2

如果我的问题措辞不当，我很抱歉。

我在 r 工作，我有一个名为 Rent 的表，它可能看起来像这样：

Rent
       ID      Invoice    Payment      Paid Date
       lucy   7/1/2018     100        9/1/2018
       lucy   7/1/2018     150        10/1/2018
       lucy   8/1/2018     100        11/1/2018

所以我想做的是，由于 Lucy 在 2018 年 7 月 1 日有两次付款，我想将它们组合在一起，然后将付款相加，并使用最新的付款日期。

到目前为止我所拥有的是

#to create a row that has the sum of the sales prices 

    Rent[,sum_late:=sum( as.numeric(("Sales Price"))),
    by= c("Id","Invoice Date")]

#take the first of the unique IDs by the max paid date
    head (SD,1) by=c("ID", "Invoice Date", max("Paid Date")

但是当我运行第一行时，所有 sum_late 列都是 N/A。我不确定我做错了什么。理想情况下，我想要一张这样的桌子。

Rent
       ID      Invoice    Payment      Paid Date
       lucy   7/1/2018     250        10/1/2018
       lucy   8/1/2018     100        11/1/2018

对不起，如果这是一个愚蠢的问题，我感谢任何帮助和反馈！谢谢大家的时间！！

4

2 回答 2

3

我们可以更改Paid_Date为日期类，group_by ID和Invoice，sum Payment然后选择max Paid_Date。

library(dplyr)
Rent %>%
  mutate_at(vars(Invoice, Paid_Date), as.Date, '%d/%m/%Y') %>%
  group_by(ID, Invoice) %>%
  summarise(Payment = sum(Payment), 
            Paid_Date = max(Paid_Date))

#  ID    Invoice    Payment Paid_Date 
#  <chr> <date>       <int> <date>    
#1 lucy  2018-01-07     250 2018-01-10
#2 lucy  2018-01-08     100 2018-01-11

或者，如果您更喜欢data.table使用相同的逻辑。

library(data.table)
setDT(Rent)[, c("Invoice", "Paid_Date") := .(as.IDate(Invoice, '%d/%m/%Y'), 
                                             as.IDate(Paid_Date, '%d/%m/%Y'))]
Rent[, .(Payment = sum(Payment), Paid_Date = max(Paid_Date)), .(ID, Invoice)]

数据

Rent <- structure(list(ID = c("lucy", "lucy", "lucy"), Invoice = c("7/1/2018", 
"7/1/2018", "8/1/2018"), Payment = c(100L, 150L, 100L), Paid_Date = c("9/1/2018", 
"10/1/2018", "11/1/2018")), class = "data.frame", row.names = c(NA, -3L))
于 2020-05-21T06:12:01.153 回答
2

有多种方法可以完成此任务，我将使用 for 循环来创建所需的输出。我使用 dplyr 方法与@Ronak Shah 呼应，这使得处理时间更短，感谢使用 for-loops

数据

Rent <- structure(list(ID = c("lucy", "lucy", "lucy"), Invoice = c("7/1/2018", 
                                                                   "7/1/2018", "8/1/2018"), Payment = c(100L, 150L, 100L), Paid_Date = c("9/1/2018", 
                                                                                                                                         "10/1/2018", "11/1/2018")), class = "data.frame", row.names = c(NA, -3L))

将支付日期转换为日期格式

Rent$Paid_Date <- as.Date(Rent$Paid_Date, "%d/%m/%Y")

For循环

for ( i in unique (Rent$ID)){
  for (j in unique(Rent$Invoice[Rent$ID == i])){
    Rent$Payment_[Rent$ID==i & Rent$Invoice ==j ] <- sum (Rent$Payment [Rent$ID==i & Rent$Invoice ==j])
    Rent$Paid_dt[Rent$ID==i & Rent$Invoice ==j ] <- max(Rent$Paid_Date[Rent$ID==i & Rent$Invoice ==j])

  }
}

Rent$Paid_dt <- as.Date(Rent$Paid_dt ,origin = "1970-01-01") # converting into date format

Rent1 <- Rent[, unique(c("ID", "Invoice", "Payment_", "Paid_dt"))]

print (Rent1)

    ID  Invoice Payment_    Paid_dt
1 lucy 7/1/2018      250 2018-01-10
2 lucy 7/1/2018      250 2018-01-10
3 lucy 8/1/2018      100 2018-01-11
于 2020-05-21T08:26:04.953 回答