How to sort an array of strings based on search keywords?

Question

I've an array of strings. I need to sort the array based on array of keywords. The string containing max. number of keywords should come first and so on. Also, the string which contains max. no. of search keywords should come first than the number of occurrences of same search keyword. testArray should ignore case of searchTerms. If possible, you can ignore the strings which doesn't contain any search keywords in the result array.

var testArray = [
    "I am",
    "I am wrong and I don't know",
    "I am right and I know",
    "I don't know",
    "I do know"
  ],
  searchTerms = ["I", "right","know"];

$.each(searchTerms, function(index, term) {
  var regX = new RegExp(term, "i");
  testArray = $.map(testArray, function(item) {
    if (regX.test(item)) {
      return item;
    } else {
      return;
    }
  });
});

console.log(testArray);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

If you observe in the above code, the keywords are "I", "right","know". So the testArray results should be as below,

testArray = [
    "I am right and I know",    
    "I am wrong and I don't know",    
    "I don't know",
    "I do know",
    "I am"
  ]

string contain all the keywords come first and the other strings contain "I","know", So they come next and string "I am" comes last as it contains only "I" keyword.

Codepen

Community · Accepted Answer · 2020-06-20 09:12:55Z

You can try something like this:

Idea:

Loop over data and criteria and create a Map that has both count and value.
Use a regex to match the string. This way you can do case-insensitive search.
Sort this Map based on count.
Return list of values.

function getMatchCountMap(data, criterias) {
  return data.reduce((countMap, curItem) => {
    var count = criterias.filter((criteria) => new RegExp(criteria, 'i').test(curItem) ).length;
    countMap.push({
      value: curItem,
      count: count
    });
    return countMap;
  }, [])
}

function sortBasedOnCount(data, criterias) {
  var map = getMatchCountMap(data, criterias);
  map.sort((a, b) => b.count - a.count);
  return map.map(x => x.value);
}

var testArray = [
    "I am",
    "I am wrong and I don't know",
    "I am right and I know",
    "I don't know",
    "I do know"
  ],
  searchTerms = ["I", "right","know"];
  
console.log(sortBasedOnCount(testArray, searchTerms))

J. Pichardo · Accepted Answer · 2018-06-04 10:33:34Z

1

You could try to create regex with those terms and sort by the number of matches, something like:

var testArray = [
    "I am",
    "i,i,i will come first i'm RIGHT i do KNOW",
    "I am wrong and I don't know",
    "I am right and I know",
    "I don't know",
    "I do know",
    "Something else"
  ],

  searchTerms = ["I", "right", "know"];


// (?:I)|(?:right)|(?:know)
const searchExp = new RegExp(searchTerms.reduce((acc, term) => acc ? `${acc}|(?:${term})` : `(?:${term})`, ''), 'gi');

const result = testArray.sort((a, b) => {
  const bMatch = b.match(searchExp) || [];
  const aMatch = a.match(searchExp) || [];

  return bMatch.length - aMatch.length;
});

console.log(result);

edited Jun 4, 2018 at 10:33

answered Jun 4, 2018 at 9:10

J. Pichardo

3,1151 gold badge23 silver badges37 bronze badges

9 Comments

imixtron Over a year ago

this fails in the case where there is no matching value. It is not able to read .length property for a null value (i.e. b.match(searchExp)).

J. Pichardo Over a year ago

@imixtron you are right I didn't considered that case

Raviteja Over a year ago

@J.Pichardo should ignore the case. Not working here

J. Pichardo Over a year ago

@Raviteja, hmm, i didn't see that in the question.

J. Pichardo Over a year ago

@Raviteja, haha, no worries, Anyway, example updated, check the i flag in the regexp

|

Nina Scholz · Accepted Answer · 2018-06-05 09:04:45Z

1

You could take an object for the counting and sort by the count of the string.

var array = [ "I am", "I am wrong and I don't know", "i am RIGHT and I know", "I don't know", "I do know", "i,i,i will come first i'm RIGHT i do KNOW"],
    search = ["I", "am","know"].map(v => v.toLowerCase()),
    count = array.reduce((c, k) => {
        var a = k.toLowerCase().split(/[^a-z0-9']+/),
            count = Object.create(null);

        a.forEach(w => count[w] = (count[w] || 0) + 1);
        c[k] = [0, 0];
        search.forEach(v => {
            c[k][0] += v in count;
            c[k][1] += count[v] || 0;
        });
        return c;
    }, Object.create(null));
  
array.sort((a, b) => count[b][0] - count[a][0] || count[b][1] - count[a][1]);

console.log(array);

edited Jun 5, 2018 at 9:04

answered Jun 4, 2018 at 9:07

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

6 Comments

Raviteja Over a year ago

Thanks for the quick response

Raviteja Over a year ago

change in case not working. codepen.io/iamraviteja/pen/pKgZRp?editors=0011

Nina Scholz Over a year ago

you could normalize the search array with all lower case characters and take a lower case string for includes.

Raviteja Over a year ago

no. When the search words are "I", "am", "know", then "I am right and I know"should come first but, i,i,i will come first i'm RIGHT i do KNOW is coming first as count of i is more.

Nina Scholz Over a year ago

what should come actually first by searching for ["I", "am", "know"]?

|

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