Some of this material originally appeared in What’s the difference between Currying and Partial Application? Here it is again, in ECMAScript 2015 syntax.

What is Partial Application? Good question!

arity

Before we jump in, let’s get some terminology straight. Functions have arity, meaning the number of arguments they accept. A “unary” function accepts one argument, a “polyadic” function takes more than one argument. There are specialized terms we can use: A “binary” function accepts two, a “ternary” function accepts three, and you can rustle about with greek or latin words and invent names for functions that accept more than three arguments.

(Some functions accept a variable number of arguments, we call them variadic, although variadic functions and functions taking no arguments aren’t our primary focus in this essay.)

partial application

Partial application is straightforward. We could start with addition or some such completely trivial example, but if you don’t mind we’ll have a look at something that is of actual use in daily programming.

As a preamble, let’s make ourselves a mapWith function that maps a function over any collection that has a .map method:

const mapWith = ( unaryFn , collection ) => collection . map ( unaryFn ); const square = ( n ) => n * n ; mapWith ( square , [ 1 , 2 , 3 ]) //=> [1, 4, 9]

mapWith is a binary function, square is a unary function. When we called mapWith with arguments square and [1, 2, 3] and square , we applied the arguments to the function and got our result.

Since mapWith takes two arguments, and we supplied two arguments, we fully applied the arguments to the function. So what’s “partial” application? Supplying fewer arguments. Like supplying one argument to mapWith .

Now what happens if we supply one argument to mapWith ? We can’t get a result without the other argument, so as we’ve written it, it breaks:

mapWith ( square ) //=> undefined is not an object (evaluating 'collection.map')

But let’s imagine that we could apply fewer arguments. We wouldn’t be fully applying mapWith , we’d be partially applying mapWith . What would we expect to get? Well, imagine we decide to buy a $2,000 bicycle. We go into the store, we give them $1,000. What do we get back? A pice of paper saying that we are doing a layaway program. The $1,000 is held in trust for us, when we come back with the other $1,000, we get the bicycle.

Putting down $1,000 on a $2,000 bicycle is partially buying a bicycle. What it gets us is the right to finish buying the bicycle later. It’s the same with partial application. If we were able to partially apply the mapWith function, we’d get back the right to finish applying it later, with the other argument.

Something like this:

const mapWith = ( unaryFn ) => ( collection ) => collection . map ( unaryFn ); const square = ( n ) => n * n ; const partiallyAppliedMapWith = mapWith ( square ); partiallyAppliedMapWith ([ 1 , 2 , 3 ]) //=> [1, 4, 9]

The thing is, we don’t want to always write our functions in this way. So what we want is a function that takes this:

const mapWith = ( unaryFn , collection ) => collection . map ( unaryFn );

And turns it into this:

partiallyAppliedMapWith ([ 1 , 2 , 3 ]) //=> [1, 4, 9]

Working backwards:

const partiallyAppliedMapWith = ( collection ) => mapWith ( unaryFn , collection );

The expression (collection) => mapWith(unaryFn, collection) has two free variables, mapWith and unaryFn . If we were using a fancy refactoring editor, we could extract a function. Let’s do it by hand:

const ____ = ( mapWith , unaryFn ) => ( collection ) => mapWith ( unaryFn , collection );

What is this _____ function? It takes the mapWith function and the unaryFn , and returns a function that takes a collection and returns the result of applying unaryFn and collection to mapWith . Let’s make the names very generic, the function works no matter what we call the arguments:

const ____ = ( fn , x ) => ( y ) => fn ( x , y );

This is a function that takes a function and an argument, and returns a function that takes another argument, and applies both arguments to the function. So, we can write this:

const mapWith = ( unaryFn , collection ) => collection . map ( unaryFn ); const square = ( n ) => n * n ; const ____ = ( fn , x ) => ( y ) => fn ( x , y ); const partiallyAppliedMapWith = ____ ( mapWith , square ); partiallyAppliedMapWith ([ 1 , 2 , 3 ]) //=> [1, 4, 9]

So what is this ____ function? It partially applies one argument to any function that takes two arguments.

We can dress it up a bit. For one thing, it doesn’t work with methods, it’s strictly a blue higher-order function. Let’s make it khaki by passing this properly:

const ____ = ( fn , x ) => function ( y ) { return fn . call ( this , x , y ); };

Another problem is that it only works for binary functions. Let’s make it so we can pass one argument and we get back a function that takes as many remaining arguments as we like:

const ____ = ( fn , x ) => function (... remainingArgs ) { return fn . call ( this , x , ... remainingArgs ); }; const add = ( verb , a , b ) => `The ${ verb } of ${ a } and ${ b } is ${ a + b } ` const summer = ____ ( add , ' sum ' ); summer ( 2 , 3 ) //=> The sum of 2 and 3 is 5

And what if we want to partially apply more than one argument?

const ____ = ( fn , ... partiallyAppliedArgs ) => function (... remainingArgs ) { return fn . apply ( this , partiallyAppliedArgs . concat ( remainingArgs )); }; const add = ( verb , a , b ) => `The ${ verb } of ${ a } and ${ b } is ${ a + b } ` const sum2 = ____ ( add , ' sum ' , 2 ); sum2 ( 3 ) //=> The sum of 2 and 3 is 5

What we have just written is a left partial application function: Given any function and some arguments, we partially apply those arguments and get back a function that applies those arguments and any more you supply to the original function.

Partial application is thus the application of part of the arguments of a function, and getting in return a function that takes the remaining arguments.

And here’s our finished function, properly named:

const leftPartialApply = ( fn , ... partiallyAppliedArgs ) => function (... remainingArgs ) { return fn . apply ( this , partiallyAppliedArgs . concat ( remainingArgs )); }; const add = ( verb , a , b ) => `The ${ verb } of ${ a } and ${ b } is ${ a + b } ` const sum2 = leftPartialApply ( add , ' sum ' , 2 ); sum2 ( 3 ) //=> The sum of 2 and 3 is 5

right partial application

It is very convenient to have a mapWith function, because you are far more likely to want to write something like:

const squareAll = leftPartialApply ( mapWith , x => x * x );

Than to write:

const map123 = leftPartialApply ( map , [ 1 , 2 , 3 ]);

But sometime you have map but not mapWith , or some other analogous situation where you want to apply the values from the right rather than the left. No problem:

const rightPartialApply = ( fn , ... partiallyAppliedArgs ) => function (... remainingArgs ) { return fn . apply ( this , remainingArgs . concat ( partiallyAppliedArgs )); };

arbitrary partial application

What if you want to apply some, but not all of the arguments, and they may not be neatly lined up at the beginning or end? This is also possible, provided we define a placeholder of some kind, and then write some code to “fill in the blanks”.

This implementation takes a “template” of values, you insert placeholder values (traditionally _ , but anything will do) where you want values to be supplied later.

const div = ( verbed , numerator , denominator ) => ` ${ numerator } ${ verbed } ${ denominator } is ${ numerator / denominator } ` div ( ' divided by ' , 1 , 3 ) //=> 1 divided by 3 is 0.3333333333333333 const arbitraryPartialApply = (() => { const placeholder = {}, arbitraryPartialApply = ( fn , ... template ) => { let remainingArgIndex = 0 ; const mapper = template . map (( templateArg ) => templateArg === placeholder ? (( i ) => ( args ) => args [ i ])( remainingArgIndex ++ ) : ( args ) => templateArg ); return function (... remainingArgs ) { const composedArgs = mapper . map ( f => f ( remainingArgs )); return fn . apply ( this , composedArgs ); } }; arbitraryPartialApply . _ = placeholder ; return arbitraryPartialApply ; })(); const _ = arbitraryPartialApply . _ ; const dividedByThree = arbitraryPartialApply ( div , ' divided by ' , _ , 3 ); dividedByThree ( 2 ) //=> 2 divided by 3 is 0.6666666666666666

Arbitrary partial application handles most of the cases for left- or right-partial application, but has more internal moving parts. It also doesn’t handle cases involving an arbitrary number of parameters.

For example, the built-in function Math.max returns the largest of its arguments, or null if no arguments are supplied:

Math . max ( 1 , 2 , 3 ) //=> 3 Math . max ( - 1 , - 2 , - 3 ) //=> -1 Math . max () //=> null

What if we want to have a default argument? For example, what if we want it tor return the largest number greater than or equal to 0 , or 0 if there aren’t any? We can do that with leftPartialApplication , but we can’t with arbitraryPartialApply , because we want to accept an arbitrary number of arguments:

const maxDefaultZero = leftPartialApply ( Math . max , 0 ); maxDefaultZero ( 1 , 2 , 3 ) //=> 3 Math . max ( - 1 , - 2 , - 3 ) //=> 0 maxDefaultZero () //=> 0

So there’s good reason to have left-, right-, and arbitrary partial application functions in our toolbox.

what’s partial application again?

“Partial application is the conversion of a polyadic function into a function taking fewer arguments arguments by providing one or more arguments in advance.” JavaScript does not have partial application built into the language (yet), but it is possible to write our own higher-order functions that perform left-, right-, or arbitrary partial application.

This post was extracted from a draft of the book, JavaScript Allongé, The “Six” Edition. The extracts so far: