Find if two arrays are repeated in array and then

问题:

I have multiple arrays in a main/parent array like this:

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];

here are the array's for simpler reading:

[1, 17]
[1, 17]
[1, 17]
[2, 12]
[5, 9]
[2, 12]
[6, 2]
[2, 12]
[2, 12]

I want to select the arrays that are repeated 3 or more times (> 3) and assign it to a variable. So in this example, var repeatedArrays would be [1, 17] and [2, 12].

So this should be the final result:

console.log(repeatedArrays);
>>> [[1, 17], [2, 12]]

I found something similar here but it uses underscore.js and lodash.

How could I it with javascript or even jquery (if need be)?

回答1:

You could take a Map with stringified arrays and count, then filter by count and restore the arrays.

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
    result = Array
        .from(array.reduce(
            (map, array) =>
                (json => map.set(json, (map.get(json) || 0) + 1))
                (JSON.stringify(array)),
            new Map
         ))
        .filter(([, count]) => count > 2)
        .map(([json]) => JSON.parse(json));
        
console.log(result);

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Filter with a map at wanted count.

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
    result = array.filter(
        (map => a => 
            (json =>
                (count => map.set(json, count) && !(2 - count))
                (1 + map.get(json) || 1)
            )
            (JSON.stringify(a))
        )
        (new Map)
    );
        
console.log(result);

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Unique!

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]],
    result = array.filter(
        (s => a => (j => !s.has(j) && s.add(j))(JSON.stringify(a)))
        (new Set)
    );
        
console.log(result);

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回答2:

Try this

array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))

Time complexity O(n) (one array pass by filter function). Inspired by Nitish answer.

Explanation

The (r={}, a=>...) will return last expression after comma (which is a=>...) (e.g. (5,6)==6). In r={} we set once temporary object where we will store unique keys. In filter function a=>... in a we have current array element . In r[a] JS implicity cast a to string (e.g 1,17). Then in !(2-(r[a]=++r[a]|0)) we increase counter of occurrence element a and return true (as filter function value) if element a occurred 3 times. If r[a] is undefined the ++r[a] returns NaN, and further NaN|0=0 (also number|0=number). The r[a]= initialise first counter value, if we omit it the ++ will only set NaN to r[a] which is non-incrementable (so we need to put zero at init). If we remove 2- as result we get input array without duplicates - or alternatively we can also get this by a=>!(r[a]=a in r). If we change 2- to 1- we get array with duplicates only.

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];

var r= array.filter(( r={}, a=>!(2-(r[a]=++r[a]|0)) ))

console.log(JSON.stringify(r));

回答3:

You can use Object.reduce, Object.entries for this like below

var array = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];


let res = Object.entries(
            array.reduce((o, d) => {
              let key = d.join('-')
              o[key] = (o[key] || 0) + 1

              return o
          }, {}))
          .flatMap(([k, v]) => v > 2 ? [k.split('-').map(Number)] : [])
  
  
console.log(res)

OR may be just with Array.filters

var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];

let temp = {}
let res = array.filter(d => {
  let key = d.join('-')
  temp[key] = (temp[key] || 0) + 1
  
  return temp[key] == 3
})

console.log(res)

回答4:

For a different take, you can first sort your list, then loop through once and pull out the elements that meet your requirement. This will probably be faster than stringifying keys from the array even with the sort:

var arr = [[1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]]
arr.sort((a, b) => a[0] - b[0] || a[1] - b[1])

// define equal for array
const equal = (arr1, arr2) => arr1.every((n, j) => n === arr2[j])

let GROUP_SIZE = 3
first = 0, last = 1, res = []

while(last < arr.length){
    if (equal(arr[first], arr[last])) last++
    else {
        if (last - first >= GROUP_SIZE)  res.push(arr[first])
        first = last
    }
}
if (last - first >= GROUP_SIZE)  res.push(arr[first])
console.log(res)

回答5:

ES6:

const repeatMap = {}

array.forEach(arr => {
  const key = JSON.stringify(arr)
  if (repeatMap[key]) {
    repeatMap[key]++
  } else {
    repeatMap[key] = 1
  }
})

const repeatedArrays = Object.keys(repeatMap)
  .filter(key => repeatMap[key] >= 3)
  .map(key => JSON.parse(key))

回答6:

You could also do this with a single Array.reduce where you would only push to a result property if the length is equal to 3:

var array = [[1, 17], [1, 17], [1, 17], [1, 17], [2, 12], [5, 9], [2, 12], [6, 2], [2, 12]];

console.log(array.reduce((r,c) => {
  let key = c.join('-')
  r[key] = (r[key] || 0) + 1
  r[key] == 3 ? r.result.push(c) : 0  // if we have a hit push to result
  return r
}, { result: []}).result)             // print the result property