tidyr重塑tibble而不扩展.grid

我想在不使用expand.grid的情况下重塑TIBLE。虽然expand.grid+delete missing obs+delete“fliped duplicates”（即a，b与b，a相同）可以工作，但如果我有许多组合，计算速度会非常慢

这是我想要实现的虚拟版本：

library(dplyr)
library(tidyr)

initial_data <- tibble(x = c("east","east","east"), y = c("a","b","c"), z = c(0.1,0.2,0.3))

> initial_data
# A tibble: 3 x 3
      x     y     z
  <chr> <chr> <dbl>
1  east     a   0.1
2  east     b   0.2
3  east     c   0.3

final_data <- tibble(x = c("east","east","east"), y1 = c("a","a","b"), y2 = c("b","c","c"), z1 = c(0.1,0.1,0.2), z2 = c(0.2,0.3,0.3))

> final_data
# A tibble: 3 x 5
      x    y1    y2    z1    z2
  <chr> <chr> <chr> <dbl> <dbl>
1  east     a     b   0.1   0.2
2  east     a     c   0.1   0.3
3  east     b     c   0.2   0.3

库（dplyr）
图书馆（tidyr）
初始数据初始数据
#一个tibble:3x3
x y z
1东a 0.1
2东b 0.2
3东c 0.3
最终数据最终数据
#一个tibble:3x5
x y1 y2 z1 z2
1东a b 0.1 0.2
2东a c 0.1 0.3
3东b c 0.2 0.3

这是可行的，但效率极低：

expand_data <- as_tibble(expand.grid(initial_data$x, initial_data$y, initial_data$y)) %>% 
  filter(Var2 != Var3) %>% 
  distinct()

index <- !duplicated(t(apply(expand_data, 1, sort)))
expand_data <- expand_data[index, ] %>% 
  left_join(initial_data, by = c("Var1" = "x", "Var2" = "y")) %>% 
  left_join(initial_data, by = c("Var1" = "x", "Var3" = "y"))

> expand_data
# A tibble: 3 x 5
   Var1  Var2  Var3   z.x   z.y
  <chr> <chr> <chr> <dbl> <dbl>
1  east     b     a   0.2   0.1
2  east     c     a   0.3   0.1
3  east     c     b   0.3   0.2

扩展_数据%
过滤器（Var2！=Var3）%>%
不同的（）
指数%
左联接（初始数据，by=c（“Var1”=“x”，“Var3”=“y”））
>扩展数据
#一个tibble:3x5
Var1 Var2 Var3 z.x z.y
1东b 0.2 0.1
2东c 0.30.1
3东c b 0.3 0.2

---

非常感谢

如何进行

内部连接

，然后筛选唯一的组合

library(dplyr)
inner_join(initial_data, initial_data,
           suffix = c('1', '2'), by = 'x') %>%
    filter(y1 < y2) %>%
    select(x, y1, y2, z1, z2)

#      x y1 y2  z1  z2
# 1 east  a  b 0.1 0.2
# 2 east  a  c 0.1 0.3
# 3 east  b  c 0.2 0.3

库（dplyr）
内部连接（初始数据、初始数据、，
后缀=c（'1'，'2'），by='x'）%>%
过滤器（y1%
选择（x、y1、y2、z1、z2）
#x y1 y2 z1 z2
#1东a b 0.1 0.2
#2东a c 0.1 0.3
#3东b c 0.2 0.3

此基本R解决方案适合您吗

data.frame(x = rep("east", 3),
           matrix(rep(initial_data$y, each = 2), 3), 
           matrix(rep(initial_data$z, each = 2), 3))

#      x X1 X2 X1.1 X2.1
# 1 east  a  b  0.1  0.2
# 2 east  a  c  0.1  0.3
# 3 east  b  c  0.2  0.3

---

我会尝试一下

combn

和

purr:：map

你的数据

initial_data <- tibble(x = c("east","east","east"), y = c("a","b","c"), z = c(0.1,0.2,0.3))

输出

# A tibble: 3 x 5
      # x    y1    y2    z1    z2
  # <chr> <chr> <chr> <dbl> <dbl>
# 1  east     a     b   0.1   0.2
# 2  east     a     c   0.1   0.3
# 3  east     b     c   0.2   0.3

#一个tible:3 x 5
#x y1 y2 z1 z2
#     
#1东a b 0.1 0.2
#2东a c 0.1 0.3
#3东b c 0.2 0.3

我会尽快检查可行性：）你能更详细地解释一下你的问题吗？我不明白你想要什么样的最终格式。

# A tibble: 3 x 5
      # x    y1    y2    z1    z2
  # <chr> <chr> <chr> <dbl> <dbl>
# 1  east     a     b   0.1   0.2
# 2  east     a     c   0.1   0.3
# 3  east     b     c   0.2   0.3