创建数字数组的功能方法

Question

How could I write the following code more functionally using ES6, without any 3rd party libraries?

// sample pager array
// * output up to 11 pages
// * the current page in the middle, if page > 5
// * don't include pager < 1 or pager > lastPage
// * Expected output using example:
//     [9,10,11,12,13,14,15,16,17,18,19]

const page = 14 // by example
const lastPage = 40 // by example
const pagerPages = page => {
  let newArray = []
  for (let i = page - 5; i <= page + 5; i++) {
    i >= 1 && i <= lastPage ? newArray.push(i) : null
  }
  return newArray
}

I would like to avoid Array.push, and possibly the for loop, but I'm not sure how I would achieve it in this situation.

Answer 1

  const pageRange = (lastPage, page) => ((start, end) => Array.from({length: end - start + 1}, (_,i) => i + start))(Math.max(1, page - 5), Math.min(lastPage, page + 5));
 const newArray = pageRange(40, 14);

This is a purely functional approach. It uses Math.max/min to achieve the boundaries and then uses an IIFE to pass these boundaries to Array.from which will create an array of end - start elements and every of these elements will be the position in the array increased by the startvalue.

---

PS: IMO your code is actually much more concise (except from that unneccessary ternary) and far more readable than mys, just saying...

Answer 2

Functional programming isn't limited to reduce , filter , and map ; it's about functions. This means we don't have to rely on perverse knowledge like Array.from ({ length: x }) where an object with a length property can be treated like an array. This kind of behavior is bewildering for beginners and mental overhead for anyone else. It think you'll enjoy writing programs that encode your intentions more clearly.

reduce starts with 1 or more values and reduces to (usually) a single value. In this case, you actually want the reverse of a reduce (or fold ), here called unfold . The difference is we start with a single value, and expand or unfold it into (usually) multiple values.

We start with a simplified example, alphabet . We begin unfolding with an initial value of 97 , the char code for the letter a . We stop unfolding when the char code exceeds 122 , the char code for the letter z .

 const unfold = (f, initState) => f ( (value, nextState) => [ value, ...unfold (f, nextState) ] , () => [] , initState ) const alphabet = () => unfold ( (next, done, char) => char > 122 ? done () : next ( String.fromCharCode (char) // value to add to output , char + 1 // next state ) , 97 // initial state ) console.log (alphabet ()) // [ a, b, c, ..., x, y, z ] 

Above, we use a single integer for our state, but other unfolds may require a more complex representation. Below, we show the classic Fibonacci sequence by unfolding a compound initial state of [ n, a, b ] where n is a decrementing counter, and a and b are numbers used to compute the sequence's terms. This demonstrates unfold can be used with any seed state, even arrays or objects.

const fib = (n = 0) =>
  unfold
    ( (next, done, [ n, a, b ]) =>
        n < 0
          ? done ()
          : next ( a                   // value to add to output
                 , [ n - 1, b, a + b ] // next state
                 )
    , [ n, 0, 1 ] // initial state
    )

console.log (fib (20))
// [ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765 ]

Now we have the confidence to write pagination . Again, our initial state is compound data [ page, count ] as we need to keep track of the page to add, and how many pages ( count ) we've already added.

Another advantage to this approach is that you can easily parameterize things like 10 or -5 or +1 and there's a sensible, semantic structure to place them in.

 const unfold = (f, initState) => f ( (value, nextState) => [ value, ...unfold (f, nextState) ] , () => [] , initState ) const pagination = (totalPages, currentPage = 1) => unfold ( (next, done, [ page, count ]) => page > totalPages ? done () : count > 10 ? done () : next (page, [ page + 1, count + 1 ]) , [ Math.max (1, currentPage - 5), 0 ] ) console.log (pagination (40, 1)) // [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ] console.log (pagination (40, 14)) // [ 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19 ] console.log (pagination (40, 38)) // [ 33, 34, 35, 36, 37, 38, 39, 40 ] console.log (pagination (40, 40)) // [ 35, 36, 37, 38, 39, 40 ] 

Above, there are two conditions which result in a call to done () . We can collapse these using || and the code reads a little nicer

const pagination = (totalPages, currentPage = 1) =>
  unfold
    ( (next, done, [ page, count ]) =>
        page > totalPages || count > 10
          ? done ()
          : next (page, [ page + 1, count + 1 ])
    , [ Math.max (1, currentPage - 5), 0 ]
    )

Answer 3

There are many ways to create an array functionally but creating an array depending on some correlated items like a math series is mostly done by unfolding. You may consider unfolding like the inverse of reducing. Your case do not necessarily require unfolding but just for the sake of proper functional programming lets see how it can be done.

JS do not have a native unfolding function but we may simply implement it. First of all what does unfold function look like..?

Array.unfold = function(p,f,t,v){
  var res = [],
   runner = d =>  p(d,res.length,res) ? [] : (res.push(f(d)),runner(t(d)), res);
  return runner(v);
};

As seen it takes 4 arguments.

1. p : This is a callback function just like we have in reduce. It gets invoked with the current seed element e to be processed before insertion, it's index i to be inserted and the currently available array a like p(e,i,a) . When it returns a true , the unfolding operation concludes and returns the created array.
2. f : Is the function that we will apply for each item to be constructed. It takes a single argument which is the current iterating value. You may consider iterating value like the index value but we have control over how to iterate it.
3. t : Is the function that we will apply the iterating value and get the next iterating value. For index like iterations this should be x => x+1 .
4. v : Is the glorious initial value.

So far so good. How are we going to use unfold to achieve this job. First of all lets find our initial value by a function which takes page as an argument.

var v = (pg => pg - 5 > 0 ? pg - 5 : 1)(page)

How about the p function which decides where to stop?

var p = (_,i) => i > 10

We will increase pages one by one but if we have a value greater than lastpage we need to feed null values instead. So f may look like

var f = (lp => v => v > lp ? null : v)(lastpage)

and finally t is the function how we increase the iterating value. It's x => x + 1 .

 Array.unfold = function(p,f,t,v){ var res = [], runner = d => p(d,res.length,res) ? [] : (res.push(f(d)),runner(t(d)), res); return runner(v); }; var v = pg => pg - 5 > 0 ? pg - 5 : 1, p = (_,i) => i > 10, f = lp => v => v > lp ? null : v, t = x => x + 1, a = Array.unfold(p,f(40),t,v(14)); console.log(a);