CODEWITHSUNDEEP
javascriptarrays
poniboy4
poniboy4

Reputation: 19

Count number of times a word from one array occurs in a second array

So you have two arrays. Search the second array and count how many times each word from the first array occurs in the second one. Display a scrollable list of all the words from the second array and their counts from the first array. I don't want to show only the duplicate values so an intersection function wouldn't help .Use only javascript.

Ex:

listOne = [Pat, cat, hat, Tat, bat, rat];
listTwo = [Pat, mat, sat, rat, cat, Tat, Pat];

And the output would be as follows:

Pat: 2
mat: 0
sat: 0
rat: 1
cat: 1
Tat: 1

I have tried multiple ways and the closest I came up with is this:

function findFreq(listOne, listTwo, concatList) {
  var a = [], b = [], lookup = {}, prev;

  for(var j = 0; j < listTwo.length; j++) {
      lookup[listTwo[j]] = listTwo[j];
  }

  for (var i = 0; i < concatList.length; i++ ) {
      if (concatList[i] !== prev) {
          a.push(lookup[concatList[i]]);
          b.push(0);
      } else {
          b[b.length - 1]++;
      }
      prev = concatList[i];
  }

      return [a, b];          
  }

var result = findFreq(listOne, listTwo, concatList);
alert('[' + result[0] + ']<br />[' + result[1] + ']');

As you can see I thought concating the two arrays would be easiest because all of the examples I found of counting the occurrence of elements only dealt with a single array. In fact I took the code from here and tried modifying it to fit my needs.

The problem is the output gives me empty or extra elements.

[Pat,Tat,,cat,,mat,rat,sat]
[2,1,0,1,0,0,1,0]

So the answer is wrong and so is the formatting. I'm probably over complicating this. It is probably easier to compare the two arrays directly, but I just can't figure it out. Any help would be appreciated.

Note: For my program the two arrays I'm using actually come from user specified files that I just use the split() function on. That's why my naming conventions are strange.

Upvotes: 0

Views: 901

Answers (5)

Pablo Souza
Pablo Souza

Reputation: 59

take a further look at the documentation:

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce

function myCounter(listOne, listTwo) {
    return [listOne, listTwo].reduce(function(a, b) {
      return a.concat(b);
    }, []).reduce( (countWords, word) => {  
        countWords[word] = ++countWords[word] || 0;     
        return countWords;
    }, {}); 
}

var listOne = ["Pat", "cat", "hat", "Tat", "bat", "rat"];
var listTwo = ["Pat", "mat", "sat", "rat", "cat", "Tat", "Pat"];

console.log(myCounter(listOne, listTwo));

Maybe it helps you!

Upvotes: 0

Justin Makeig
Justin Makeig

Reputation: 2137

var listOne = 'pat, cat, hat, tat, bat, rat'.split(', ');
var listTwo = 'pat, mat, sat, rat, cat, tat, pat'.split(', ');

// “Count how many times each word from the first array occurs in the second one.”
var results = listOne
  .reduce(function(prev, current) {
    if(!prev[current]) {
      prev[current] = (prev[current] || 0) 
        + listTwo.filter(function(item) { return current === item; }).length;
    }
    return prev;
  }, {});

Returns

{
  "pat": 2,
  "cat": 1,
  "hat": 0,
  "tat": 1,
  "bat": 0,
  "rat": 1
}

Then you could build HTML as a string, for example, 

var htmlStr = Object.keys(results)
  .map(function(key) {
    return '<tr><td>'+key+'</td><td>'+results[key]+'</td></tr>';
  })
  .join('\r');

which would yield

'<tr><td>pat</td><td>2</td></tr>
<tr><td>cat</td><td>1</td></tr>
<tr><td>hat</td><td>0</td></tr>
<tr><td>tat</td><td>1</td></tr>
<tr><td>bat</td><td>0</td></tr>
<tr><td>rat</td><td>1</td></tr>'

(I’ve been fast and loose with my line breaks, but you get the idea.)

Upvotes: 0

dexhering
dexhering

Reputation: 422

You can create a cycle for the first list and compare all the elements of the second list to each element of the first list to see if they exist and if so, how often repeated.

var listOne = ["Pat", "cat", "hat", "Tat", "bat", "rat"];
var listTwo = ["Pat", "mat", "sat", "rat", "cat", "Tat", "Pat"];    

function findFreq(listOne, listTwo) {
    var a = [], b = [], flag = false, cache = 0;
    for(var i=0; i<listOne.length; i++){
        var temp = listOne[i];
        for(var j=0; j<listTwo.length; j++){
            if(temp === listTwo[j]){
                cache++;
            }
        }
        if(cache!=0){
            a.push(temp);
            b.push(cache);
            cache=0;
        }
    }
    return [a,b];          
}
var result = findFreq(listOne, listTwo);
    for(var k=0; k<result[0].length; k++){
    console.log(result[0][k] + ': ' + result[1][k]);
}

Upvotes: 0

charlietfl
charlietfl

Reputation: 171679

Using Array.prototype.map() to create array of counts, Array.prototype.reduce() to do counting

listOne = ["Pat", "cat", "hat", "Tat", "bat", "rat"];
listTwo = ["Pat", "mat", "sat", "rat", "cat", "Tat", "Pat"];

var res = listOne.map(function(word) {
  return listTwo.reduce(function(count, test) {
    return word === test ? ++count : count;
  }, 0);
});


document.getElementById('pre').innerHTML =JSON.stringify(res)
console.log(res)//[2,1,0,1,0,1]
<pre id="pre"></pre>

Upvotes: 0

ak_
ak_

Reputation: 2815

Just loop over both arrays. Something like this should work:

listOne = ["Pat", "cat", "hat", "Tat", "bat", "rat"];
listTwo = ["Pat", "mat", "sat", "rat", "cat", "Tat", "Pat"];


for(var i = 0; i < listOne.length; i++) {
    matches = 0;
    for(var j = 0; j < listTwo.length; j++) {
    if(listOne[i] == listTwo[j]) {
        matches++;
    }
  }
  console.log(listOne[i] + ' has '+ matches+ ' occurance(s)')
}

You can see it working here: https://jsfiddle.net/igor_9000/9n883kse/

Hope that helps!

Upvotes: 2

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