Question

这里有一些播放数据

df = data.frame(ID = c(1,1,1,2,2,2,2,3,3),
                food = c("bacon","bacon","bacon","bacon","bacon","cheese","sausage","avocado","ham"),
                enjoyment = c(20,20,20,20,20,20,20,20,20))

导致

 ID    food enjoyment
1  1   bacon        20
2  1   bacon        20
3  1   bacon        20
4  2   bacon        20
5  2   bacon        20
6  2  cheese        20
7  2 sausage        20
8  3 avocado        20
9  3     ham        20

我想做的是，对于每个人（身份证），总结他们对培根和奶酪的享受

到目前为止，我的代码是

library(data.table)
setDT(df)
df[,id_enjoyment_sum := sum(enjoyment), by =.(ID,food == "bacon"|food == "cheese")]

导致

 ID    food enjoyment id_enjoyment_sum
1:  1   bacon        20               60
2:  1   bacon        20               60
3:  1   bacon        20               60
4:  2   bacon        20               60
5:  2   bacon        20               60
6:  2  cheese        20               60
7:  2 sausage        20               20
8:  3 avocado        20               40
9:  3     ham        20               40

这已经完成了我想要它做的事情，但它也总结了每个人，他们享受非培根和非奶酪食品。请注意，ID 3不会吃培根或奶酪，但我的代码仍然总结了他对他吃的东西的享受。

理想情况下，代码应该导致

ID    food enjoyment id_enjoyment_sum
1:  1   bacon        20               60
2:  1   bacon        20               60
3:  1   bacon        20               60
4:  2   bacon        20               60
5:  2   bacon        20               60
6:  2  cheese        20               60
7:  2 sausage        20               60
8:  3 avocado        20               0
9:  3     ham        20               0

所以我的问题是，我如何设置BY子句来总结，对于每个id，只有培根和奶酪的享受？

Answer 1

在一个班轮里，我会这样做：

df[,
    id_enjoyment_sum := sum(
        ifelse(food %in% c("bacon", "cheese"), enjoyment, 0)
    )
    , by =.(ID)]

如果覆盖享受栏不是问题，你可以考虑这个：

df[! food %in% c("bacon", "cheese"), enjoyment := 0]
df[, id_enjoyment_sum := sum(enjoyment), by = .(ID)]

当您按多个变量分组时，每个组合都会有一组，并且聚合将在这些组中发生。所以在你的情况下，

有一组行

ID == 1 and (food == "bacon"|food == "cheese") == TRUE，
ID == 2 and (food == "bacon"|food == "cheese") == TRUE，
ID == 2 and (food == "bacon"|food == "cheese") == FALSE等等。

带有多个条件的data.table聚合BY语句

1 个答案: