Reduce array of arrays and remove null from Javascript data

Question

I am having difficulty "compressing" the results of my grammar by stripping out null values and embedded arrays using Javascript. I am using the Nearley grammar checker which can run JS functions after a sentence is matched. Unfortunately you get the results of the full parse as a series of arrays. The following is an example of the output

[
   [
      [
         [
            [
               [
                  [ [ 'climb'], [ [ null, 'to' ] ] ],
                  [ [ null, [ 'alt' ] ] ],
                  [ 332, [ null, [ 'km' ] ] ]
               ]
            ],
            [ null ]
         ]
      ]
   ]
]

I would like to remove all those null values and possibly "compress" all those arrays to something of the form:

[
   [ 'climb', 'to' ],
   [ 'alt', 332, 'km' ]
]

Or something similar. I have tried using various filter methods without success.

what is happening with 'alt'? why not only a sinlge array? what have you tried? — Nina Scholz
– Nina Scholz, Commented Nov 30, 2020 at 18:18
Sorry. Forgot about 'alt'. Single array is great too just provided an option. Added alt — NastyDiaper
– NastyDiaper, Commented Nov 30, 2020 at 18:21
Sure, whatever you'd like to produce. If I can simplify it, I can more easily put it in a JSON structure, which is my ultimate goal. climb to may be considered verbs and alt 332 km may be considered the target or subject — NastyDiaper
– NastyDiaper, Commented Nov 30, 2020 at 18:25
From what was presented until now only [[ 'climb', 'to' ], [ 'alt' ], [ 332, 'km' ]] was the most logic result for a generically working approach. The edited output of [[ 'climb', 'to' ], [ 'alt', 332, 'km' ]] does not make any sense at all. — Peter Seliger
– Peter Seliger, Commented Nov 30, 2020 at 18:32

Nina Scholz · Accepted Answer · 2020-11-30 19:46:17Z

2

You could filter an map with the same approach and flat all arrays.

const
    filter = array => array.flatMap(v => Array.isArray(v)
        ? filter(v)
        : v === null
            ? []
            : v
    ),
    array = [[[[[[[['climb'], [[null, 'to']]], [[null, ['alt']]], [332, [null, ['km']]]]], [null]]]]],
    result = filter(array);

console.log(result);

An even shorter approach

const
    filter = array => array
        .flat(Infinity)
        .filter(v => v !== null),
    array = [[[[[[[['climb'], [[null, 'to']]], [[null, ['alt']]], [332, [null, ['km']]]]], [null]]]]],
    result = filter(array);

console.log(result);

edited Nov 30, 2020 at 19:46

answered Nov 30, 2020 at 18:57

Nina Scholz

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Comments

Peter Seliger · Accepted Answer · 2020-11-30 19:42:13Z

2

Really?.. how about ['climb', 'to', 'alt', 332, 'km'] then? – Peter Seliger

That's wonderful. Some sort of logical grouping would be preferable but hey if not...

...

const sample = [[[[
  [[
    [ [ 'climb'], [ [ null, 'to' ] ] ],
    [ [ null, [ 'alt' ] ] ],
    [ 332, [ null, [ 'km' ] ] ]
  ]],
  [ null ]
]]]];

function flatOut(list, item) {
  if (Array.isArray(item)) {
    item = item.reduce(flatOut, []);
  }
  return list.concat(item);
}
const result = sample
  .reduce(flatOut, [])
  .filter(elm => (elm != null)); // non strict equality
  //...in order to skip both values, undefined and null.

console.log('result :', result);

edited Nov 30, 2020 at 19:42

answered Nov 30, 2020 at 18:58

Peter Seliger

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Comments

54ka · Accepted Answer · 2020-11-30 20:15:41Z

2

One line solution:

1.Convert to string 2.Split to flat array 3. Remove empty values

Example:

var arr = [
    [
        [
            [
                [
                    [
                        [['climb'], [[null, 'to']]],
                        [[null, ['alt']]],
                        [332, [null, ['km']]]
                    ]
                ],
                [null]
            ]
        ]
    ]
];



var res = arr.toString().split(",").filter(item => item);


console.log(res);

edited Nov 30, 2020 at 20:15

answered Nov 30, 2020 at 19:21

54ka

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1 Comment

Peter Seliger Over a year ago

... for the OP's simple grammar check which deals with string, number and null values only, this is the most elegant approach.

Sunil Kumar · Accepted Answer · 2020-11-30 20:17:02Z

let resArr = [];
  const findLoc = (arr, loc = []) => {
    arr.map((d, i) => {
      if (Array.isArray(d)) {
        findLoc(d, [ ...loc, i ]);
      } else {
        if (d !== null) {
          // console.log(d, loc);
          resArr.push([...loc, d])
        }
      }
    })
}

  const a = [
     [
        [
           [
              [
                 [
                    [ [ 'climb'], [ [ null, 'to' ] ] ],
                    [ [ null, [ 'alt' ] ] ],
                    [ 332, [ null, [ 'km' ] ] ],
                    [ 56, [ null, null, [ [ [8] ] ] ] ]
                 ]
              ],
              [ null ]
           ]
        ]
     ]
  ];
  findLoc(a);
  let finalIndex = [...resArr.reverse()[0]];
  finalIndex.splice(finalIndex.length -1 , 1);
  finalIndex = resArr[0].indexOf(Math.max(...finalIndex));
  const finalObj = {};
  resArr.forEach((d) => {
    finalObj[d[finalIndex]] = finalObj[d[finalIndex]] ? [...finalObj[d[finalIndex]], d[d.length -1]] : [d[d.length -1]]
  });
  console.log(Object.values(finalObj));
  
  // [ [ "to", "climb" ], [ "alt" ], [ "km", 332 ], [ 8, 56 ] ]

change the input and test,

using matrix logic we can active this

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