snorfalorpagus's solution to Accumulate on the TypeScript track

Instructions

Test suite

Solution

Implement the accumulate operation, which, given a collection and an operation to perform on each element of the collection, returns a new collection containing the result of applying that operation to each element of the input collection.

Given the collection of numbers:

1, 2, 3, 4, 5

And the operation:

square a number (x => x * x)

Your code should be able to produce the collection of squares:

1, 4, 9, 16, 25

Check out the test suite to see the expected function signature.

Restrictions

Keep your hands off that collect/map/fmap/whatchamacallit functionality provided by your standard library! Solve this one yourself using other basic tools instead.

Setup

Go through the setup instructions for TypeScript to install the necessary dependencies:

https://exercism.io/tracks/typescript/installation

Requirements

Install assignment dependencies:

$ yarn install

Making the test suite pass

Execute the tests with:

$ yarn test

In the test suites all tests but the first have been skipped.

Once you get a test passing, you can enable the next one by changing xit to it.

Source

Conversation with James Edward Gray II https://twitter.com/jeg2

Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

accumulate.test.ts

import accumulate from './accumulate'

describe('accumulate()', () => {
  it('accumulation empty', () => {
    const accumulator = (e: number): number => e * e
    expect(accumulate([], accumulator)).toEqual([])
  })

  it('accumulate squares', () => {
    const accumulator = (n: number): number => n * n
    const result = accumulate([1, 2, 3], accumulator)
    expect(result).toEqual([1, 4, 9])
  })

  it('accumulate upcases', () => {
    const accumulator = (word: string): string => word.toUpperCase()
    const result = accumulate('hello world'.split(/\s/), accumulator)
    expect(result).toEqual(['HELLO', 'WORLD'])
  })

  it('accumulate reversed strings', () => {
    const accumulator = (word: string): string => word.split('').reverse().join('')
    const result = accumulate('the quick brown fox etc'.split(/\s/), accumulator)
    expect(result).toEqual(['eht', 'kciuq', 'nworb', 'xof', 'cte'])
  })

  it('accumulate recursively', () => {
    const result = accumulate('a b c'.split(/\s/), (char: string) => accumulate('1 2 3'.split(/\s/), (digit: string) => char + digit))

    expect(result).toEqual([['a1', 'a2', 'a3'], ['b1', 'b2', 'b3'], ['c1', 'c2', 'c3']])
  })
})

function accumulate<T, U>(arr: Array<T>, f: (v: T) => U): Array<U> {
    return arr.map((v: T, _: number) => f(v));
}

export default accumulate;

What can you learn from this solution?

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Here are some questions to help you reflect on this solution and learn the most from it.

What compromises have been made?
Are there new concepts here that you could read more about to improve your understanding?

snorfalorpagus's solution

to Accumulate in the TypeScript Track