如何通过唯一 ID 将 R 中的列中的某些行加在一起？

Question

I'm new and sorry if my question is badly worded.

I'm working in r and I have table called Rent that might look like this:

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

So what I want to do is that since Lucy has two payments on 7/1/2018, I want to combine them together and then sum the payment, and use the latest Paid Date.

What I have so far is that

#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") 

But when I run the first line all the sum_late column is N/A. I'm not sure what I did wrong. Ideally, I would want a table just like this.

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

Sorry if this is a stupid question, I appreciate any help and feedback!! Thank you all for your time!!

Answer 1

We can change Paid_Date to date class, group_by ID and Invoice , sum Payment and select 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

Or if you prefer data.table using the same logic.

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)]

data

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))

Answer 2

There are multiple ways of doing this task, I will be using for-loops for creating desired output. I echo with @Ronak Shah using dplyr method which make lesser processing time thank using for-loops

Data

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))

Converting Paid_date into date formats

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

For-loops

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