Grouping data based on repetitive records using R - r

Question

I have a dataset which contains repetitive records/common records. It looks something like this:
| Vendor | Buyer | Amount |
|--------|:-----:|-------:|
| A | P | 100 |
| B | P | 150 |
| C | Q | 300 |
| A | P | 290 |
I need to group similar records together but I do not want to summarize my amount. I want to have the amount value being represented individually. The output should like something like this:
| Vendor | Buyer | Amount |
|--------|:-----:|-------:|
| A | P | 100 |
| A | P | 290 |
| | | |
| B | P | 150 |
| | | |
| C | Q | 300 |
I thought of using split(), but since my original data has too many records, the split function creates too many lists and it becomes tedious to create new datasets from them. How can I achieve the above stated output with any other method?
EDIT:
Let us assume that we have an additional column called date and the dataset now looks like this:
| Vendor | Buyer | Amount | Date |
|--------|:-----:|-------:|-----------|
| A | P | 100 | 3/6/2019 |
| B | P | 150 | 7/6/2018 |
| C | Q | 300 | 4/21/2018 |
| A | P | 290 | 6/5/2018 |
Once, each buyer and vendor is grouped together, I need to arrange the dates in ascending order for each buyer and vendor such that it looks something like the below one:
| Vendor | Buyer | Amount | Date |
|--------|:-----:|-------:|-----------|
| A | P | 290 | 6/5/2018 |
| A | P | 100 | 3/6/2019 |
| | | | |
| B | P | 150 | 7/6/2018 |
| | | | |
| C | Q | 300 | 4/21/2018 |
and then remove the single transactions to get the final table containing only
| Vendor | Buyer | Amount | Date |
|--------|:-----:|-------:|----------|
| A | P | 290 | 6/5/2018 |
| A | P | 100 | 3/6/2019 |

Answer 1

In the following we sort the data frame and add a group column which allows easy subsequent processing of individual groups. For example, to process the groups without creating a large split of DF:
for(g in unique(DFout$group)) {
DFsub <- subset(DFout, group == g)
... process DFsub ...
}
1) Base R Sort the data and then assign the group column using cumsum on the non-duplicated elements.
library(data.table)
o <- with(DF, order(Vendor, Buyer))
DFo <- DF[o, ]
DFout <- transform(DFo, group = cumsum(!duplicated(data.frame(Vendor, Buyer))))
DFout
giving:
Vendor Buyer Amount group
1 A P 100 1
4 A P 290 1
2 B P 150 2
3 C Q 300 3
I am not sure this is such a good idea to do in the first place but if you really want to add a row of NAs after each group:
ix <- unname(unlist(tapply(DFout$group, DFout$group, function(x) c(x, NA))))
ix[!is.na(ix)] <- seq_len(nrow(DFout))
DFout[ix, ]
2) data.table Convert to data.table, set the key (which sorts it) and use rleid to assign the group number.
library(data.table)
DT <- data.table(DF)
setkey(DT, Vendor, Buyer)
DT[, group := rleid(Vendor, Buyer)]
3) sqldf Another approach is to use SQL. This requires the development version of RSQLite on github. Here dense_rank acts similarly to rleid above.
library(sqldf)
sqldf("select *, dense_rank() over (order by Vendor, Buyer) as [group]
from DF
order by Vendor, Buyer")
giving:
Vendor Buyer Amount group
1 A P 100 1
2 A P 290 1
3 B P 150 2
4 C Q 300 3
Note
DF <- structure(list(Vendor = structure(c(1L, 2L, 3L, 1L), .Label = c("A",
"B", "C"), class = "factor"), Buyer = structure(c(1L, 1L, 2L,
1L), .Label = c("P", "Q"), class = "factor"), Amount = c(100L,
150L, 300L, 290L)), class = "data.frame", row.names = c(NA, -4L
))