Joining aggregated values back to the original data frame [duplicate]

Question

One of the design patterns I use over and over is performing a "group by" or "split, apply, combine (SAC)" on a data frame and then joining the aggregated data back to the original data. This is useful, for example, when calculating each county's deviation from the state mean in a data frame with many states and counties. Rarely is my aggregate calculation only a simple mean, but it makes a good example. I often solve this problem the following way:

require(plyr)
set.seed(1)

## set up some data
group1 <- rep(1:3, 4)
group2 <- sample(c("A","B","C"), 12, rep=TRUE) 
values <- rnorm(12)
df <- data.frame(group1, group2, values)

## got some data, so let's aggregate

group1Mean <- ddply( df, "group1", function(x) 
                     data.frame( meanValue = mean(x$values) ) )
df <- merge( df, group1Mean )
df

Which produces nice aggregate data like the following:

> df
   group1 group2   values meanValue
1       1      A  0.48743 -0.121033
2       1      A -0.04493 -0.121033
3       1      C -0.62124 -0.121033
4       1      C -0.30539 -0.121033
5       2      A  1.51178  0.004804
6       2      B  0.73832  0.004804
7       2      A -0.01619  0.004804
8       2      B -2.21470  0.004804
9       3      B  1.12493  0.758598
10      3      C  0.38984  0.758598
11      3      B  0.57578  0.758598
12      3      A  0.94384  0.758598

This works, but are there alternative ways of doing this which improve on readability, performance, etc?

Answer 1

One line of code does the trick:

new <- ddply( df, "group1", transform, numcolwise(mean))
new

group1 group2      values    meanValue
1       1      A  0.48742905 -0.121033381
2       1      A -0.04493361 -0.121033381
3       1      C -0.62124058 -0.121033381
4       1      C -0.30538839 -0.121033381
5       2      A  1.51178117  0.004803931
6       2      B  0.73832471  0.004803931
7       2      A -0.01619026  0.004803931
8       2      B -2.21469989  0.004803931
9       3      B  1.12493092  0.758597929
10      3      C  0.38984324  0.758597929
11      3      B  0.57578135  0.758597929
12      3      A  0.94383621  0.758597929

identical(df, new)
[1] TRUE

Answer 2

I think ave() is more useful here than the plyr call you show (I'm not familiar enough with plyr to know if you can do what you want with plyr directly or not, I would be surprised if you can't!) or the other base R alternatives (aggregate(), tapply()).:

> with(df, ave(values, group1, FUN = mean))
 [1] -0.121033381  0.004803931  0.758597929 -0.121033381  0.004803931
 [6]  0.758597929 -0.121033381  0.004803931  0.758597929 -0.121033381
[11]  0.004803931  0.758597929

You can use within() or transform() to embed this result directly into df:

> df2 <- within(df, meanValue <- ave(values, group1, FUN = mean))
> head(df2)
  group1 group2     values    meanValue
1      1      A  0.4874291 -0.121033381
2      2      B  0.7383247  0.004803931
3      3      B  0.5757814  0.758597929
4      1      C -0.3053884 -0.121033381
5      2      A  1.5117812  0.004803931
6      3      C  0.3898432  0.758597929
> df3 <- transform(df, meanValue = ave(values, group1, FUN = mean))
> all.equal(df2,df3)
[1] TRUE

And if the ordering is important:

> head(df2[order(df2$group1, df2$group2), ])
   group1 group2      values    meanValue
1       1      A  0.48742905 -0.121033381
10      1      A -0.04493361 -0.121033381
4       1      C -0.30538839 -0.121033381
7       1      C -0.62124058 -0.121033381
5       2      A  1.51178117  0.004803931
11      2      A -0.01619026  0.004803931