simplify multiple one to one paired aggregation - r

Question

I'd like to calculate mean for multiple one to one paired aggregation. For example, I have a data.frame below. I'd like to calculate the mean for for column b1 by sym & a1, and b2 by sym & a2 simultaneously.
sym a1 a2 b1 b2
1 a 1 2 1 1
2 a 2 2 2 2
3 a 1 2 3 3
4 a 2 2 4 4
5 b 1 1 5 5
6 b 2 1 6 6
7 b 1 1 7 7
8 b 2 1 8 8
Here is my code which uses lapply to iterate over each pair. Is there any more efficient way than this?
df <- data.frame(sym=c(rep('a', 4), rep('b', 4)), a1=rep(1:2, 4),
a2=rep(2:1, each=4), b1=rep(1:8), b2=rep(1:8))
tmp <- ddply(df, "sym", function(x) {
temp.ls <- lapply(1:2, function(i) {
t2 <- aggregate(x = x[3+i], by=x[1+i], FUN=function(.){mean(., na.rm = T)})
colnames(t2) <- c("a", "b")
t2
})
temp.all <- Reduce(function(x, y) merge(x, y, by=c("a"), all=T, sort=T),
temp.ls)
})

Answer 1

dplyr makes this pretty straightforward:
library(dplyr)
inner_join(df %>% group_by(sym, a1) %>% summarise(b1.mean=mean(b1)),
df %>% group_by(sym, a2) %>% summarise(b2.mean=mean(b2)))
# Joining by: "sym"
# Source: local data frame [4 x 5]
# Groups: sym
#
# sym a1 b1.mean a2 b2.mean
# 1 a 1 2 2 2.5
# 2 a 2 3 2 2.5
# 3 b 1 6 1 6.5
# 4 b 2 7 1 6.5
If you want a single column for a, and want to fill non-appearing combinations with NA as in your example solution, then left_join is an option:
left_join(df %>% group_by(sym, a=a1) %>% summarise(b1.mean=mean(b1)),
df %>% group_by(sym, a=a2) %>% summarise(b2.mean=mean(b2)),
by=c('sym', 'a'))
# Source: local data frame [4 x 4]
# Groups: sym
#
# sym a b1.mean b2.mean
# 1 a 1 2 NA
# 2 a 2 3 2.5
# 3 b 1 6 6.5
# 4 b 2 7 NA
Hat-tip to #beginnerR for reminding me about dplyr join operations.
EDIT
In response to the comments, if you have more than two groupings, and want to join all the resulting tables together, then here's one way to do this:
# Example data
set.seed(1)
(d <- data.frame(sym=sample(letters[1:4], 10, replace=T),
a1=sample(5, 10, replace=TRUE),
a2=sample(5, 10, replace=TRUE),
a3=sample(5, 10, replace=TRUE),
b1=runif(10), b2=runif(10), b3=runif(10)))
# sym a1 a2 a3 b1 b2 b3
# 1 b 2 5 3 0.8209463 0.47761962 0.91287592
# 2 b 1 2 3 0.6470602 0.86120948 0.29360337
# 3 c 4 4 3 0.7829328 0.43809711 0.45906573
# 4 d 2 1 1 0.5530363 0.24479728 0.33239467
# 5 a 4 2 5 0.5297196 0.07067905 0.65087047
# 6 d 3 2 4 0.7893562 0.09946616 0.25801678
# 7 d 4 1 4 0.0233312 0.31627171 0.47854525
# 8 c 5 2 1 0.4772301 0.51863426 0.76631067
# 9 c 2 5 4 0.7323137 0.66200508 0.08424691
# 10 a 4 2 3 0.6927316 0.40683019 0.87532133
L <- mapply(function(x, y) {
grpd <- eval(substitute(group_by(d, sym, a=x), list(x=as.name(x))))
eval(substitute(summarise(grpd, mean(y)), list(y=as.name(y))))
}, paste0('a', 1:3), paste0('b', 1:3), SIMPLIFY=FALSE)
Reduce(function(...) left_join(..., all=T), L)
# Source: local data frame [9 x 5]
# Groups: sym
#
# sym a mean(b1) mean(b2) mean(b3)
# 1 a 4 0.6112256 NA NA
# 2 b 1 0.6470602 NA NA
# 3 b 2 0.8209463 0.86120948 NA
# 4 c 2 0.7323137 0.51863426 NA
# 5 c 4 0.7829328 0.43809711 0.08424691
# 6 c 5 0.4772301 0.66200508 NA
# 7 d 2 0.5530363 0.09946616 NA
# 8 d 3 0.7893562 NA NA
# 9 d 4 0.0233312 NA 0.36828101