我有以下数据：
dataset <- structure(list(id = structure(c(2L, 3L, 1L, 3L, 1L, 9L), .Label = c("215101",
"215559", "216566", "217284", "219435", "220209", "220249", "220250",
"225678", "225679", "225687", "225869", "228420", "228435", "230621",
"230623", "233063", "233097", "233098", "235546", "235560", "235567",
"236379"), class = "factor"), cat1 = c("A", "B", "B", "A", "A",
"A"), cat2 = c("item 1", "item 1", "item 2", "item 5", "item 3",
"item 28"), cat3 = c("theme 2", "theme 2", "theme 1", "theme 4",
"theme 10", "theme 40")), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -6L))
我想创建一种具有从列cat2和cat3创建的热编码列特征的模型矩阵。因此，我的输出将如下所示：
structure(list(id = structure(c(1L, 1L, 2L, 3L, 3L, 9L), .Label = c("215101",
"215559", "216566", "217284", "219435", "220209", "220249", "220250",
"225678", "225679", "225687", "225869", "228420", "228435", "230621",
"230623", "233063", "233097", "233098", "235546", "235560", "235567",
"236379"), class = "factor"), cat1 = c("A", "B", "A", "A", "B",
"A"), `item 1` = c(0, 0, 1, 0, 1, 0), `item 2` = c(0, 1, 0, 0,
0, 0), `item 28` = c(0, 0, 0, 0, 0, 1), `item 3` = c(1, 0, 0,
0, 0, 0), `item 5` = c(0, 0, 0, 1, 0, 0), `theme 1` = c(0, 1,
0, 0, 0, 0), `theme 10` = c(1, 0, 0, 0, 0, 0), `theme 2` = c(0,
0, 1, 0, 1, 0), `theme 4` = c(0, 0, 0, 1, 0, 0), `theme 40` = c(0,
0, 0, 0, 0, 1)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-6L))
但是，我在此数据集中没有自变量，我想保留id和cat1列。我该怎么办？
答案 0 :(得分：1)
您可以使用employee.belongsTo(models.employee, {
foreignKey: 'manager_id'
}),
和merge两次。
dcast如果要分布的变量更多，则可以library(reshape2)
merge(dcast(dataset, id + cat1 ~ cat2, fun.aggregate = length),
dcast(dataset, id + cat1 ~ cat3, fun.aggregate = length),
by = c("id", "cat1"))
# id cat1 item 1 item 2 item 28 item 3 item 5 theme 1 theme 10 theme 2 theme 4 theme 40
#1 215101 A 0 0 0 1 0 0 1 0 0 0
#2 215101 B 0 1 0 0 0 1 0 0 0 0
#3 215559 A 1 0 0 0 0 0 0 1 0 0
#4 216566 A 0 0 0 0 1 0 0 0 1 0
#5 216566 B 1 0 0 0 0 0 0 1 0 0
#6 225678 A 0 0 1 0 0 0 0 0 0 1
首先进行数据处理。这样可以节省您一些输入。
melt