# Descending sort in Haskell

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When confronted with a problem of sorting a list in descending order in Haskell, it is tempting to reach for a “lazy” solution `reverse . sort`

.

An obvious issue with this is efficiency. While sorting in descending order should in theory be exactly as efficient as sorting in ascending order, the above solution requires an extra list traversal after the sorting itself is done.

This argument can be dismissed on the grounds that `reverse`

’s run time, \(\Theta(n)\), is, in general, less than `sort`

’s run time, \(O(n \log n)\), so it’s not a big deal. Additionally, one could argue that, unlike more complex solutions, this one is “obviously correct”.

As the rest of this article explains, *neither* of these claims holds universally.

## Proper solutions

Here are the two ways to sort a list in descending order that I am aware of. Both require the more general `sortBy`

function

`sortBy :: (a -> a -> Ordering) -> [a] -> [a]`

The first argument to `sortBy`

is the comparison function. For each pair of arguments it returns a value of type

`data Ordering = LT | EQ | GT`

which describes the ordering of those arguments.

The “standard” ordering is given by the `compare`

function from the `Ord`

typeclass. Thus, `sort`

is nothing more than

`sort = sortBy compare`

The first solution to the descending sort problem exploits the fact that, to get the opposite ordering, we can simply swap around the two arguments to `compare`

:

`= sortBy (flip compare) sortDesc `

The second solution relies on the `comparing`

function from `Data.Ord`

, which gives a particular way to compare two values: map them to other values which are then compared using the standard `Ord`

ordering.

`comparing :: Ord a => (b -> a) -> b -> b -> Ordering`

This trick is often used in mathematics: to maximize a function \(x\mapsto f(x)\), it suffices to minimize the function \(x \mapsto -f(x)\). In Haskell, we could write

`= sortBy (comparing negate) sortDesc `

and it would work most of the time. However,

```
> sortDesc [1,minBound::Int]
-9223372036854775808,1] [
```

Besides, negation only works on numbers; what if you want to sort a list of pairs of numbers?

Fortunately, `Data.Ord`

defines a `Down`

newtype which does exactly what we want: it reverses the ordering between values that it’s applied to.

```
> 1 < 2
True
> Down 1 < Down 2
False
```

Thus, the second way to sort the list in descending order is

`= sortBy (comparing Down) sortDesc `

## sortOn

Christopher King points out that the last example may be simplified with the help of the `sortOn`

function introduced in base 4.8 (GHC 7.10):

`sortOn :: Ord b => (a -> b) -> [a] -> [a]`

Thus, we can write

`= sortOn Down sortDesc `

Very elegant, but let’s look at how `sortOn`

is implemented:

```
=
sortOn f map snd .
fst) .
sortBy (comparing map (\x -> let y = f x in y `seq` (y, x))
```

This is somewhat more complicated than `sortBy (comparing Down)`

, and it does the extra work of first allocating \(n\) cons cells and \(n\) tuples, then allocating another \(n\) cons cells for the final result.

Thus, we might expect that `sortOn`

performs worse than `sortBy`

. Let’s check our intuition:

```
import Criterion
import Criterion.Main
import Data.List
import Data.Ord
list :: [Int]
= [1..10000]
list
= defaultMain
main "sort" $ nf sort (reverse list)
[ bench "sortBy" $ nf (sortBy (comparing Down)) list
, bench "sortOn" $ nf (sortOn Down) list
, bench ]
```

```
benchmarking sort
time 134.9 μs (134.3 μs .. 135.4 μs)
1.000 R² (0.999 R² .. 1.000 R²)
mean 134.8 μs (134.2 μs .. 135.7 μs)
std dev 2.677 μs (1.762 μs .. 3.956 μs)
variance introduced by outliers: 14% (moderately inflated)
benchmarking sortBy
time 131.0 μs (130.6 μs .. 131.4 μs)
1.000 R² (1.000 R² .. 1.000 R²)
mean 131.1 μs (130.8 μs .. 131.4 μs)
std dev 965.1 ns (766.5 ns .. 1.252 μs)
benchmarking sortOn
time 940.5 μs (928.6 μs .. 958.1 μs)
0.998 R² (0.997 R² .. 0.999 R²)
mean 950.6 μs (940.9 μs .. 961.1 μs)
std dev 34.88 μs (30.06 μs .. 44.19 μs)
variance introduced by outliers: 27% (moderately inflated)
```

As we see, `sortOn`

is 7 times slower than `sortBy`

in this example. I also included `sort`

in this comparison to show that `sortBy (comparing Down)`

has no runtime overhead.

There is a good reason why `sortOn`

is implemented in that way. To quote the documentation:

`sortOn f`

is equivalent to`sortBy (comparing f)`

, but has the performance advantage of only evaluating f once for each element in the input list. This is called the decorate-sort-undecorate paradigm, or Schwartzian transform.

Indeed, if `f`

performed any non-trivial amount of work, it would be wise to cache its results—and that’s what `sortOn`

does.

But `Down`

is a newtype constructor—it performs literally no work at all—so the caching effort is wasted.

## sortWith

Yuriy Syrovetskiy points out that there is also `sortWith`

defined in `GHC.Exts`

, which has the same type as `sortOn`

but does no caching. So if you want to abbreviate `sortBy (comparing Down)`

, you can say `sortWith Down`

—but you need to import `GHC.Exts`

first.

Wonder why `sortWith`

lives in `GHC.Exts`

and not in `Data.List`

or `Data.Ord`

? It was originally added to aid writing SQL-like queries in Haskell, although I haven’t seen it used a single time in my career.

## Asymptotics

Thanks to Haskell’s laziness in general and the careful implementation of `sort`

in particular, `sort`

can run in linear time when only a fixed number of first elements is requested.

So this function will return the 10 largest elements in \(\Theta(n)\) time:

`take 10 . sortBy (comparing Down)`

While our “lazy” solution

`take 10 . reverse . sort`

(which, ironically, turns out not to be lazy enough—in the technical sense of the word “lazy”) will run in \(O(n \log n)\) time. This is because it requests *the last* 10 elements of the sorted list, and in the process of doing so needs to traverse the whole sorted list.

This may appear paradoxical if considered outside of the context of lazy evaluation. Normally, if two linear steps are performed sequentially, the result is still linear. Here we see that adding a linear step upgrades the overall complexity to \(O(n \log n)\).

## Semantics

As I mentioned in the beginning, the simplicity of `reverse . sort`

may be deceptive. The semantics of `reverse . sort`

and `sortBy (comparing Down)`

differ in a subtle way, and you *probably* want the semantics of `sortBy (comparing Down)`

.

This is because `sort`

and `sortBy`

are *stable* sorting functions. They preserve the relative ordering of “equal” elements within the list. Often this doesn’t matter because you cannot tell equal elements apart anyway.

It starts to matter when you use `comparing`

to sort objects by a certain feature. Here we sort the list of pairs by their first elements in descending order:

```
> sortBy (comparing (Down . fst)) [(1,'a'),(2,'b'),(2,'c')]
2,'b'),(2,'c'),(1,'a')] [(
```

According to our criterion, the elements `(2,'b')`

and `(2,'c')`

are considered equal, but we can see that their ordering has been preserved.

The `revese`

-based solution, on the other hand, reverses the order of equal elements, too:

```
> (reverse . sortBy (comparing fst)) [(1,'a'),(2,'b'),(2,'c')]
2,'c'),(2,'b'),(1,'a')] [(
```

Sort with care!

## A note on sorted lists

The original version of this article said that `sort`

runs in \(\Theta(n \log n)\) time. @obadzz points out that this is not true: `sort`

is implemented in such a way that it will run linearly when the list is already almost sorted in any direction. Thus I have replaced \(\Theta(n \log n)\) with \(O(n \log n)\) when talking about `sort`

’s complexity.

## Why reverse in the benchmark?

Jens Petersen asks why in the benchmark above I’m comparing

`Down)) list nf (sortBy (comparing `

to

`sort (reverse list) nf `

I include `sort`

here as a baseline, to demonstrate that `sortBy (comparing Down)`

does not have any significant overhead comparing to a simple sort. But why have `reverse`

there? Is that a typo?

First of all, it’s important to understand what

`sort (reverse list) nf `

does. It may look as if it’s measuring the time it takes to reverse and then sort a list—and why would anyone want to do such a thing?

But that’s not what’s happening. In criterion, the `nf`

combinator takes two separate values: the function (here, `sort`

) and its argument (here, `(reverse list)`

). Both the function and the argument are precomputed, and what’s measured repeatedly is the time it takes to evaluate the application of the function to its argument.

The reason I measure the time it takes to sort the *reversed* list is to have more of an apples-to-apples comparison. If I compared

`Down)) list nf (sortBy (comparing `

to

`sort (reverse list) nf `

… I would be comparing the same algorithm but on the arguments on which it takes two different code paths: in one case it ends up reversing the list, in the other keeping it intact. (Counterintuitively, it looks like it’s faster for `sort`

/`sortBy`

to reverse the list than to keep it as-is.)

Therefore, by reversing the argument, I make sure that in all three cases (`sort`

, `sortBy`

, and `sortOn`

) the input list “looks” the same as seen by the comparison function, and in all three cases the sorting algorithm needs to reverse the list.