计算 data.frame 中两个变量的组合

Question

我有两个 data.frames - g 包含两个变量的所有可能（这里：8）组合，h 包含 8 个组合中任何一个的 62 个观察值（dput() 在底部）。

我在g 中添加了第三列，它应该对h 中的每个组合进行观察计数：

> g
  where what days
1    sg free    0
2    in free    0
3    hk free    0
4    de free    0
5    sg work    0
6    in work    0
7    hk work    0
8    de work    0

我想计算g 中的每个组合出现在h 中的频率，我现在使用运行良好的老式嵌套循环来计算：

for( i in seq( nrow( g ) ) )
    for( j in seq( nrow( h ) ) )
        if( all( g[ i, 1:2 ] == h[ j, ] ) ) g[ i, 3 ] <- g[ i, 3 ] + 1

这给了我想要的：

> g
  where what days
1    sg free   10
2    in free    0
3    hk free    4
4    de free    4
5    sg work   18
6    in work   10
7    hk work    6
8    de work   10

但我想知道是否有不那么神秘、更简洁的方法来做到这一点；我特别好奇base R 是否提供了我还没有发现的工具。

数据：

g <- structure(list(where = structure(c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 
4L), .Label = c("sg", "in", "hk", "de"), class = "factor"), what = structure(c(1L, 
1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c("free", "work"), class = "factor"), 
days = c(0, 0, 0, 0, 0, 0, 0, 0)), .Names = c("where", "what", 
"days"), out.attrs = structure(list(dim = c(4L, 2L), dimnames = structure(list(
Var1 = c("Var1=sg", "Var1=in", "Var1=hk", "Var1=de"), Var2 = c("Var2=free", 
"Var2=work")), .Names = c("Var1", "Var2"))), .Names = c("dim", "dimnames")), 
row.names = c(NA, -8L), class = "data.frame")

h <- structure(list(values = c("sg", "sg", "sg", "sg", "sg", "sg", 
"sg", "sg", "sg", "sg", "sg", "sg", "sg", "sg", "in", "in", "in", 
"in", "in", "hk", "hk", "hk", "hk", "hk", "de", "de", "de", "de", 
"de", "de", "de", "sg", "sg", "sg", "sg", "sg", "sg", "sg", "sg", 
"sg", "sg", "sg", "sg", "sg", "sg", "in", "in", "in", "in", "in", 
"hk", "hk", "hk", "hk", "hk", "de", "de", "de", "de", "de", "de", 
"de"), values.1 = c("free", "work", "work", "work", "work", "free", 
"free", "work", "work", "work", "work", "work", "free", "free", 
"work", "work", "work", "work", "work", "free", "free", "work", 
"work", "work", "work", "work", "free", "free", "work", "work", 
"work", "free", "work", "work", "work", "work", "free", "free", 
"work", "work", "work", "work", "work", "free", "free", "work", 
"work", "work", "work", "work", "free", "free", "work", "work", 
"work", "work", "work", "free", "free", "work", "work", "work"
)), .Names = c("values", "values.1"), row.names = c(NA, -62L), class = "data.frame")

Answer 1

有一个简单整洁的解决方案。我更改了 h 中的列名以匹配 g 中的内容（位置和内容）。按两个值分组并总结 - 这将给出组合的计数。然后，将left_join 返回到 g，然后您就知道了。

library(dplyr)

h_s = h %>% 
  group_by(where,what) %>% 
  summarise(days=n())

g %>% 
  left_join(h_s,by=c("where","what")) %>% 
  select(where,what,days=days.y) %>%
  mutate(days = ifelse(is.na(days),0,days))

编辑

左连接的原因是为了确保在 h 中找不到任何情况。我添加了一个 mutate 来将缺失值转换为 0。