我有一个 df 有这样的数据:
sub = c("X001","X002", "X001","X003","X002","X001","X001","X003","X002","X003","X003","X002")
month = c("201506", "201507", "201506","201507","201507","201508", "201508","201507","201508","201508", "201508", "201508")
tech = c("mobile", "tablet", "PC","mobile","mobile","tablet", "PC","tablet","PC","PC", "mobile", "tablet")
brand = c("apple", "samsung", "dell","apple","samsung","apple", "samsung","dell","samsung","dell", "dell", "dell")
revenue = c(20, 15, 10,25,20,20, 17,9,14,12, 9, 11)
df = data.frame(sub, month, brand, tech, revenue)
我想使用 sub 和 month 作为键,并为每个订阅者每月获得一行显示该订阅者在技术和品牌方面的独特值(value)的收入总和。这个例子是简化的,列更少,因为我有一个巨大的数据集,我决定尝试用
data.table
来做。 .我已经成功地为一个分类专栏做到了这一点,无论是技术还是品牌,都使用这个:
df1 <- dcast(df, sub + month ~ tech, fun=sum, value.var = "revenue")
但我想为两个或更多 caqtogorical 列执行此操作,到目前为止我已经尝试过:
df2 <- dcast(df, sub + month ~ tech+brand, fun=sum, value.var = "revenue")
它只是连接了分类列和总和的唯一值,但我不想要那样。我想为所有分类列的每个唯一值单独列。
我是 R 的新手,非常感谢任何帮助。
最佳答案
(我假设 df
是一个 data.table
而不是一个 data.frame
就像你的例子一样。)
一个可能的解决方案是首先 melt
数据同时保持sub
, month
和 revenue
作为 key 。这样,brand
和 tech
将转换为单个变量,其值对应于每个现有的键组合。这样我们就可以轻松dcast
它回来了,因为我们将针对单个列进行操作 - 就像在您的第一个示例中一样
dcast(melt(df, c(1:2, 5)), sub + month ~ value, sum, value.var = "revenue")
# sub month PC apple dell mobile samsung tablet
# 1: X001 201506 10 20 10 20 0 0
# 2: X001 201508 17 20 0 0 17 20
# 3: X002 201507 0 0 0 20 35 15
# 4: X002 201508 14 0 11 0 14 11
# 5: X003 201507 0 25 9 25 0 9
# 6: X003 201508 12 0 21 9 0 0
根据 OP 评论,您可以通过添加
variable
轻松添加前缀。到公式。这样,列也将被正确排序dcast(melt(df, c(1:2, 5)), sub + month ~ variable + value, sum, value.var = "revenue")
# sub month brand_apple brand_dell brand_samsung tech_PC tech_mobile tech_tablet
# 1: X001 201506 20 10 0 10 20 0
# 2: X001 201508 20 0 17 17 0 20
# 3: X002 201507 0 0 35 0 20 15
# 4: X002 201508 0 11 14 14 0 11
# 5: X003 201507 25 9 0 0 25 9
# 6: X003 201508 0 21 0 12 9 0
关于r - 扩大数据框以获取 R 中分类列的所有唯一值的每月收入总和,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40482456/