从R中的两组列创建所有采样组合
我有下面的数据框，在这两个“组”中，A列和B列以及D列和E列。我希望找到所有组合，然后通过在A列和B列以及D列和E列应用不同过滤器的所有组合进行分组，但此时仅从每组中选择一列。我不知道做这件事的正确公式，事实上问题更大 df= 所以要过滤的组合应该是 过滤器1:A=1和D=1 过滤器2:A=1和D=0 过滤器3:A=1和E=1 过滤器4:A=1和E=0 过滤器5:A=0和D=1 过滤器6:A=0和D=0 过滤器7:A=0和E=1 过滤器8:A=0和E=0 过滤器9:B=1和D=1 过滤器10:B=1和D=0 过滤器11:B=1和E=1 过滤器12:B=1和E=0 过滤器13:B=0和D=1 过滤器14:B=0和D=0 过滤器15:B=0和E=1 过滤器16:B=0和E=0 我想找到一种方法来高效地创建这些过滤器组（总是从列a&B或D&E中绘制一个过滤器），然后找到每个过滤器设置的大小列的平均值和计数。我只是设法做到了这一点，没有不同的小组来采样过滤器 我尝试的形式是：从R中的两组列创建所有采样组合,r,dataframe,group-by,combinations,permutation,R,Dataframe,Group By,Combinations,Permutation,我有下面的数据框，在这两个“组”中，A列和B列以及D列和E列。我希望找到所有组合，然后通过在A列和B列以及D列和E列应用不同过滤器的所有组合进行分组，但此时仅从每组中选择一列。我不知道做这件事的正确公式，事实上问题更大 df= 所以要过滤的组合应该是 过滤器1:A=1和D=1 过滤器2:A=1和D=0 过滤器3:A=1和E=1 过滤器4:A=1和E=0 过滤器5:A=0和D=1 过滤器6:A=0和D=0 过滤器7:A=0和E=1 过滤器8:A=0和E=0 过滤器9:B=1和D=1 过滤器10:B
groupNames%摘要（n=length（Size），avgVar1=mean（Size）））库（tidyverse）
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"))
df根据评论中的讨论，我认为我们可以将组视为变量。因此，我们需要重塑数据帧，使每个因子有一列，然后我们可以使用标准的tidyverse方法。我假设这些组是由列名（A1…Ak，B1…Bk，…）定义的
库（tidyverse）
df%
突变（A=因子（映射chr（A，获取水平）），
B=系数（映射系数（B，获取等级）））
#现在看看因子组合
df_系数%>%
（A，B）组%>%
总结（n=n（），
平均值=平均值（尺寸））
#一个tibble:8x4
#分组：A[3]
#平均数
#
#1“B1”15
#2“B1、B2”1 2
#3“B2”155
#4“A1”“1”3
#5“A1”B1，B2”1 5
#6“A1，A2”“1 1
#7“A1，A2”“B1”2 7
#8英寸A1、A2英寸B1、B2英寸2 1.5
我明确地称之为“A”和“B”。6组似乎仍然可以做到这一点。如果您有更多，则有必要实现自动化，但我不确定如何轻松做到这一点。评论不适用于扩展讨论；这段对话已经结束了，谢谢你，阿列克斯洛。我试着用6个组和每组3个过滤器运行它，但计算起来需要很长时间。我意识到它还测试设置为0的过滤器的情况（例如，A1=1、B1=0、C3=0、D1=0、E2=0、F=0。我如何删除它，以便只测试过滤器为1的所有过滤器组合（当然还有根本不使用相应过滤器的NAs）？我不知道为什么它仍然慢。你能找出一行慢一点吗？要删除设置为0的过滤器，你可以在重新编码后使用filter（stru长度（a）>1）
。要删除所有过滤器为0的行，你可以使用df_factors%>%rowwise（）%%filter（sum（跨越（a:B，stru长度））>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"))
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