從 R 中的兩組列創建所有采樣組合

Question

我有 dataframe 下面和這“兩組”，列 A&B 和 D&E。 我想找到所有組合，然后按在 A&B 和 D&E 列應用不同過濾器的所有組合進行分組，但形式是當時只從每個組中選擇 1 列。 我不知道執行此操作的正確公式，而實際上問題要大得多。

df=

     Size    A     B     D     E
       1     1     1     0     0
       5     0     0     1     0
       10    1     1     1     0
       3     1     0     0     0
       2     1     1     1     1
       55    0     0     0     1
       5     1     0     1     1
       2     0     0     1     1
       1     1     1     1     1
       4     1     1     1     0

所以過濾的組合應該是

過濾器 1：A=1 和 D=1

過濾器 2：A=1 和 D=0

過濾器 3：A=1 AND E=1

過濾器 4：A=1 AND E=0

過濾器 5：A=0 AND D=1

過濾器 6：A=0 和 D=0

過濾器 7：A=0 AND E=1

過濾器 8：A=0 和 E=0

過濾器 9：B=1 且 D=1

過濾器 10：B=1 且 D=0

過濾器 11：B=1 且 E=1

過濾器 12：B=1 且 E=0

過濾器 13：B=0 且 D=1

過濾器 14：B=0 且 D=0

過濾器 15：B=0 AND E=1

過濾器 16：B=0 AND E=0

我想找到一種方法來有效地創建這些過濾器組（始終從 A&B 或 D&E 列中繪制 1 個過濾器），然后找到每個過濾器設置的 Size 列的平均值和計數。 我只設法在沒有不同組的情況下做到這一點來對過濾器進行采樣。

我嘗試的是這樣的形式：

groupNames <- names(df)[2:5]

myGroups <- Map(combn,list(groupNames),seq_along(groupNames),simplify = FALSE) %>% unlist(recursive = FALSE)

results = lapply(myGroups, FUN = function(x) {do.call(what = group_by_, args = c(list(df), x)) %>% summarise( n = length(Size), avgVar1 = mean(Size))})

它平等對待四列，不考慮從 2 組中抽樣。 我可以對代碼做些什么來完成這項工作？

非常感謝。

Answer 1

library(tidyverse)
df <- tribble(~Size, ~A, ~B, ~D, ~E,
              1, "1", "1", "0", "0",
              5, "0", "0", "1", "0",
              10, "1", "1", "1", "0",
              3, "1", "0", "0", "0",
              2, "1", "1", "1", "1",
              55, "0", "0", "0", "1",
              5, "1", "0", "1", "1",
              2, "0", "0", "1", "1",
              1, "1", "1", "1", "1",
              4, "1", "1", "1", "0")
p <- function(...) paste0(...) # for legibility, should rather use glue

all_filtering_groups <- list(c("A", "B"), c("D", "E")) # assuming these are known
all_combns <- map(1:length(all_filtering_groups), ~ combn(all_filtering_groups, .))
res <- list(length(all_combns))

#microbenchmark::microbenchmark({
for(comb_length in seq_along(all_combns)){
  res[[comb_length]] <- list(ncol(all_combns[[comb_length]]))
  for(col_i in seq_len(ncol(all_combns[[comb_length]]))){
    
    filtering_groups <- all_combns[[comb_length]][,col_i]
    group_names <- as.character(seq_along(filtering_groups))
    
    
    # prepare grid of all combinations
    filtering_combs <- c(filtering_groups, rep(list(0:1), length(filtering_groups)))
    names(filtering_combs) <- c(p("vars_", group_names), p("vals_", group_names))
    full_grid <- expand.grid(filtering_combs)
    
    for(ll in 1:nrow(full_grid)){ # for each line in the full_grid
      # find df lines that correspond
      cond <- as.logical(rep(TRUE, nrow(df)))
      for(grp in group_names){
        cond <- cond & df[[full_grid[p("vars_", grp)][ll,]]] == full_grid[p("vals_", grp)][ll,]
      }
      # and compute whatever
      full_grid$lines[ll] <- paste(which(cond), collapse = ", ") #for visual verification
      full_grid$n[ll] <- length(df$Size[cond])
      full_grid$sum[ll] <- sum(df$Size[cond])
      full_grid$mean[ll] <- mean(df$Size[cond])
    }
    res[[comb_length]][[col_i]] <- full_grid
    
  }
}
#}, times = 10) #microbenchmark

bind_rows(res) %>% relocate(starts_with("vars") | starts_with("vals"))

Answer 2

在評論中的討論之后，我認為我們可以將組視為變量。 因此，我們需要將 dataframe 重塑為每個因子一列，然后我們可以使用標准的 tidyverse 方法。 我假設這些組是由列名（A1...Ak，B1...Bk，...）定義的。

library(tidyverse)
df <- tribble(~Size, ~A1, ~A2, ~B1, ~B2,
              1, "1", "1", "0", "0",
              5, "0", "0", "1", "0",
              10, "1", "1", "1", "0",
              3, "1", "0", "0", "0",
              2, "1", "1", "1", "1",
              55, "0", "0", "0", "1",
              5, "1", "0", "1", "1",
              2, "0", "0", "1", "1",
              1, "1", "1", "1", "1",
              4, "1", "1", "1", "0")

get_levels <- function(col){
  paste(names(col)[col == "1"], collapse = ",")
}
# Rewrite with groups as factors
df_factors <- df %>%
  mutate(id = row_number()) %>%  #to avoid aggregating same Size
  nest(A = starts_with("A"), B = starts_with("B")) %>%
  mutate(A = factor(map_chr(A, get_levels)),
         B = factor(map_chr(B, get_levels)))

# Now look at factor combinations
df_factors %>%
  group_by(A, B) %>%
  summarize(n = n(),
            mean = mean(Size))

# A tibble: 8 x 4
# Groups:   A [3]
#   A       B           n  mean
#   <fct>   <fct>   <int> <dbl>
# 1 ""      "B1"        1   5  
# 2 ""      "B1,B2"     1   2  
# 3 ""      "B2"        1  55  
# 4 "A1"    ""          1   3  
# 5 "A1"    "B1,B2"     1   5  
# 6 "A1,A2" ""          1   1  
# 7 "A1,A2" "B1"        2   7  
# 8 "A1,A2" "B1,B2"     2   1.5

我明確地稱為“A”和“B”。 使用 6 組似乎仍然可行。 如果您有更多，則有必要進行自動化，但我不確定如何輕松做到這一點。