("A", "B", "C", 
"D", "E"), class = "factor"), col_B = c(37L, 28L, 10L, 11L, 99L
), col_C = c(2L, 7L, 5L, 5L, 4L)), class = "data.frame", row.names = c(NA, 
-5L))

nested_subgroup_agg <- function(choice=2, mydf) {
    ans <- setDT(mydf)[.(g=rep(seq(choose(.N, choice)), each=choice), col_A=c(combn(col_A, choice))), on=.(col_A)][, 
        .(toString(col_A), sum(col_B) / sum(col_C)), g]
    setnames(ans, names(ans)[-1L], c(paste0("Grp_", choice), "val"))[]
}

nested_subgroup_agg(3, DT)

     g   Grp_3       val
 1:  1 A, B, C  5.357143
 2:  2 A, B, D  5.428571
 3:  3 A, B, E 12.615385
 4:  4 A, C, D  4.833333
 5:  5 A, C, E 13.272727
 6:  6 A, D, E 13.363636
 7:  7 B, C, D  2.882353
 8:  8 B, C, E  8.562500
 9:  9 B, D, E  8.625000
10: 10 C, D, E  8.571429

library(data.table)
DT <- fread("col_A   col_B   col_C
A       37      2
B       28      7
C       10      5
D       11      5
E       99      4")