Question

使用R，我需要为每个部门的前2名员工编制一份费用最高的报告，并为该部门的其他员工添加“其他”。例如，我需要一份类似的报告。

Dept.      EmployeeId     Expense
Marketing       12345         100
Marketing       12346          90
Marketing      Others         200
Sales           12347          50 <-- There's just one employee with expenses
Research        12348        2000
Research        12349         900
Research       Others       10000

换句话说，我需要总结数据，重点关注费用最高的前2名员工。费用栏的总和应该是公司费用的总额。

employeIds <- sample(1000:9999, 20)
depts <- sample(c('Sales', 'Marketing', 'Research'), 20, replace = TRUE)
expenses <- sample(1:1000, 20, replace = TRUE)

df <- data.frame(employeIds, depts, expenses)

# Based on that data, how do I build a table with the top 2 employees with the most expenses in each department, including an "Other" employee per department.

我是R的新手，我不知道如何处理这个问题。在SQL中，我可以使用RANK（）函数和JOIN，但它不是一个选项。

Answer 1

这是一个data.table解决方案：

创建数据：我还提出了“其他”不会发生的情况（该部门的条目数为：1＆lt; = entries＆lt; = 2）

set.seed(45)
employeIds <- sample(1000:9999, 20)
depts <- sample(c('Sales', 'Marketing', 'Research'), 20, replace = TRUE)
expenses <- sample(1:1000, 20, replace = TRUE)

df <- data.frame(employeIds, depts, expenses)
df <- df[-c(6,10,12,18,19), ]

data.table解决方案：

require(data.table)
dt <- data.table(df, key=c("depts", "expenses"))
k <- 2
dt[, if(.N > k) {
        idx <- (seq_len(.N)-1) %/% max(k, (.N - k)) == 1
        list(EmployeeIds = c(employeIds[idx], "Others"), 
           Expenses = c(expenses[idx], sum(expenses[!idx])))
     } else {
        list(EmployeeIds = as.character(employeIds), Expenses = expenses)
     }, by = depts]

#        depts EmployeeIds Expenses
# 1: Marketing        4870      567
# 2: Marketing        3167      591
# 3: Marketing      Others     2285
# 4:  Research        5989      878
# 5:  Research        9667      930
# 6:  Research      Others     1301
# 7:     Sales        6700      129
# 8:     Sales        3857      714

创意：使用dt创建key = depts, expenses的第一步可确保expenses按递增顺序排序。然后，根据每个dept的条目数，我们要么创建一个“其他”条目。

Answer 2

可能不是最优雅的，但这是一个解决方案：

func <- function(data) {
 data1 <- aggregate(data$expenses, list(employeIds=data$employeIds), sum)
 # rank without ties.method = "first" will screw things up with identical values
 data1$employeIds[!(rank(data1$x, ties.method="first") %in% 1:2)] <- 'Others'
 data1 <- aggregate(data.frame(expenses=data1$x), list(employeIds=data1$employeIds), sum)
}

do.call(rbind, by(df, df$depts, func))

Answer 3

另一种data.table方法（可能更接近你所知道的SQL风格）：

dt <- data.table(employeIds, depts, expenses)
dt[, rank:=rank(-expenses), by=depts][,
    list("Expenses"=sum(expenses)),
    keyby=list(depts, "Employee"=ifelse(rank<=2,employeIds,"Other"))
]
       depts Employee Expenses
1: Marketing     6988      986
2: Marketing     7011      940
3: Marketing    Other     2614
4:  Research     2434      763
5:  Research     9852      731
6:  Research    Other     3397
7:     Sales     3120      581
8:     Sales     6069      868

Answer 4

df <- split(df, df$depts)
df <- lapply(df, FUN=function(x){
  x <- x[order(x$expenses, decreasing=TRUE), ]
  x$total.expenses <- sum(x$expenses)
  x$group <- 1:nrow(x)
  x$group <- ifelse(x$group <= 2, x$group, "Other")
  x
})
df <- do.call(rbind, df)

总结和排名数据框架

4 个答案: