How to count property different values in json object

Question

I am getting a json objects from database. What is the Best practice to count number of ClassDef values A , B , C.
In the json object for eg: I have in this json " 2 A and 2 B and 3 c "

My json object:

[{
    "ItemCode": "200500303800356",
    "LastCost": 8,
    "OnHand": 593889,
    "InventoryValue": 4751112,
    "PercOfTotalInventory": 18.294517,
    "CumerativePercTotal": 18.29,
    "ClassDef": "A"
}, {
    "ItemCode": "201600701800197",
    "LastCost": 400,
    "OnHand": 300,
    "InventoryValue": 120000,
    "PercOfTotalInventory": 0.462069,
    "CumerativePercTotal": 75.68,
    "ClassDef": "A"
}, {
    "ItemCode": "200701507000107",
    "LastCost": 75,
    "OnHand": 239,
    "InventoryValue": 17925,
    "PercOfTotalInventory": 0.069022,
    "CumerativePercTotal": 91.75,
    "ClassDef": "B"
}, {
    "ItemCode": "200501303400308",
    "LastCost": 3515,
    "OnHand": 5,
    "InventoryValue": 17575,
    "PercOfTotalInventory": 0.067674,
    "CumerativePercTotal": 91.81,
    "ClassDef": "B"
}, {
    "ItemCode": "200200106701035",
    "LastCost": 80,
    "OnHand": 27,
    "InventoryValue": 2160,
    "PercOfTotalInventory": 0.008317,
    "CumerativePercTotal": 99.28,
    "ClassDef": "C"
}, {
    "ItemCode": "200200902700248",
    "LastCost": 10,
    "OnHand": 213,
    "InventoryValue": 2130,
    "PercOfTotalInventory": 0.008202,
    "CumerativePercTotal": 99.29,
    "ClassDef": "C"
}, {
    "ItemCode": "200601302001093",
    "LastCost": 0.3,
    "OnHand": 6,
    "InventoryValue": 1.8,
    "PercOfTotalInventory": 0.000007,
    "CumerativePercTotal": 100,
    "ClassDef": "C"
}]

Does any have an idea how to do this?

which value do you loke to count? just how many occurences or a special value? — Nina Scholz
– Nina Scholz, Commented May 18, 2017 at 8:34
Not sure to well understand your demand. You want to count the number of ClassDef in your whole array ? Or the number of value that contains ClassDef in a single object ? — Zooly
– Zooly, Commented May 18, 2017 at 8:35

Nina Scholz · Accepted Answer · 2017-05-18 08:40:55Z

You could use an object for the result and use ClassDef as key.

var data = [{ ItemCode: "200500303800356", LastCost: 8, OnHand: 593889, InventoryValue: 4751112, PercOfTotalInventory: 18.294517, CumerativePercTotal: 18.29, ClassDef: "A" }, { ItemCode: "201600701800197", LastCost: 400, OnHand: 300, InventoryValue: 120000, PercOfTotalInventory: 0.462069, CumerativePercTotal: 75.68, ClassDef: "A" }, { ItemCode: "200701507000107", LastCost: 75, OnHand: 239, InventoryValue: 17925, PercOfTotalInventory: 0.069022, CumerativePercTotal: 91.75, ClassDef: "B" }, { ItemCode: "200501303400308", LastCost: 3515, OnHand: 5, InventoryValue: 17575, PercOfTotalInventory: 0.067674, CumerativePercTotal: 91.81, ClassDef: "B" }, { ItemCode: "200200106701035", LastCost: 80, OnHand: 27, InventoryValue: 2160, PercOfTotalInventory: 0.008317, CumerativePercTotal: 99.28, ClassDef: "C" }, { ItemCode: "200200902700248", LastCost: 10, OnHand: 213, InventoryValue: 2130, PercOfTotalInventory: 0.008202, CumerativePercTotal: 99.29, ClassDef: "C" }, { ItemCode: "200601302001093", LastCost: 0.3, OnHand: 6, InventoryValue: 1.8, PercOfTotalInventory: 0.000007, CumerativePercTotal: 100, ClassDef: "C" }],
    result = data.reduce(function (r, o) {
        r[o.ClassDef] = (r[o.ClassDef] || 0) + 1;
        return r;
    }, Object.create(null));

console.log(result);

Aron · Accepted Answer · 2017-05-18 08:40:11Z

1

Through some clever use of .reduce() and Set this is a piece of cake.

const objArray = [{
    "ItemCode": "200500303800356",
    "LastCost": 8,
    "OnHand": 593889,
    "InventoryValue": 4751112,
    "PercOfTotalInventory": 18.294517,
    "CumerativePercTotal": 18.29,
    "ClassDef": "A"
}, ...];

const classDefSet = objArray.reduce((set, obj) => {
    set.add(obj.ClassDef); // add value to the set. If it already exists nothing will happen
    return set;
}, new Set());

const count = [...classDefSet].length; // convert set to plain array
console.log(count); // 3

answered May 18, 2017 at 8:40

Aron

9,3187 gold badges37 silver badges60 bronze badges

Comments

barbsan · Accepted Answer · 2017-05-18 08:45:12Z

Yet another reduce example:

let arr = [{
"ItemCode": "200500303800356",
"LastCost": 8,
"OnHand": 593889,
"InventoryValue": 4751112,
"PercOfTotalInventory": 18.294517,
"CumerativePercTotal": 18.29,
"ClassDef": "A"
}, {
"ItemCode": "201600701800197",
"LastCost": 400,
"OnHand": 300,
"InventoryValue": 120000,
"PercOfTotalInventory": 0.462069,
"CumerativePercTotal": 75.68,
"ClassDef": "A"
 }, {
"ItemCode": "200701507000107",
"LastCost": 75,
"OnHand": 239,
"InventoryValue": 17925,
"PercOfTotalInventory": 0.069022,
"CumerativePercTotal": 91.75,
"ClassDef": "B"
 }, {
"ItemCode": "200501303400308",
"LastCost": 3515,
"OnHand": 5,
"InventoryValue": 17575,
"PercOfTotalInventory": 0.067674,
"CumerativePercTotal": 91.81,
"ClassDef": "B"
 },  {
"ItemCode": "200200106701035",
"LastCost": 80,
"OnHand": 27,
"InventoryValue": 2160,
"PercOfTotalInventory": 0.008317,
"CumerativePercTotal": 99.28,
"ClassDef": "C"
 }, {
"ItemCode": "200200902700248",
"LastCost": 10,
"OnHand": 213,
"InventoryValue": 2130,
"PercOfTotalInventory": 0.008202,
"CumerativePercTotal": 99.29,
"ClassDef": "C"
 },  {
"ItemCode": "200601302001093",
"LastCost": 0.3,
"OnHand": 6,
"InventoryValue": 1.8,
"PercOfTotalInventory": 0.000007,
"CumerativePercTotal": 100,
"ClassDef": "C"
}];

let result = arr.reduce((acc, el)=>{
  if (el.hasOwnProperty("ClassDef")){
    if (!acc.hasOwnProperty(el.ClassDef)){
      acc[el.ClassDef] = 0;
    }
    acc[el.ClassDef]++;
  }
  return acc;
}, {});

console.log(result);

Stephane Janicaud · Accepted Answer · 2017-05-18 09:03:46Z

Try this to find how many times

ClassDef equals C
ClassDef equals C or B

var arr=[{ItemCode:"200500303800356",LastCost:8,OnHand:593889,InventoryValue:4751112,PercOfTotalInventory:18.294517,CumerativePercTotal:18.29,ClassDef:"A"},{ItemCode:"201600701800197",LastCost:400,OnHand:300,InventoryValue:12e4,PercOfTotalInventory:.462069,CumerativePercTotal:75.68,ClassDef:"A"},{ItemCode:"200701507000107",LastCost:75,OnHand:239,InventoryValue:17925,PercOfTotalInventory:.069022,CumerativePercTotal:91.75,ClassDef:"B"},{ItemCode:"200501303400308",LastCost:3515,OnHand:5,InventoryValue:17575,PercOfTotalInventory:.067674,CumerativePercTotal:91.81,ClassDef:"B"},{ItemCode:"200200106701035",LastCost:80,OnHand:27,InventoryValue:2160,PercOfTotalInventory:.008317,CumerativePercTotal:99.28,ClassDef:"C"},{ItemCode:"200200902700248",LastCost:10,OnHand:213,InventoryValue:2130,PercOfTotalInventory:.008202,CumerativePercTotal:99.29,ClassDef:"C"},{ItemCode:"200601302001093",LastCost:.3,OnHand:6,InventoryValue:1.8,PercOfTotalInventory:7e-6,CumerativePercTotal:100,ClassDef:"C"}];

function countOccurences(name, value) {
  var count = 0;
  // If name and values are defined
  if (name && value) {
    // Check either a single value or an array of values
    var values = (value instanceof Array ? value : [value])
    // For each object in array
    arr.forEach(function(v) {
      // If values array contains object[name] value
      if (v[name] && values.indexOf(v[name]) !== -1) {
        // Increment counter
        count++;
      }
    });
  }
  return count;
}

console.log("ClassDef === 'C' :", countOccurences("ClassDef", "C"));
console.log("ClassDef === 'C' or 'B' :", countOccurences("ClassDef", ["C", "B"]));

Community · Accepted Answer · 2017-05-23 11:47:20Z

0

This may useful for you. you can count the values assigned to the key "ClassDef" as show here the count of the duplicate value assigned to the same key in JSON array

edited May 23, 2017 at 11:47

CommunityBot

11 silver badge

answered May 18, 2017 at 8:43

Ramana V V K

1,24915 silver badges25 bronze badges

Comments

Mubi · Accepted Answer · 2017-05-18 09:11:55Z

0

Using your sample json object, try this :

var countArr = [];
for(i = 0; i < obj.length; i++)
{
    var val =  obj[i]["ClassDef"];
    if(countArr[val] != null)
    {
        countArr[val] += 1;
    }
    else
    {
        countArr[val] = 1;
    }

}

this will return you

[A: 2, B: 2, C: 3]

answered May 18, 2017 at 9:11

Mubi

541 silver badge8 bronze badges

Collectives™ on Stack Overflow

How to count property different values in json object

6 Answers 6

Comments

Comments

Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

Comments

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related