Javascript count number of occurrence

Question

I have the following array data and i am calculating the number of records that match using the includes criteria.

var final = 
[
  [
    "61131",
    "NISSAN",
    "BOLTON",
    "MAINT"
  ],
  [
    "61132",
    "NISSAN",
    "BOLTON",
    "MAINT"
  ],
  [
    "61133",
    "TOYOTA",
    "STOCKPORT",
    "STORED"
  ],
  [
    "61134",
    "TOYOTA",
    "STOCKPORT",
    "MAINT"
  ],
  [
    "61135",
    "NISSAN",
    "STOCKPORT",
    "MAINT"
  ],
  [
    "61136",
    "NISSAN",
    null,
    null
  ]
]

and the code is this one :

var obj = {};
var num1 = 0;
var num2 = 0;
for (var i=0; i<final.length; i++){
    if(final[i].join(';').includes('BOLTON') && final[i].join(';').includes('MAINT') && final[i].join(';').includes('NISSAN')) {
        num1++;
    }
}

for (var i=0; i<final.length; i++){
    if(final[i].join(';').includes('STOCKPORT') && final[i].join(';').includes('STORED') && final[i].join(';').includes('TOYOTA')) {
        num2++;
    }
}


obj['BOLTON_MAINT_NISSAN'] = num1
obj['STOCKPORT_STORED_TOYOTA'] = num2

console.log(obj)

output

  {  "BOLTON_MAINT_NISSAN": 2, "STOCKPORT_STORED_TOYOTA": 1}

I am getting the desired result, is there a more efficient way of writing the above code that is minimal ?

definitely not the most efficient. consider that final[i].join(';') always returns the same thing, but it's being run six (6) times per element in your array. — Mulan
– Mulan, Commented Oct 21, 2020 at 13:18
Consider using reduce developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — Mike Cheel
– Mike Cheel, Commented Oct 21, 2020 at 13:19
Well the code is very inefficient in the fact you are joining on every check. — epascarello
– epascarello, Commented Oct 21, 2020 at 13:19
it looks like the arrays have positional data. ie, part number always appears in the first position, and maker Nissan or Toyota always appears in the second position, etc. You probably don't need to use .join at all — Mulan
– Mulan, Commented Oct 21, 2020 at 13:20

iAmOren · Accepted Answer · 2020-10-23 03:08:49Z

3

const count = (list, keys) => {
  return keys.reduce(
    (obj, key) => ({
      ...obj,
      [key]: list.filter(sublist =>
        key.split('_').every(keyPart => sublist.includes(keyPart))
      ).length
    }),
    {}
  );
};

count(final, ['BOLTON_MAINT_NISSAN', 'STOCKPORT_STORED_TOYOTA'])

const final = [
  [ "61131", "NISSAN", "BOLTON", "MAINT" ],
  [ "61132", "NISSAN", "BOLTON", "MAINT" ],
  [ "61133", "TOYOTA", "STOCKPORT", "STORED" ],
  [ "61134", "TOYOTA", "STOCKPORT", "MAINT" ],
  [ "61135", "NISSAN", "STOCKPORT", "MAINT" ],
  [ "61136", "NISSAN", null, null ]
];

const count = (list, keys) => {
  return keys.reduce(
    (obj, key) => ({
      ...obj,
      [key]: list.filter(sublist =>
        key.split('_').every(keyPart => sublist.includes(keyPart))
      ).length
    }),
    {}
  );
};

var bmtNsst = count(final, ['BOLTON_MAINT_NISSAN', 'STOCKPORT_STORED_TOYOTA']);

console.log(bmtNsst);

edited Oct 23, 2020 at 3:08

iAmOren

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answered Oct 21, 2020 at 13:26

peter554

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Comments

epascarello · Accepted Answer · 2020-10-21 13:28:29Z

3

Using reduce, slice to get rid of the number, filter to get rid of the nulls, and join to make it the key.

var final = [
  ["61131", "NISSAN", "BOLTON", "MAINT"],
  ["61132", "NISSAN", "BOLTON", "MAINT"],
  ["61133", "TOYOTA", "STOCKPORT", "STORED"],
  ["61134", "TOYOTA", "STOCKPORT", "MAINT"],
  ["61135", "NISSAN", "STOCKPORT", "MAINT"],
  ["61136", "NISSAN", null, null]
];

var grouped = final.reduce(function(obj, data) {
  var key = data.slice(1).filter(Boolean).join("_");
  obj[key] = (obj[key] || 0) + 1;
  return obj;
}, {});

console.log(grouped)

answered Oct 21, 2020 at 13:28

epascarello

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Comments

Derek Wang · Accepted Answer · 2020-10-21 13:24:45Z

There is no need to use String.join. Array has also got function Array.prototype.includes to check if item is existed in array or not.

Using that, you can find the subArray that satisfies your condition. And using Array.prototype.filter, you can extract the sub-arrays to satisfy the conditions as follows.

const final = [
  [ "61131", "NISSAN", "BOLTON", "MAINT" ],
  [ "61132", "NISSAN", "BOLTON", "MAINT" ],
  [ "61133", "TOYOTA", "STOCKPORT", "STORED" ],
  [ "61134", "TOYOTA", "STOCKPORT", "MAINT" ],
  [ "61135", "NISSAN", "STOCKPORT", "MAINT" ],
  [ "61136", "NISSAN", null, null ]
];

const num1Arr = final.filter((item) => item.includes('BOLTON') && item.includes('MAINT') && item.includes('NISSAN'));

const num2Arr = final.filter((item) => item.includes('STOCKPORT') && item.includes('STORED') && item.includes('TOYOTA'));

const output = {
  BOLTON_MAINT_NISSAN: num1Arr.length,
  STOCKPORT_STORED_TOYOTA: num2Arr.length
};

console.log(output);

Or simply, you can do it using Array.prototype.reduce.

const final = [
  [ "61131", "NISSAN", "BOLTON", "MAINT" ],
  [ "61132", "NISSAN", "BOLTON", "MAINT" ],
  [ "61133", "TOYOTA", "STOCKPORT", "STORED" ],
  [ "61134", "TOYOTA", "STOCKPORT", "MAINT" ],
  [ "61135", "NISSAN", "STOCKPORT", "MAINT" ],
  [ "61136", "NISSAN", null, null ]
];

const output = final.reduce((acc, cur) => {
  if (cur.includes('BOLTON') && cur.includes('MAINT') && cur.includes('NISSAN')) {
    acc['BOLTON_MAINT_NISSAN'] ++;
  }
  
  if (cur.includes('STOCKPORT') && cur.includes('STORED') && cur.includes('TOYOTA')) {
    acc['STOCKPORT_STORED_TOYOTA'] ++;
  }
  
  return acc;
}, {
  BOLTON_MAINT_NISSAN: 0,
  STOCKPORT_STORED_TOYOTA: 0
});

console.log(output);

Obzi · Accepted Answer · 2020-10-21 13:25:37Z

2

const final = 
[
  [
    "61131",
    "NISSAN",
    "BOLTON",
    "MAINT"
  ],
  [
    "61132",
    "NISSAN",
    "BOLTON",
    "MAINT"
  ],
  [
    "61133",
    "TOYOTA",
    "STOCKPORT",
    "STORED"
  ],
  [
    "61134",
    "TOYOTA",
    "STOCKPORT",
    "MAINT"
  ],
  [
    "61135",
    "NISSAN",
    "STOCKPORT",
    "MAINT"
  ],
  [
    "61136",
    "NISSAN",
    null,
    null
  ]
];

console.log({
  BOLTON_MAINT_NISSAN: final.filter((value) => {
    return value[1] === 'NISSAN' && value[2] === 'BOLTON' && value[3] === 'MAINT';
  }).length,
  STOCKPORT_STORED_TOYOTA: final.filter((value) => {
    return value[1] === 'TOYOTA' && value[2] === 'STOCKPORT' && value[3] === 'STORED';
  }).length,
});

answered Oct 21, 2020 at 13:25

Obzi

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Comments

iAmOren · Accepted Answer · 2020-10-21 13:33:56Z

2

Using Array.prototype.reduce():

const final = [
  [ "61131", "NISSAN", "BOLTON", "MAINT" ],
  [ "61132", "NISSAN", "BOLTON", "MAINT" ],
  [ "61133", "TOYOTA", "STOCKPORT", "STORED" ],
  [ "61134", "TOYOTA", "STOCKPORT", "MAINT" ],
  [ "61135", "NISSAN", "STOCKPORT", "MAINT" ],
  [ "61136", "NISSAN", null, null ]
];

let occurrence = final.reduce((obj, item)=>{
  var key=item[2]+"_"+item[3]+"_"+item[1];
  if(!obj[key]) obj[key]=0;
  obj[key]++;
  return obj;
},{});

console.log(occurrence);

answered Oct 21, 2020 at 13:33

iAmOren

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Comments

Lorenzo Felletti · Accepted Answer · 2020-10-21 13:35:11Z

In your solution, you iterate twice through the array, which is unnecessary as if an object is one of the two you're searching, then it is not the other, so you can do all in one loop.

var final = [
  ["61131", "NISSAN", "BOLTON", "MAINT"],
  ["61132", "NISSAN", "BOLTON", "MAINT"],
  ["61133", "TOYOTA", "STOCKPORT", "STORED"],
  ["61134", "TOYOTA", "STOCKPORT", "MAINT"],
  ["61135", "NISSAN", "STOCKPORT", "MAINT"],
  ["61136", "NISSAN", null, null]
];

var obj = {
  BOLTON_MAINT_NISSAN: 0,
  STOCKPORT_STORED_TOYOTA: 0
};

final.forEach(y => {
  let x = y.join(';');
  if(x.includes('BOLTON') &&
     x.includes('MAINT') &&
     x.includes('NISSAN'))
    obj.BOLTON_MAINT_NISSAN++;
  else if (
    x.includes('STOCKPORT') &&
    x.includes('STORED') &&
    x.includes('TOYOTA')
  )
    obj.STOCKPORT_STORED_TOYOTA++;
})


console.log(obj)

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