Finding Common Values While Looping Over Two Arrays

Question

I have a situation where I need to compare and find common values over two arrays. I am clear on how to do it with one, but not sure how to do it in this case.

My first array looks like this:

[ { kind: 'E',
    path: [ 'short_name' ],
    lhs: 'testing',
    rhs: 'testing1' },
  { kind: 'E',
    path: [ 'agent_name' ],
    lhs: 'testing',
    rhs: 'testing2' } ]

The array above represents information pertaining to what changed on a document.

My second array looks like this:

[ { lhs: 'legacyId', rhs: 'id_number' },
  { lhs: 'name.short', rhs: 'short_name' },
  { lhs: 'name.long', rhs: 'agent_name' },
  { lhs: 'gender', rhs: 'gender' },
  { lhs: 'dob', rhs: 'date_of_birth' } ]

What I need to do is loop through and find common values for "path" in the first array's elements, and the second array's "rhs" value.

So according to my examples here, I should end up with these values being found: short_name and agent_name.

How could I write a loop to do this over the two arrays?

Kobe · Accepted Answer · 2019-08-15 12:50:17Z

6

You can reduce the 1st array and use a forEach loop over the second array to see if each of the values are equal, then push the value to the accumulator:

const arr1 = [{ kind: 'E', path: [ 'short_name' ], lhs: 'testing', rhs: 'testing1' }, { kind: 'E', path: [ 'agent_name' ], lhs: 'testing', rhs: 'testing2' }]
const arr2 = [{ lhs: 'legacyId', rhs: 'id_number' }, { lhs: 'name.short', rhs: 'short_name' }, { lhs: 'name.long', rhs: 'agent_name' }, { lhs: 'gender', rhs: 'gender' }, { lhs: 'dob', rhs: 'date_of_birth' }]

const common = arr1.reduce((a, o1) => {
  const match = arr2.find(o2 => o1.path[0] === o2.rhs)
  match && a.push(match.rhs)
  return a
}, [])

console.log(common)

If you truly wanted to, you could write this in one line with a find instead of a second reduce:

const a = [{ kind: 'E', path: [ 'short_name' ], lhs: 'testing', rhs: 'testing1' }, { kind: 'E', path: [ 'agent_name' ], lhs: 'testing', rhs: 'testing2' }]
const b = [{ lhs: 'legacyId', rhs: 'id_number' }, { lhs: 'name.short', rhs: 'short_name' }, { lhs: 'name.long', rhs: 'agent_name' }, { lhs: 'gender', rhs: 'gender' }, { lhs: 'dob', rhs: 'date_of_birth' }]

const common = a.reduce((a, o1) => (a.push(b.find(o2 => o1.path[0] === o2.rhs).rhs), a), [])

console.log(common)

Or, for a more performant solution _;) You could use a set:

const a = [{ kind: 'E', path: [ 'short_name' ], lhs: 'testing', rhs: 'testing1' }, { kind: 'E', path: [ 'agent_name' ], lhs: 'testing', rhs: 'testing2' }]
const b = [{ lhs: 'legacyId', rhs: 'id_number' }, { lhs: 'name.short', rhs: 'short_name' }, { lhs: 'name.long', rhs: 'agent_name' }, { lhs: 'gender', rhs: 'gender' }, { lhs: 'dob', rhs: 'date_of_birth' }]
var commonValues = []
var set = new Set([])

for (let i = 0; i < a.length; i++) { 
  const value = a[i].path[0]
  if (!set.has(value)) set.add(value)
}
for (let i = 0; i < b.length; i++) {
  const val = b[i].rhs
  if (set.has(val)) commonValues.push(val)
}

console.log(commonValues)

edited Aug 15, 2019 at 12:50

answered Jul 12, 2019 at 13:43

Kobe

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10 Comments

Rey Over a year ago

I really like how succinct this solution is.

Cal Irvine Over a year ago

@Ademo definitely the most readable solution, but keep in mind that the time complexity is O(n^2) when using nested loops (both reduce and .forEach are loops) so if your data is large this solution will run pretty slowly.

Kobe Over a year ago

Not sure why I got downvoted? Please explain why so I can fix my code :)

Kobe Over a year ago

@CalIrvine It works? It was accepted? It has solved the problem. OP didn't ask for a performance-first solution, otherwise I would've tailored my answer accordingly. I could pick on other answers because it wasn't written in one line, but that wasn't what was asked :P

Kobe Over a year ago

@CalIrvine It will work at a large scale, maybe not as fast as you would like it, but computers are very, very fast these days :) It's a shame you don't see that it doesn't matter. Sure, I should strive to make it as efficient as possible, which I did under the circumstances of using reduce, but I made a readable and short version. If this data was to grow, a database and query would be the solution here, not a script.

|

Calvin Nunes · Accepted Answer · 2019-07-12 13:47:23Z

You can use a simply for inside a for...

This is the most simple (maybe not the most performatic) way you can do that, getting the value from first array, then looping the second one. If a value is found, then push it to another array (results) and break the second loop (there's no need to still running if a value already match).

var a = [ { kind: 'E',
    path: [ 'short_name' ],
    lhs: 'testing',
    rhs: 'testing1' },
  { kind: 'E',
    path: [ 'agent_name' ],
    lhs: 'testing',
    rhs: 'testing2' } ]


var b =[ { lhs: 'legacyId', rhs: 'id_number' },
  { lhs: 'name.short', rhs: 'short_name' },
  { lhs: 'name.long', rhs: 'agent_name' },
  { lhs: 'gender', rhs: 'gender' },
  { lhs: 'dob', rhs: 'date_of_birth' } ]
  
var results = [];  
for (var i = 0; i < a.length; i++){
  var path = a[i].path;
  
  for (var j = 0; j < b.length; j++){
    var rhs = b[j].rhs;
    if (rhs == path){
      results.push(b[j].rhs)
      break;
    }
    
  }
}

console.log(results)

Cal Irvine · Accepted Answer · 2019-07-12 15:15:12Z

2

You can do it in O(n^2) time by nesting a loop and searching through the second loop for every value of the first loop, or you can use a hashmap to do it in O(n) time. In JavaScript we use objects for that because their values can be accessed in O(1) time.

Example:

const array1 = [
  { kind: "E", path: ["short_name"], lhs: "testing", rhs: "testing1" },
  { kind: "E", path: ["agent_name"], lhs: "testing", rhs: "testing2" }
];

const array2 = [
  { lhs: "legacyId", rhs: "id_number" },
  { lhs: "name.short", rhs: "short_name" },
  { lhs: "name.long", rhs: "agent_name" },
  { lhs: "gender", rhs: "gender" },
  { lhs: "dob", rhs: "date_of_birth" }
];

const hashMap = {};
const commonValues = [];

for (let i = 0; i < array1.length; i++) {
  const currentValue = array1[i].path[0];
  hashMap[currentValue] = true;
}

for (let i = 0; i < array2.length; i++) {
  const currentValue = array2[i].rhs;
  if (hashMap[currentValue]) commonValues.push(currentValue);
}

//now commonValues contains all of the common between them. You may want to filter out duplicate matches, depends on your use case.

While it may seem like a lot more code, the time complexity savings are huge as your data gets arbitrarily large by doing it this way. Imagine your data is 100 items long, already an O(n^2) solution would require 10,000 passes to find all of the matches. Conversely with the above solution, it would require 200 passes. These savings in time add up quickly as your data gets larger.

edited Jul 12, 2019 at 15:15

answered Jul 12, 2019 at 13:45

Cal Irvine

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8 Comments

Kobe Over a year ago

I would like you to see this jsperf.com/spread-vs-return I tried testing my two methods with a bigger data set, turns out your solution is actually 99% slower :/ If you could make your solution 99% faster, I'll remove my downvote :)

Cal Irvine Over a year ago

@Kobe the dramatic difference is because I forgot to remove a console.log from one of the loops. I've edited my answer and it's gone now.

Cal Irvine Over a year ago

@Kobe Your answer keeps the lead with the data provided, but I then tripled the size of one array and my answer took the lead.

Cal Irvine Over a year ago

@Kobe jsperf.com/spread-vs-return/5 May I have that downvote removed, as promised? :D

Kobe Over a year ago

Last time I checked, that's more than triple the size :P True my answer doesn't handle duplicates well, but then using that size of data i would use a db query :) Sure, ill remove it, your answer is better :)

|

charlietfl · Accepted Answer · 2019-07-12 14:06:02Z

1

Make a Set out of the rhs values in second array and reduce first array and filter paths by checking against the set

const arr1= [ { kind: 'E',
    path: [ 'short_name' ],
    lhs: 'testing',
    rhs: 'testing1' },
  { kind: 'E',
    path: [ 'agent_name' ],
    lhs: 'testing',
    rhs: 'testing2' } ],
    
    arr2= [ { lhs: 'legacyId', rhs: 'id_number' },
  { lhs: 'name.short', rhs: 'short_name' },
  { lhs: 'name.long', rhs: 'agent_name' },
  { lhs: 'gender', rhs: 'gender' },
  { lhs: 'dob', rhs: 'date_of_birth' } ],
  
  arr2Set = new Set(arr2.map(({rhs}) => rhs)),
  matches = arr1.reduce((a,{path}) =>[...a, ...path.filter(p => arr2Set.has(p))],[]);
  
  
  console.log(matches)

edited Jul 12, 2019 at 14:06

answered Jul 12, 2019 at 13:52

charlietfl

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1 Comment

Kobe Over a year ago

Spreading out a here heavily reduces speed. It would be best to push the values from the filter to a and then return a.

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