计算关于 R 中列的唯一观测值对之间的关系答案

【问题标题】：Calculating a relationship between unique pairs of observations about a colum in R计算关于 R 中列的唯一观测值对之间的关系
【发布时间】：2021-07-20 01:37:34
【问题描述】：

我正在尝试根据比例计算两种商品出现在同一组中的概率。

我有以下数据，

data <- data.frame(group = c(1,1,1,1,2,2,2,2,3,3,3,3),
                   commodity = c("Wheat", "Coal", "Steel", "Iron", "Wheat", "Coal", "Steel", "Iron", "Wheat", "Coal", "Steel", "Iron"),
                   quantity = c(5,10,0,5,20,5,10,0,0,10,15,15),
                   proportion = c(0.25,0.5,0,0.25,0.57,0.14,0.29,0,0,0.25,0.375,0.375))

我想计算每个唯一可能的商品对的比例（产品总和除以 2）。

结果应该是这样的，

result <- data.frame(commodity1 = c("Wheat", "Wheat", "Wheat", "Coal", "Coal", "Steel"),
                     commodity2 = c("Coal", "Steel", "Iron", "Steel", "Iron", "Iron"),
                     result = c(0.103,0.082,0.031,0.067,0.109,0.070))

例如，小麦 - 煤的结果将计算为 (0.25 * 0.5/2)+(0.57 * 0.14/2)+(0 * 0.25/2)=0.103

我已将商品对隔离到单独的 data.frame 中以将结果变异为并尝试进行 rowwise() 操作。

任何建议将不胜感激。

【问题讨论】：

组对中的每个比例乘积除以 2
没有一个答案达到预期目的吗？您还没有接受任何答案！

标签： r dplyr statistics rowwise

【解决方案1】：

虽然不是很干净，但看起来很有效

library(tidyverse)

#make an intermediate data.frame say `dd`
data %>% select(-quantity) %>%
  pivot_longer(proportion) %>%
  select(-name, -group) %>%
  group_by(commodity) %>%
  nest(data = c(value)) -> dd


t(combn(unique(data$commodity), 2)) %>% as.data.frame() %>%
  mutate(result = map2_dbl(V1, V2, 
                        ~ sum(unlist(dd$data[match(.x, dd$commodity)]) * unlist(dd$data[match(.y, dd$commodity)]))/2
                      )
         )

     V1    V2    result
1 Wheat  Coal 0.1024000
2 Wheat Steel 0.0826500
3 Wheat  Iron 0.0312500
4  Coal Steel 0.0671750
5  Coal  Iron 0.1093750
6 Steel  Iron 0.0703125

【讨论】：

感谢 Anil，这是一个非常棒的解决方案

【解决方案2】：

library(tidyverse)
crossing(commodity1 = data$commodity, commodity2 = data$commodity, group = data$group) %>%
  filter(commodity1 < commodity2) %>%
  left_join(data, by = c("commodity1" = "commodity", "group")) %>%
  left_join(data, by = c("commodity2" = "commodity", "group")) %>%
  mutate(avg = proportion.x * proportion.y / 2) %>%
  group_by(commodity1, commodity2) %>%
  summarize(result = sum(avg), .groups = "drop")


# A tibble: 6 x 3
  commodity1 commodity2 result
* <chr>      <chr>       <dbl>
1 Coal       Iron       0.109 
2 Coal       Steel      0.0672
3 Coal       Wheat      0.102 
4 Iron       Steel      0.0703
5 Iron       Wheat      0.0312
6 Steel      Wheat      0.0826

【讨论】：

谢谢，这真是一个很好的解决方案，我一直想解决这个问题。