从 R 中的两组列创建所有采样组合

Question

I have the dataframe below and within this "two groups", the columns A&B and D&E. I would like to find all combinations and then to group by all combinations of applying different filters at columns A&B and D&E but in the form of only choosing 1 column from each group at the time. I dont know the right formula to do this and the problem is way bigger in reality.

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

So the combinations to filter should be

Filter 1: A=1 AND D=1

Filter 2: A=1 AND D=0

Filter 3: A=1 AND E=1

Filter 4: A=1 AND E=0

Filter 5: A=0 AND D=1

Filter 6: A=0 AND D=0

Filter 7: A=0 AND E=1

Filter 8: A=0 AND E=0

Filter 9: B=1 AND D=1

Filter 10: B=1 AND D=0

Filter 11: B=1 AND E=1

Filter 12: B=1 AND E=0

Filter 13: B=0 AND D=1

Filter 14: B=0 AND D=0

Filter 15: B=0 AND E=1

Filter 16: B=0 AND E=0

I want to find a way to efficiently create these filter groups (drawing always 1 filter from either columns A&B or D&E) and then to find the average and count of the Size column for each filter setting. I only managed to do this without different groups to sample the filter from.

What I tried was in the form of this:

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))})

It treats the four columns equally and does not consider sampling from the 2 groups. What could I do to the code to make this work?

Thank you very much.

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

Following the discussion in the comments, I think we can treat the groups as variables. So we need to reshape the dataframe to have one column per factor, then we can use standard tidyverse approaches. I'm assuming the groups are defined by the column names (A1...Ak, B1...Bk, ...).

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

I called "A" and "B" explicitly. It seems still doable to do that with 6 groups. If you have more, it would become necessary to automatize, but I'm not sure how to do that easily.