# R Language Using split in the split-apply-combine paradigm

## Example

A popular form of data analysis is split-apply-combine, in which you split your data into groups, apply some sort of processing on each group, and then combine the results.

Let's consider a data analysis where we want to obtain the two cars with the best miles per gallon (mpg) for each cylinder count (cyl) in the built-in mtcars dataset. First, we split the `mtcars` data frame by the cylinder count:

``````(spl <- split(mtcars, mtcars\$cyl))
# \$`4`
#                 mpg cyl  disp  hp drat    wt  qsec vs am gear carb
# Datsun 710     22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1
# Merc 240D      24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2
# Merc 230       22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2
# Fiat 128       32.4   4  78.7  66 4.08 2.200 19.47  1  1    4    1
# ...
#
# \$`6`
#                 mpg cyl  disp  hp drat    wt  qsec vs am gear carb
# Mazda RX4      21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4
# Mazda RX4 Wag  21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
# Hornet 4 Drive 21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1
# Valiant        18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1
# ...
#
# \$`8`
#                      mpg cyl  disp  hp drat    wt  qsec vs am gear carb
# Hornet Sportabout   18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2
# Duster 360          14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4
# Merc 450SE          16.4   8 275.8 180 3.07 4.070 17.40  0  0    3    3
# Merc 450SL          17.3   8 275.8 180 3.07 3.730 17.60  0  0    3    3
# ...
``````

This has returned a list of data frames, one for each cylinder count. As indicated by the output, we could obtain the relevant data frames with `spl\$`4``, `spl\$`6``, and `spl\$`8`` (some might find it more visually appealing to use `spl\$"4"` or `spl[["4"]]` instead).

Now, we can use `lapply` to loop through this list, applying our function that extracts the cars with the best 2 mpg values from each of the list elements:

``````(best2 <- lapply(spl, function(x) tail(x[order(x\$mpg),], 2)))
# \$`4`
#                 mpg cyl disp hp drat    wt  qsec vs am gear carb
# Fiat 128       32.4   4 78.7 66 4.08 2.200 19.47  1  1    4    1
# Toyota Corolla 33.9   4 71.1 65 4.22 1.835 19.90  1  1    4    1
#
# \$`6`
#                 mpg cyl disp  hp drat    wt  qsec vs am gear carb
# Mazda RX4 Wag  21.0   6  160 110 3.90 2.875 17.02  0  1    4    4
# Hornet 4 Drive 21.4   6  258 110 3.08 3.215 19.44  1  0    3    1
#
# \$`8`
#                    mpg cyl disp  hp drat    wt  qsec vs am gear carb
# Hornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2
# Pontiac Firebird  19.2   8  400 175 3.08 3.845 17.05  0  0    3    2
``````

Finally, we can combine everything together using `rbind`. We want to call `rbind(best2[["4"]], best2[["6"]], best2[["8"]])`, but this would be tedious if we had a huge list. As a result, we use:

``````do.call(rbind, best2)
#                      mpg cyl  disp  hp drat    wt  qsec vs am gear carb
# 4.Fiat 128          32.4   4  78.7  66 4.08 2.200 19.47  1  1    4    1
# 4.Toyota Corolla    33.9   4  71.1  65 4.22 1.835 19.90  1  1    4    1
# 6.Mazda RX4 Wag     21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
# 6.Hornet 4 Drive    21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1
# 8.Hornet Sportabout 18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2
# 8.Pontiac Firebird  19.2   8 400.0 175 3.08 3.845 17.05  0  0    3    2
``````

This returns the result of `rbind` (argument 1, a function) with all the elements of `best2` (argument 2, a list) passed as arguments.

With simple analyses like this one, it can be more compact (and possibly much less readable!) to do the whole split-apply-combine in a single line of code:

``````do.call(rbind, lapply(split(mtcars, mtcars\$cyl), function(x) tail(x[order(x\$mpg),], 2)))
``````

It is also worth noting that the `lapply(split(x,f), FUN)` combination can be alternatively framed using the `?by` function:

``````by(mtcars, mtcars\$cyl, function(x) tail(x[order(x\$mpg),], 2))
do.call(rbind, by(mtcars, mtcars\$cyl, function(x) tail(x[order(x\$mpg),], 2)))
``````