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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，显示该订阅者在该月的技术和品牌独特价值的收入总和。这个例子被简化了，并且列更少，因为我有一个庞大的数据集，我决定尝试使用data.table.

我已经设法为一个分类专栏做到了这一点，无论是技术还是品牌，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，但我不希望这样。我希望为所有分类列的每个唯一值单独列。

我是 R 新手，非常感谢任何帮助。

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1 回答 1

5

（我会假设这在你的例子中很像df。data.table）data.frame

一种可能的解决方案是在保留和作为键的同时首先melt获取数据。这样，和将被转换为单个变量，其值对应于每个现有的键组合。这样，我们将能够轻松地将其恢复，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submonthrevenuebrandtechdcast

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 OPs 的评论，您可以通过添加variable到公式中来轻松添加前缀。这样，该列也将正确排序

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
于 2016-11-08T09:17:04.690 回答