Reputation: 13
** Sample data added after comment**
What I have:
pmts <- data.frame(stringsAsFactors=FALSE,
name = c("johndoe", "johndoe", "janedoe", "foo", "foo", "foo"),
pmt_amount = c(550L, 550L, 995L, 375L, 375L, 375L),
pmt_date = c("9/1/16", "11/1/16", "12/15/16", "1/5/17", "3/5/17", "5/5/17")
)
#> name pmt_amount pmt_date
#> 1 johndoe 550 9/1/16
#> 2 johndoe 550 11/1/16
#> 3 janedoe 995 12/15/16
#> 4 foo 375 1/5/17
#> 5 foo 375 3/5/17
#> 6 foo 375 5/5/17
What I am looking to achieve:
read.table(header = T, text =
"name pmt_amount first_pmt second_pmt third_pmt
johndoe 550 9/1/16 11/1/16 NA
janedoe 995 12/15/16 NA NA
foo 375 1/5/17 3/5/17 5/5/17"
)
#> name pmt_amount first_pmt second_pmt third_pmt
#> 1 johndoe 550 9/1/16 11/1/16 <NA>
#> 2 janedoe 995 12/15/16 <NA> <NA>
#> 3 foo 375 1/5/17 3/5/17 5/5/17
** End of update**
I have a large dataset with payment information for different products. Some of these products have a pay-in-full option as well as a two-pay and three-pay option. I need to create fields that would be First_Payment, Second_Payment, and Third_Payment and would populate NA in the respective fields if there was only one or two payments.
I've tried a couple options and the best workaround I have thus far is this:
pmts %>%
group_by(Email, Name, Amount, Form.Title) %>%
summarise(First_Payment = min(Payment.Date),
Second_Payment = median(Payment.Date),
Last_Payment = max(Payment.Date)) -> pmts
This obviously is not ideal as is making up a payment date for the 2-pay plans and I would have to instruct the end-user to ignore this field and just look at the 1st and 3rd fields.
I also tried to summarise with partial sorts like this:
n <- length(pmts$Payment.Date)
sort(pmts$Payment.Date,partial=n-1)[n-1]
However, if there wasn't three payments for the person, it would take the n-1 date from the entire data set and apply to all other fields.
Ideally, I would have it so if it was a pay-in-full the the First_Payment field would have the date and the 2nd/3rd fields would say NA. The 2-pay would have 1st and 2nd dates and the 3rd field would say NA. And finally the 3 pay would have all 3 dates.
The end users here are not super data savvy so I'm trying to make this as easy to interpret as possible. Any suggestions would be tremendously appreciated. Thank you!
Upvotes: 1
Views: 74
Reputation: 92300
Using data.table this is a simple one-liner
library(data.table) #v1.9.8+
dcast(setDT(pmts), name + pmt_amount ~ rowid(pmt_amount))
# Using 'pmt_date' as value column. Use 'value.var' to override
# name pmt_amount 1 2 3
# 1: foo 375 1/5/17 3/5/17 5/5/17
# 2: janedoe 995 12/15/16 NA NA
# 3: johndoe 550 9/1/16 11/1/16 NA
dcast
converts from long to wide and it accepts expressions. rowid
is just adding a row counter per pmt_amount
.
Upvotes: 1
Reputation: 3017
You can use tidyr
for this.
library(dplyr)
library(tidyr)
pmts <- tibble(
name = c("johndoe", "johndoe", "janedoe", "foo", "foo", "foo"),
pmt_amount = c(550L, 550L, 995L, 375L, 375L, 375L),
pmt_date = lubridate::mdy(c("9/1/16", "11/1/16", "12/15/16", "1/5/17", "3/5/17", "5/5/17"))
)
pmts
#> # A tibble: 6 x 3
#> name pmt_amount pmt_date
#> <chr> <int> <date>
#> 1 johndoe 550 2016-09-01
#> 2 johndoe 550 2016-11-01
#> 3 janedoe 995 2016-12-15
#> 4 foo 375 2017-01-05
#> 5 foo 375 2017-03-05
#> 6 foo 375 2017-05-05
pmts_long <- pmts %>%
group_by(name) %>%
arrange(name, pmt_date) %>%
mutate(pmt = row_number()) %>%
ungroup() %>%
complete(name, nesting(pmt)) %>%
fill(pmt_amount, .direction = "down")
pmts_long
#> # A tibble: 9 x 4
#> name pmt pmt_amount pmt_date
#> <chr> <int> <int> <date>
#> 1 foo 1 375 2017-01-05
#> 2 foo 2 375 2017-03-05
#> 3 foo 3 375 2017-05-05
#> 4 janedoe 1 995 2016-12-15
#> 5 janedoe 2 995 NA
#> 6 janedoe 3 995 NA
#> 7 johndoe 1 550 2016-09-01
#> 8 johndoe 2 550 2016-11-01
#> 9 johndoe 3 550 NA
pmts_wide <- pmts_long %>%
gather("key", "val", -name, -pmt_amount, -pmt) %>%
unite(pmt_number, key, pmt) %>%
spread(pmt_number, val)
pmts_wide
#> # A tibble: 3 x 5
#> name pmt_amount pmt_date_1 pmt_date_2 pmt_date_3
#> * <chr> <int> <date> <date> <date>
#> 1 foo 375 2017-01-05 2017-03-05 2017-05-05
#> 2 janedoe 995 2016-12-15 NA NA
#> 3 johndoe 550 2016-09-01 2016-11-01 NA
Upvotes: 1