计算数据框中特定列（SUM、AVG、STDEV）的所有嵌套级别聚合

Posted 2023-02-16

【中文标题】计算数据框中特定列（SUM、AVG、STDEV）的所有嵌套级别聚合

【英文标题】：Calculating ALL nested level aggregations of specific Column ( SUM, AVG, STDEV) in dataframe

【发布时间】：2020-05-09 21:09:54

【问题描述】：

我有一个如下所示的表格（简化）：

col_A   col_B   col_C
A       37      2
B       28      7
C       10      5
D       11      5
E       99      4

我想得到一个表，其中包含 col_A 的每个级别的所有嵌套组合，并计算子组内的平均值：例如，choose-any-2 表看起来像（10 个唯一级别组合）：

Grp_2   AVG (col_B/col_C)
A,B     7.76
A,C     6.61
A,D     7.55
…       …
D,E     12.99

Choose-any-4 看起来像（5 个独特的关卡组合）：

Grp_4   AVG (col_B/col_C)
A,B,C,D 7.84
A,B,C,E 6.68
A,C,D,E 7.63
…       …
B,C,D,E 13.12 

(order od preference) R, SQL(postgres, ANSI) , Python.; 我目前在 R 中的解决方案（如下）不能很好地扩展col_A 的级别数增长：

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)

你能帮忙改进吗？

【问题讨论】：

我删除了 SQL 标签，因为您的问题是关于 R 中的数据框。

【参考方案1】：

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)

输出：

     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

数据：

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")

【参考方案2】：

一个想法是使用combn 来获取行的所有组合（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))

数据

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))