How to match data in different columns in R? - r

Question

I have a dataset simplified as below: there are multiple customers and each CUSTOMER may have several loans. A CUSTOMER with at least 1 LOAN_DEFAULT is marked as CUSTOMER_DEFAULT, and the DEFAULT_DATE is the first time of default.
CUSTOMER LOAN DATE AMOUNT LOAN_DEFAULT CUSTOMER_DEFAULT DEFAULT_DATE CLASSIFICATION
1 101 201601 100 Y Y 201501 S
1 102 201603 100 N Y 201501 S
1 103 201501 100 Y Y 201501 S
1 104 201501 200 N Y 201501 S
2 201 201601 100 N N - M
2 202 201603 100 N N - M
How can I calculate the loan amount for a customer at the first default date, e.g. 201501 for customer 1 that equals to the total AMOUNT at that month, that I should get a number of 300?
What I think of is compare the DATE and DEFAULT_DATE and if they are the same, use sum function. But my code didn't work.
I want to summarise the number of default customers by Classification, but by using summarise function it seems not working properly?

Answer 1

We can sum AMOUNT value where DATE is equal to first DEFAULT_DATE for each CUSTOMER.
library(dplyr)
df %>%
group_by(CUSTOMER) %>%
summarise(total_sum = sum(AMOUNT[DATE == first(DEFAULT_DATE)]))
# CUSTOMER total_sum
# <int> <int>
#1 1 300
#2 2 0
To get number of default customers for each CLASSIFICATION we can do :
df %>%
group_by(CLASSIFICATION) %>%
summarise(no_default_cust = n_distinct(CUSTOMER[CUSTOMER_DEFAULT == "Y"]))
data
df <- structure(list(CUSTOMER = c(1L, 1L, 1L, 1L, 2L, 2L), LOAN = c(101L,
102L, 103L, 104L, 201L, 202L), DATE = c(201601L, 201603L, 201501L,
201501L, 201601L, 201603L), AMOUNT = c(100L, 100L, 100L, 200L,
100L, 100L), LOAN_DEFAULT = structure(c(2L, 1L, 2L, 1L, 1L, 1L
), .Label = c("N", "Y"), class = "factor"), CUSTOMER_DEFAULT = structure(c(2L,
2L, 2L, 2L, 1L, 1L), .Label = c("N", "Y"), class = "factor"),
DEFAULT_DATE = structure(c(2L, 2L, 2L, 2L, 1L, 1L), .Label = c("-",
"201501"), class = "factor"), CLASSIFICATION = structure(c(2L,
2L, 2L, 2L, 1L, 1L), .Label = c("M", "S"), class = "factor")),
class = "data.frame", row.names = c(NA, -6L))

Answer 2

In base you can use aggregate to get the sum of AMOUNT per CUSTOMER. With x[x$DATE == x$DEFAULT_DATE,] you can subset to those lines where DATE equals to DEFAULT_DATE.
aggregate(AMOUNT ~ CUSTOMER, x[x$DATE == x$DEFAULT_DATE,], sum)
# CUSTOMER AMOUNT
#1 1 300
To get the number of default customers by Classification you can use table in combination with unique:
table(unique(x[x$CUSTOMER_DEFAULT=="Y",c("CUSTOMER", "CLASSIFICATION")])[,2])
#M S
#0 1
Data:
x <- read.table(header=TRUE, text="CUSTOMER LOAN DATE AMOUNT LOAN_DEFAULT CUSTOMER_DEFAULT DEFAULT_DATE CLASSIFICATION
1 101 201601 100 Y Y 201501 S
1 102 201603 100 N Y 201501 S
1 103 201501 100 Y Y 201501 S
1 104 201501 200 N Y 201501 S
2 201 201601 100 N N - M
2 202 201603 100 N N - M")

Answer 3

Another option with data.table
library(data.table)
setDT(df)[, .(total_sum = sum(AMOUNT[DATE == first(DEFAULT_DATE)])), CUSTOMER]