Question

我不确定这个问题甚至被称为。假设我计算了2列的不同组合，但我希望两列的 order 区别开来。这就是我的意思：

df = data.frame(fruit1 = c("apple", "orange", "orange", "banana", "kiwi"),
                fruit2 = c("orange", "apple", "banana", "orange", "apple"),
                stringsAsFactors = FALSE)

# What I want: total number of fruit combinations, regardless of 
# which fruit comes first and which second.
# Eg 2 apple-orange, 2 banana-orange, 1 kiwi-apple

# What I know *doesn't* work:

table(df$fruit1, df$fruit2) 

# What *does* work:
library(dplyr)
df %>% group_by(fruit1, fruit2) %>% 
  transmute(fruitA = sort(c(fruit1, fruit2))[1],
            fruitB = sort(c(fruit1, fruit2))[2]) %>%
  group_by(fruitA, fruitB) %>%
  summarise(combinations = n())

正如你所看到的，我有办法让这项工作成功，但这个一般问题是否有名称？这是一种组合问题，但是计算，而不是生成组合。如果我有三列或四列相似类型怎么办？上述方法很难推广。 Tidyverse最受欢迎！

Answer 1

使用apply和sort订购您的数据框，我们只使用group_by count

data.frame(t(apply(df,1,sort)))%>%group_by_all(.)%>%count()
# A tibble: 3 x 3
# Groups:   X1, X2 [3]
      X1     X2     n
  <fctr> <fctr> <int>
1  apple   kiwi     1
2  apple orange     2
3 banana orange     2

Answer 2

以下是使用pmap和count

的选项

library(tidyverse)
library(rlang)
pmap_df(df, ~ sort(c(...)) %>%
                 as.list %>%
                 as_tibble %>%
                 set_names(names(df))) %>% 
    count(!!! rlang::syms(names(.)))
# A tibble: 3 x 3
#  fruit1 fruit2     n
#   <chr>  <chr>  <int>
#1 apple  kiwi       1
#2 apple  orange     2
#3 banana orange     2

R：计算在列可互换的数据框中找到的不同组合

2 个答案: