Writing a Functional Programming-Style map Function

Many thanks to Helen Durrant for reviewing this post and offering stellar suggestions. Originally posted on https://robertwpearce.com/javascript-writing-a-functional-programming-style-map-function.html.

In this post, we will write a functional programming-style implementation of JavaScript's map function that not only works with Array but any data structure that implements a map method. Such data structures are known as Functors. Some examples of Functors are the algebraic data types¹ Maybe and Async (prior knowledge of them is not required, and out of the two, we'll only use Maybe).

By the end of this post, you will:

• know how to implement a generic map function that includes functions for mapping Arrays, Objects, and Functors
• understand how to use map in a variety of scenarios
• know how to write a simple compose function and use composition
• know how to reliably test values for their types
• have received a small introduction to algebraic data types via the crocks library

This is a big post, so buckle up! If you want to see the final product, check out this CodeSandbox: https://codesandbox.io/s/bitter-grass-tknwb.

---

Note: if you're not familiar with Array.prototype.map already, check out my video on Using JavaScript's Array.prototype.map Method or my post on JavaScript: Understand Array.prototype.map by Reimplementing It.

We will use the implementation of the map function in crocks as our template, so if you want to skip this article entirely, you can go and view its source.

Overview

1. The Goal: map All the Things
2. Defining Our map Function
3. map an Array
4. map an Object
5. map a Function
6. map a Functor
7. throwing Out Bad Data

The Goal: map All the Things

Today we are going to write a map function that does the following:

• accepts a transformation function that takes in some argument of type a and transforms it into a value of type b; i.e., (a -> b)
• accepts and handles any of the following data types:
  • Array
  • Object
  • Function
  • Functors (e.g., Maybe)

Sounds easy, right? We'll see!

Defining Our map Function

There are some things we already know about our map function:

• it's called map (yay! nailed it!)
• it takes a function (fn) and then some datum (m²)³
• it returns the datum as transformed by said function

Let's sketch it out:

const map = (fn, m) => {
  // ???
}

Okay, it's a start. This could conceivably be used like this:

map(x => x.id, [{ id: 1 }, { id: 2 }])     // [1, 2]
map(x => x.id, [{ id: 'a' }, { id: 'b' }]) // ['a', 'b']

Note the repetition of the x => x.id. Let's try pulling it out into a
variable:

const propId = x => x.id
map(propId, [{ id: 1 }, { id: 2 }])     // [1, 2]
map(propId, [{ id: 'a' }, { id: 'b' }]) // ['a', 'b']

Alas, that's not much better – now we're just repeating the variable!

Instead, what if we could store our combination of function and map in a variable and then use that to call with our different data? By partially applying the function to map, we can!

const mapId = map.bind(null, x => x.id)
mapId([{ id: 1 }, { id: 2 }])     // [1, 2]
mapId([{ id: 'a' }, { id: 'b' }]) // ['a', 'b']

Nice! Now, let's go back to our sketch. Let's turn our binary function (which takes two parameters) to instead be a series of unary functions (which take one parameter⁴).

const map = fn => m => {
  // ???
}

Wow, that was easy. By default, languages like Haskell and Elm automatically curry all of their function parameters. There are ways to automate that in JavaScript, but for today, we will manually curry functions by using arrow functions to simulate it: const sum = a => b => a + b, for example.

Lastly, on the function definition side, it would be helpful for readers of our code to understand more about the intended types. In lieu of JavaScript not having a static type checker and me not knowing TypeScript yet, we'll do this using a Haskell-style pseudo-type signature:

map :: Functor f => (a -> b) -> f a -> f b

And we can place that as a comment above our function:

// map :: Functor f => (a -> b) -> f a -> f b
const map = fn => m => {
  // ???
}

Woah, woah, woah! What's all this? Let's break it down.

map :: Functor f => (a -> b) -> f a -> f b
--  |     |            |     |   |      |
--  1     2            3     4   5      6

1. Can be read, "has the type of"
2. Anything after :: and before => in a signature is a class constraint. This says we're going to use something in the type signature that obeys the Functor Laws⁵, identity and composition. The lowercase f represents what the Functor will be in the signature.
3. Our mapping function; e.g., x => x.id, like we did above.
4. -> Arrows are used in type signatures to say "then return...". In our map signature, we say, "We accept a function from a to b then return a function that accepts f of a and then return f of b". If we were summing three numbers, sum3 :: Number -> Number -> Number -> Number, this would read, "sum3 has the type of an expression that accepts a Number that returns a function that accepts a Number then returns a function that accepts a Number and then returns a Number."
5. f a says that a Functor, f, wraps some other type, a. A concrete example of this is [Number], which is a list (or Array) of Numbers.
6. f b says that a Functor, f, wraps some other type, b. Why isn't it a? This signifies that when we take in the Functor of any type a, it's totally cool if you want to change the return type inside the Functor. For example, when we take [{ id: 'a' }, { id: 'b' }] and use map to turn that into ['a', 'b'], we're taking [Object] (a list of Objects) and turning that into [String] (a list of Strings).

All together now! "map has the type of an expression where f is a Functor, and it accepts a function from a to b, then returns a function that accepts f of a, and then returns f of b."

map an Array

Let's map an Array!

Remember our Functor class constraint?

map :: Functor f => (a -> b) -> f a -> f b

Guess what? Array is a Functors! How? It adheres to the laws of identity and composition:

// identity
[1,2,3].map(x => x) // [1,2,3]

// composition
const add10 = x => x + 10
const mult2 = x => x * 2
[1,2,3].map(add10).map(mult2)     // [ 22, 24, 26 ]
// is equivalent to...
[1,2,3].map(x => mult2(add10(x))) // [ 22, 24, 26 ]

// another example of the composition law
const compose = (f, g) => x => f(g(x))
mult2(add10(2)) === compose(mult2, add10)(2) // true

// and applied back to our prior example
[1,2,3].map(add10).map(mult2)      // [ 22, 24, 26 ]
[1,2,3].map(x => mult2(add10(x)))  // [ 22, 24, 26 ]
[1,2,3].map(compose(mult2, add10)) // [ 22, 24, 26 ]

Through map, Array is a Functor. A way to quickly determine if something is a Functor is to ask, "Does it implement map / is it mappable?"

Since we know that Array is mappable, we can use our map function to check if the f a parameter is an Array and then use the build in Array.prototype.map function to get from a to b:

// map :: Functor f => (a -> b) -> f a -> f b
const map = fn => m => {
  if (isArray(m)) {
    return mapArray(fn, m)
  }
}

// isArray :: a -> Bool
const isArray = x => Array.isArray(x)

// mapArray :: ((a -> b), Array a) -> Array b
const mapArray = (fn, m) => m.map(x => fn(x))

Here, we use Array.isArray()⁶ to see if the argument, m, is an Array, then we call a function, mapArray, that handles the mapping of the Array.

You might be thinking: why m.map(x => fn(x)) and not m.map(fn)? As you might remember from my article on re-implementing Array.prototype.map, there are a few other arguments that the native implementation of map provide, as well as some potential changes to the this keyword in your callback function scope. Instead of allowing those to pass through, we simply take the first argument, the currently iterated value, and send that to the callback function⁷.

Now that we've seen the easy way to do map with Array, let's see what this would look like if we felt like implementing mapArray ourselves:

// mapArray :: ((a -> b), Array a) -> Array b
const mapArray = (fn, m) => {
  const newArray = []

  for (let i = 0; i < m.length; i++) {
    newArray[i] = fn(m[i])
  }

  return newArray
}

Not too shabby! All we do is create a new Array and set the results of calling the callback function with each item to its index in the new Array and then return that Array.

Do you think our map function can handle an Array of Arrays?

map(x => x * 2)([ [1,2], [3,4], [5,6] ])
// Array(3) [ NaN, NaN, NaN ]

While we can successfully iterate over the 3 items in the top-level Array, our callback function can't perform operations like [1,2] * 2! We need to do another map on the nested Arrays:

map(map(x => x * 2))([ [1,2], [3,4], [5,6] ])
// [ [2,4], [6,8], [10,12] ]

Well done! What else can you map? We're now going to leave charted waters and venture into the unknown.

map an Object

Let's say we have an i18n (short for "internationalization") object that we've been given that has a terribly annoying issue: every translation is prefixed and suffixed with an underscore (_)!

const i18n = {
  'en-US': {
    dayMode: '_Day mode_',
    greeting: '_Hello!_',
    nightMode: '_Night Mode_'
  },
  'es-ES': {
    dayMode: '_Modo día_',
    greeting: '_¡Hola!_'
    nightMode: '_Modo nocturno_'
  }
}

We could manually delete each one, or we could find and replace with our text editor, or we could write a for loop to do this, but because we're super awesome functional programmers, we'll try to map over the Object and write a function that removes the prefixed & suffixed underscores (...then we copy and paste that? work with me here!).

Before we can do this, we need to see what happens when we call .map() on an Object:

i18n['en-US'].map(x => x.slice(1))
// TypeError: i18n['en-US'].map is not a function

Oh no! If we can't even fix the en-US Object, how are we supposed to fix all of them? Let's update our map function to handle Objects.

// map :: Functor f => (a -> b) -> f a -> f b
const map = fn => m => {
  if (isArray(m)) {
    return mapArray(fn, m)
  }

  if (isObject(m)) {
    return mapObject(fn, m)
  }
}

// isObject :: a -> Bool
const isObject = x =>
  !!x && Object.prototype.toString.call(x) === '[object Object]'

// mapObject :: ((a -> b), { k: a }) -> { k: b }
const mapObject = (fn, m) => {
  const obj = {}

  for (const [k, v] of Object.entries(m)) {
    obj[k] = fn(v)
  }

  return obj
}

Here, we test if something is an object by using Object.prototype.toString and make sure to .call(x) instead of just .toString(x), for this reason:

Object.prototype.toString(null)
// "[object Object]"

Object.prototype.toString.call(null)
// "[object Null]"

Object.prototype.toString([])
// "[object Object]"

Object.prototype.toString.call([])
// "[object Array]"

Object.prototype.toString.call({})
// "[object Object]"

We then use our new mapObject function, whose signature is

mapObject :: ((a -> b), { k: a }) -> { k: b }

mapObject takes a function from a to b and an Object with a key(s) and some value(s), a, and returns an Object with a key(s) and some value(s) b. In short, it maps the values of an Object. Our mapObject function is nothing more than a for loop over each value returned from Object.entries()! It calls the callback function with each value and returns a new object with the same key and a new, updated value.

Let's try it out:

const i18n = {
  'en-US': {
    dayMode: '_Day mode_',
    greeting: '_Hello!_',
    nightMode: '_Night Mode_'
  },
  'es-ES': {
    dayMode: '_Modo día_',
    greeting: '_¡Hola!_'
    nightMode: '_Modo nocturno_'
  }
}
map(x => x.slice(1, -1))(i18n['en-US'])
// {
//   dayMode: 'Day mode',
//   greeting: 'Hello!',
//   nightMode: 'Night Mode'
// }

Okay – what about our entire i18n object?

map(map(x => x.slice(1, -1)))(i18n)
// {
//  'en-US': {
//    dayMode: 'Day mode',
//    greeting: 'Hello!',
//    nightMode: 'Night Mode'
//  },
//  'es-ES': {
//    dayMode: 'Modo día',
//    greeting: '¡Hola!',
//    nightMode: 'Modo nocturno'
//  }
// }

Since we're dealing with nested objects, we need to use map on an Object inside an Object. We pass a nested mapping function, and our little underscore problem is gone!

map a Function

Remember our functions mult2 and add10 from before?

const add10 = x => x + 10
const mult2 = x => x * 2

What would happen if we used those as the arguments to our map function and wanted them to be automatically composed together so that we can then provide a value later?

map(add10)(mult2)     // undefined
map(add10)(mult2)(12) // TypeError: map(...)(...) is not a function

Time for our map function to handle a Function as the second argument and compose the two functions together:

// map :: Functor f => (a -> b) -> f a -> f b
const map = fn => m => {
  if (isArray(m)) {
    return mapArray(fn, m)
  }

  if (isObject(m)) {
    return mapObj(fn, m)
  }

  if (isFunction(m)) {
    return compose(fn, m)
  }
}

// isFunction :: a -> Bool
const isFunction = x => typeof x === 'function'

// compose :: ((b -> c), (a -> b)) -> a -> c
const compose = (f, g) => x => f(g(x))

And when we run our previously failed code again,

map(add10)(mult2)     // function compose(x)
map(add10)(mult2)(12) // 44

we can see that calling map with two functions returns a composition of those two functions, and calling that result with a primitive value (12) gives us back our result, 44.

map a Functor

When we learned about mapping Arrays before, we learned that Arrays are Functors because they adhere to the laws of identity and composition; i.e., they are mappable.

There are all sorts of other data structures that implement a map method, just like Array.prototype does, and we want to be able to handle those, too!

We currently have all the tools required to implement map for Functors without even knowing how they might work! All we need to know is, "Does it implement map as a Function?" Let's see what we can come up with!

// map :: Functor f => (a -> b) -> f a -> f b
const map = fn => m => {
  if (isFunction(m)) {
    return compose(fn, m)
  }

  if (isArray(m)) {
    return mapArray(fn, m)
  }

  if (isFunctor(m)) {
    return mapFunctor(fn, m)
  }

  if (isObject(m)) {
    return mapObj(fn, m)
  }
}

// isFunction :: a -> Bool
const isFunction = x => typeof x === 'function'

// isFunctor :: a -> Bool
const isFunctor  = x => !!x && isFunction(x['map'])

// mapFunctor :: Functor f => ((a -> b), f a) -> f b
const mapFunctor = (fn, m) => m.map(fn)

That is surprisingly simple, isn't it? We use our isFunction check from before to test if m has a map property that is a Function, then we call map on m and pass it the callback Function in mapFunctor.

You might be thinking that mapArray and mapFunctor could use the same handler because Arrays are Functors, and you are correct; however, because of the extra implementation bits that come back from Array.prototype.map, we'll keep them separate and only call the callback to Array.prototype.map with the currently iterated item. Here's the difference:

// mapArray :: ((a -> b), Array a) -> Array b
const mapArray = (fn, m) => m.map(x => (fn(x))

// mapFunctor :: Functor f => ((a -> b), f a) -> f b
const mapFunctor = (fn, m) => m.map(fn)

If you don't care about this, it's totally acceptable to not include the Array bits at all and use the Functor map⁸ to handle the mapping of Arrays, since they're Functors.

To test our Functor mapping, we'll use crocks to provide us access to an algebraic data type called Maybe.

import { compose, option, prop } from 'crocks'

const company = {
  name: 'Pearce Software, LLC',
  locations: [
    'Charleston, SC, USA',
    'Auckland, NZ',
    'London, England, UK'
  ]
}

prop('foo', company)       // Nothing
prop('locations', company) // Just [String]

option([], prop('foo', company))
// []

option([], prop('locations', company))
// [
//   'Charleston, SC, USA',
//   'Auckland, NZ',
//   'London, England, UK'
// ]

const getLocations = compose(option([]), prop('locations'))
getLocations(company)
// [
//   'Charleston, SC, USA',
//   'Auckland, NZ',
//   'London, England, UK'
// ]

Pump the breaks! What's all this Just and Nothing stuff? We're not going to focus on Maybes today⁹, but the short version is that the locations property may or may not be present in the object, so we encapsulate that uncertainty inside of a Maybe algebraic data type via the prop function, and we provide a default value via the option function that the Maybe can fall back to in the event of not being able to find locations.

Why does this matter? We want to map a Maybe, and the prop function will give us access to one. Let's see what it looks like:

import { compose, option, prop } from 'crocks'

const upcase = x => x.toUpperCase()

const getLocations =
  compose(option([]), map(map(upcase)), prop('locations'))

getLocations({}) // []

getLocations(company)
// [
//   'CHARLESTON, SC, USA',
//   'AUCKLAND, NZ',
//   'LONDON, ENGLAND, UK'
// ]

Okay, cool! But why are we mapping twice?

When we work with algebraic data types like Maybe, instead of writing if (dataIsValid) doSomething, the map method on a Maybe gives us access to the value inside the Maybe (our locations), but it does so only if the data is available.

Once we have access to the locations, we then use map again to uppercase each location.

throwing Out Bad Data

What happens if the arguments passed to map aren't a Function and a Functor?

map(null)([1,2,3])    // TypeError: fn is not a function
map(x => x * 2)(null) // undefined
map(null)(null)       // undefined

I think we can provide some more helpful messaging to guide users of our map tool on how to use it correctly.

// map :: Functor f => (a -> b) -> f a -> f b
const map = fn => m => {
  if (!isFunction(fn)) {
    throw new TypeError(`map: Please provide a Function for the first argument`)
  }

  // ...our other handlers...

  throw new TypeError(`map: Please provide a Functor or Object for the second argument`)
}

map(null)([1,2,3])    // TypeError: map: Please provide a Function for the first argument
map(x => x * 2)(null) // TypeError: map: Please provide a Functor or Object for the second argument
map(null)(null)       // TypeError: map: Please provide a Function for the first argument

Now, when we provide bad arguments, we're told exactly what we need to do.

Wrapping Up

Congratulations and thank you for making it to the end! If you want to play around with what we created, check out this CodeSandbox: https://codesandbox.io/s/bitter-grass-tknwb.

Here is our code from today in its entirety:

const { compose, option, prop } = require('crocks')

// map :: Functor f => (a -> b) -> f a -> f b
const map = fn => m => {
  if (!isFunction(fn)) {
    throw new TypeError(`map: Please provide a Function for the first argument`)
  }

  if (isFunction(m)) {
    return compose(fn, m)
  }

  if (isArray(m)) {
    return mapArray(fn, m)
  }

  if (isFunctor(m)) {
    return mapFunctor(fn, m)
  }

  if (isObject(m)) {
    return mapObj(fn, m)
  }

  throw new TypeError(`map: Please provide a Functor or Object for the second argument`)
}

// we're opting for crocks' compose, instead
// compose :: ((b -> c), (a -> b)) -> a -> c
// const compose = (f, g) => x => f(g(x))

// isArray :: a -> Bool
const isArray = x => Array.isArray(x)

// isFunction :: a -> Bool
const isFunction = x => typeof x === 'function'

// isFunctor :: a -> Bool
const isFunctor  = x => !!x && isFunction(x['map'])

// isObject :: a -> Bool
const isObject = x =>
  !!x && Object.prototype.toString.call(x) === '[object Object]'

// mapArray :: ((a -> b), Array a) -> Array b
const mapArray = (fn, m) => {
  const newArray = []

  for (let i = 0; i < m.length; i++) {
    newArray.push(fn(m[i]))
  }

  return newArray
}
// realistically, you should use this mapArray:
// const mapArray = (fn, m) => m.map(x => fn(x))

// mapObj :: (a -> b) -> { k: a } -> { k: b }
const mapObj = (fn, m) => {
  const obj = {}

  for (const [k, v] of Object.entries(m)) {
    obj[k] = fn(v)
  }

  return obj
}

// mapFunctor :: Functor f => ((a -> b), f a) -> f b
const mapFunctor = (fn, m) => m.map(fn)

---

Thank you for reading!


Robert

---

1. https://github.com/hemanth/functional-programming-jargon#algebraic-data-type   ↩

2. m for Monoid   ↩

3. Wondering why the data comes last? Check out Brian Lonsdorf's "Hey Underscore, You're Doing It Wrong!" talk. The tl;dr is that you should arrange your arguments from least likely to change to most likely to change in order to pave the way for partial application and greater code reuse.   ↩

4. https://github.com/hemanth/functional-programming-jargon#arity   ↩

5. https://github.com/hemanth/functional-programming-jargon#functor   ↩

6. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/isArray   ↩

7. Check out ramda.js' addIndex function to see a different pattern for working with indices and Arrays.   ↩

8. Functor map is also known as fmap.   ↩

9. If you're an egghead.io subscriber, Andy Van Slaars has a great course, Safer JavaScript with the Maybe Type, or you can check out a Haskell article on The Functor class.   ↩

Comments

[Brent C]

Maybe I'm mistaken, but shouldn't the "partially applying" example,

const mapId = map(x => x.id)

be something like this instead?

const mapId = (m) => map(x => x.id, m)

[Robert Pearce]

Hi, Brent! Thanks for reading and taking the time to ask a question.

To answer your question: not quite! Check out this example:

const add  = (a, b) => a + b
const add2 = add.bind(null, 2)
add2(3) // 5
add2(5) // 7

In this example, we save the result of partially applying the first argument to add in a variable called add2 and state that it should be 2. Because we did not apply all of the arguments that add needs and did this partial application using the .bind method, we get back a function that we can reuse over and over again (add2).

Here's the same function, but it is written using a manually curried function:

const add  = a => b => a + b
const add3 = add(3)
add3(2) // 5
add3(4) // 7

In this example, add takes 2 arguments before it returns a result, but the arguments are curried. add(3) applies the argument 3 to add, but since add needs to be called twice (depending on how you've curried it...see the article below on currying), add(3) will return a function that expects to be called with the final argument. That's what we do on the last two lines; i.e., partial application allows us to store the result of calling add with 3 and then add any number we want to with 3 over and over and over.

The mapId function above could work like const mapId = (m) => map(x => x.id)(m), but since calling map(x => x.id) already returns a function, and we are merely forwarding on the argument to another function, we can do an eta reduction (see presentation below) and simplify the code. Check it out:

const mapId = m => map(x => x.id)(m)
const mapId =      map(x => x.id)

This presentation has a bit of stuff on this subject (specfically slides 36-43):
slideshare.net/robertwpearce/a-pat...

I've also written some posts on currying and composition that might be helpful, too:

• robertwpearce.com/ramda-chops-func...
• robertwpearce.com/ramda-chops-func...

[Brent C]

but since calling map(x => x.id) already returns a function

Yes, that example works if map is returning a function, but I guess my read at that point in the article was that map wasn't returning a function.

• it takes a function (fn) and then some datum (m2)3
• it returns the datum as transformed by said function

In which case the example that follows that presupposes it's already a curried implementation was confusing.

[Robert Pearce]

...aaaaand I feel like an idiot. I totally see it now! We're still working with (fn, m) => ... and not fn => m => ..., so that doesn't make sense yet. I'll update ASAP. Thanks for your patience, and nice 👀!

[Robert Pearce]

Quick fix was to change it to:

const mapId = map.bind(null, x => x.id)

Thanks again.

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