JavaScript - find unique values in array of objects, with count and create a new array of objects

Question

I have an array of objects as below:

var arr =[
 {
  price:20,
  rule:a
 },
 {
  price:10,
  rule:b
 },
 {
  price:5,
  rule:a
 },
 {
  price:50,
  rule:b
 },
 {
  price:70,
  rule:b
 }
]

I want to extract an array of objects out of this as below:

var filteredArr = [
 {
  rule:a,
  countOfRule:2,
  minPriceForThisRule:5
 },
 {
  rule:b,
  countOfRule:3,
  minPriceForThisRule:10
 }
]

This means:

1) I want to create new array with no. of objects as unique no. of rules in first array "arr"

2) Need to count the unique rule repetition and add as property in new array objects - given as "countOfRule"

3) Find the minimum price for in a category of unique rule - given as "minPriceForThisRule"

I have read similar answers on SO and was able to get first 2 conditions only, and that too were not in the format as i need.

What I tried, referring to links on SO:

var ff = {},e;
for (var i = 0,l=arr.length; i < l; i++) {
   e = arr[i];
   ff[e.rule] = (ff[e.rule] || 0) + 1;
}

But this gives only a single object as

{
 a : 2,
 b: 3
}

@mplungjan - Sorry, I didn't get that. Could you please explain a little more. — whyAto8
– whyAto8, Commented Jun 14, 2016 at 12:40

Nenad Vracar · Accepted Answer · 2016-06-14 12:40:05Z

2

You can do this with forEach and thisArg optional parameter.

var arr = [{"price":20,"rule":"a"},{"price":10,"rule":"b"},{"price":5,"rule":"a"},{"price":50,"rule":"b"},{"price":70,"rule":"b"}], 
    r = [];

arr.forEach(function(e) {
  if(!this[e.rule]) {
    this[e.rule] = {rule: e.rule, countOfRule: 0, minPriceForThisRule: e.price}
    r.push(this[e.rule]);
  }
  this[e.rule].countOfRule++;
  if(this[e.rule].minPriceForThisRule > e.price) this[e.rule].minPriceForThisRule = e.price;
}, {});

console.log(r)

answered Jun 14, 2016 at 12:40

Nenad Vracar

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6 Comments

whyAto8 Over a year ago

Thanks, let me try this out.

whyAto8 Over a year ago

Hey, just out of curiosity, how do you come up with such an answer and that too so fast. Building this logic and all. Would you have some recommendations for me on this, as to any javascript algo books, references etc. which could help me in such cases.

Nina Scholz Over a year ago

its just a standard pattern, which is not completely copied.

Nenad Vracar Over a year ago

I learned to solve problems like this from smarter and more experienced developers then me here on SO, but some good places to learn also are eloquentjavascript.net, developer.mozilla.org/en-US/docs/Web/JavaScript

whyAto8 Over a year ago

Thanks again for the references. And yes, am also doing same, as far as SO goes :-)

|

terpinmd · Accepted Answer · 2016-06-14 12:54:52Z

0

I would use reduce. Something like:

var reduced = arr.reduce(function(memo, obj){
   var rule = memo[obj.rule];
   rule.countOfRule++;
   if(!rule.minPriceForThisRule){
     rule.minPriceForThisRule = obj.price;
   } else{
     if(obj.price < rule.minPriceForThisRule){
       rule.minPriceForThisRule = obj.price;
     }
   }
   return memo;
}, map);

where the initial map looks like:

var map = {
  1: {
    rule: 1,
    countOfRule: 0,
    minPriceForThisRule: null      
  },
  2: {
    rule: 2,
    countOfRule: 0,
    minPriceForThisRule: null        
  }
}

Of course you could dynamically create the map if needed.

https://plnkr.co/edit/wLw3tEx2SMXmYE7yOEHg?p=preview

answered Jun 14, 2016 at 12:54

terpinmd

1,0189 silver badges16 bronze badges

Comments

Redu · Accepted Answer · 2016-06-14 13:16:17Z

This is how i would do this job,

var arr =[
 {
  price:20,
  rule:"a"
 },
 {
  price:10,
  rule:"b"
 },
 {
  price:5,
  rule:"a"
 },
 {
  price:50,
  rule:"b"
 },
 {
  price:70,
  rule:"b"
 }
],
reduced = arr.reduce((p,c) => { var fo = p.find(f => f.rule == c.rule);
                                fo ? (fo.countOfRule++,
                                      fo.minPriceForThisRule > c.price && (fo.minPriceForThisRule = c.price))
                                   :  p.push({rule:c.rule, countOfRule:1, minPriceForThisRule: c.price});
                                return p;
                              },[]);
console.log(reduced);

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