Is there a short Lambda expression for a->a->Ordering?

问题

Lambda Expressions can make life easier and I love the way, you can shorten expressions like (\x -> x + 1) to (+1).

That's why I'm wondering if there is anything similar for anonymous functions that take two arguments and return a Ordering.

For example: Can you shorten this code:

sortByFirst :: Ord a => [(a, b)] -> [(a, b)]
sortByFirst = sortBy (\x y -> fst x `compare` fst y)

回答1:

You could use the on function, from the Data.Function module.

import Data.List

sortByFirst = sortBy (compare `on` fst)

The definition of on is roughly

on f g = \x y -> f (g x) (g y)

回答2:

With comparing from Data.Ord, it can be spelled as...

sortByFirst = sortBy (comparing fst)

... which is equivalent to is equivalent to the compare `on` fst in chepner's answer.

(Data.List also offers a sortOn :: Ord b => (a -> b) -> [a] -> [a], but that is a special purpose function which is only worth using if the involved a -> b function is somewhat expensive, something that fst is not.)

来源：https://stackoverflow.com/questions/53491403/is-there-a-short-lambda-expression-for-a-a-ordering