Proper way to split data frames at multiple levels using ddply

Question

Let's say I have a data frame like the following:

  year stint  ID  W
1 2003     1 abc 10
2 2003     2 abc  3
3 2003     1 def 16
4 2004     1 abc 15
5 2004     1 def 11
6 2004     2 def  7

I would like to combine the data so that it looks like

  year  ID  W
1 2003 abc 13
3 2003 def 16
4 2004 abc 15
5 2004 def 18

I found a way to combine the data as desired, but I'm very sure that there's a better way.

combinedData = unique(ddply(data, "ID", function(x) {
    ddply(x, "year", function(y) {
        data.frame(ID=x$ID, W=sum(y$W))
    })
}))
combinedData[order(combinedData$year),]

This produces the following output:

   year  ID  W
1  2003 abc 13
7  2003 def 16
4  2004 abc 15
10 2004 def 18

Specifically I don't like that I had to use unique (otherwise I get each unique combo of year,ID,W three times in the outputted data), and I don't like that the row numbers aren't sequential. How can I do this more cleanly?

Try with `dplyr` i.e. `data %>% group_by(year, ID) %>% summarise(W= sum(W))`. If you are using `ddply` then `ddply(data, .(year, ID), summarise, W=sum(W))` — akrun, Sep 27 '16 at 12:19

score 2 · Answer 1 · answered Sep 27 '16 at 12:25

Do this with base R:

aggregate(W~year+ID, df, sum)

#  year  ID  W
#1 2003 abc 13
#2 2004 abc 15
#3 2003 def 16
#4 2004 def 18

data

df <- structure(list(year = c(2003L, 2003L, 2003L, 2004L, 2004L, 2004L
), stint = c(1L, 2L, 1L, 1L, 1L, 2L), ID = structure(c(1L, 1L, 
2L, 1L, 2L, 2L), .Label = c("abc", "def"), class = "factor"), 
    W = c(10L, 3L, 16L, 15L, 11L, 7L)), .Names = c("year", "stint", 
"ID", "W"), class = "data.frame", row.names = c("1", "2", "3", 
"4", "5", "6"))

Proper way to split data frames at multiple levels using ddply

1 Answers1