Removing repetitive strings from an array in the most efficient manner

Question

Removing repetitive strings from an array in the most efficient manner

We've successfully developed a script to eliminate duplicate strings from an array while preserving the order necessary for angular's ng-repeat loop. It's also important that the remaining elements maintain their original index.

scope.feedback = _.map(_.pluck(item.possibleAnswers, 'feedback'), function (element, index, collection) {
    return collection.slice(0, index).indexOf(element) === -1 ? element : '';
});

Although this code is functional, we can't help but think there may be a simpler solution out there. Has anyone encountered a similar issue and found a more straightforward workaround?

javascript arrays angularjs lodash

Answer 1

Answer №1

In case the browsers you are targeting have compatibility with the spread operator, you can test it in your console by running:

[...new Set(['3','1','1','5'])]
// ['3','1','5']

Alternatively, if the browser supports Array.from, you could use the following code as well:

Array.from(new Set(['3','1','1','5']))
// ['3','1','5']

Answer 2

In case the browsers you are targeting have compatibility with the spread operator, you can test it in your console by running:

[...new Set(['3','1','1','5'])]
// ['3','1','5']

Alternatively, if the browser supports Array.from, you could use the following code as well:

Array.from(new Set(['3','1','1','5']))
// ['3','1','5']

Answer 3

Answer №2

Here is a different approach using the reduce method https://jsfiddle.net/58z7nrfy/1/

var numbers = [1,2,3,1,2,3,2,2,3,4,5,5,12,1,23,4,1];

var uniqueNumbers = numbers.reduce(function(previousValue, currentValue, index, array){
  if (previousValue.indexOf(currentValue) == -1) 
    previousValue.push(currentValue);
  else 
    previousValue.push('');
  
  return previousValue;
}, [])
console.log(uniqueNumbers)

[1, 2, 3, "", "", "", "", "", "", 4, 5, "", 12, "", 23, "", ""]

Answer 4

Here is a different approach using the reduce method https://jsfiddle.net/58z7nrfy/1/

var numbers = [1,2,3,1,2,3,2,2,3,4,5,5,12,1,23,4,1];

var uniqueNumbers = numbers.reduce(function(previousValue, currentValue, index, array){
  if (previousValue.indexOf(currentValue) == -1) 
    previousValue.push(currentValue);
  else 
    previousValue.push('');
  
  return previousValue;
}, [])
console.log(uniqueNumbers)

[1, 2, 3, "", "", "", "", "", "", 4, 5, "", 12, "", 23, "", ""]

Answer 5

Answer №3

In addition to previously mentioned solutions, an alternative method is utilizing the lodash union function in the following manner:

const duplicates = ['Apple', 'Banana', 'Apple'];
const uniques = _.union(duplicates);

The unique values will be: ["Apple", "Banana"]

Answer 6

In addition to previously mentioned solutions, an alternative method is utilizing the lodash union function in the following manner:

const duplicates = ['Apple', 'Banana', 'Apple'];
const uniques = _.union(duplicates);

The unique values will be: ["Apple", "Banana"]

Answer 7

Answer №4

To optimize your code, consider utilizing a Map. This data structure is type-safe and avoids repetitive iterations using Array#indexOf.

var array = [1, 2, 3, 1, 2, 3, 2, 2, 3, 4, 5, 5, 12, 1, 23, 4, 1, '23'],
    uniqueValues = array.filter(function (value) {
        if (!this.has(value)) {
            this.set(value, true);
            return true;
        }
    }, new Map);

console.log(uniqueValues);

Answer 8

To optimize your code, consider utilizing a Map. This data structure is type-safe and avoids repetitive iterations using Array#indexOf.

var array = [1, 2, 3, 1, 2, 3, 2, 2, 3, 4, 5, 5, 12, 1, 23, 4, 1, '23'],
    uniqueValues = array.filter(function (value) {
        if (!this.has(value)) {
            this.set(value, true);
            return true;
        }
    }, new Map);

console.log(uniqueValues);

Answer 9

Answer №5

The current accepted answer may not be the most efficient solution. Utilizing reduce in conjunction with a hash table or map object could greatly enhance performance. In my opinion, using a map instead of reduce would be preferable. Expanding on @Nina Scholz's method by utilizing double maps, the optimal response to the original poster's question is as follows;

var   a = [1, 2, 3, 1, 2, 3, 2, 2, 3, 4, 5, 5, 12, 1, 23, 4, 1, '23'],
unified = a.map(function(e) {
                  return this.has(e) ? void 0 : (this.set(e,"Cheap Thrills"),e);
                }, new Map());
console.log(unified);

If this were to be used in a production environment with an array of arbitrary length, I would opt for implementing the map method with standard functions rather than functors. This approach can introduce a significant performance boost when dealing with large arrays, especially those exceeding a size of 10K+

Answer 10

The current accepted answer may not be the most efficient solution. Utilizing reduce in conjunction with a hash table or map object could greatly enhance performance. In my opinion, using a map instead of reduce would be preferable. Expanding on @Nina Scholz's method by utilizing double maps, the optimal response to the original poster's question is as follows;

var   a = [1, 2, 3, 1, 2, 3, 2, 2, 3, 4, 5, 5, 12, 1, 23, 4, 1, '23'],
unified = a.map(function(e) {
                  return this.has(e) ? void 0 : (this.set(e,"Cheap Thrills"),e);
                }, new Map());
console.log(unified);

If this were to be used in a production environment with an array of arbitrary length, I would opt for implementing the map method with standard functions rather than functors. This approach can introduce a significant performance boost when dealing with large arrays, especially those exceeding a size of 10K+

Answer 11

Answer №6

I find the lodash method uniq() to be incredibly useful

let numbers = [1, 2, 3, 1, 2, 3, 2, 2, 3, 4, 5, 5, 12, 1, 23, 4, 1];
let uniqueNumbers = _.uniq(numbers, (element) => {
 return element;
});

The resulting output is:

[1, 2, 3, 4, 5, 12, 23]

Answer 12

I find the lodash method uniq() to be incredibly useful

let numbers = [1, 2, 3, 1, 2, 3, 2, 2, 3, 4, 5, 5, 12, 1, 23, 4, 1];
let uniqueNumbers = _.uniq(numbers, (element) => {
 return element;
});

The resulting output is:

[1, 2, 3, 4, 5, 12, 23]

Answer 13

Answer №7

Here is the script that I incorporated:

const uniqueWords = ['alpha', 'beta', 'gamma', 'gamma', 'delta', 'beta'];
    const finalResult = [];
    for(let j=0; j<uniqueWords.length; j++){
      if(finalResult.indexOf(uniqueWords[j]) == -1){
        finalResult.push(uniqueWords[j])
      }
    }

Answer 14

Here is the script that I incorporated:

const uniqueWords = ['alpha', 'beta', 'gamma', 'gamma', 'delta', 'beta'];
    const finalResult = [];
    for(let j=0; j<uniqueWords.length; j++){
      if(finalResult.indexOf(uniqueWords[j]) == -1){
        finalResult.push(uniqueWords[j])
      }
    }

Removing repetitive strings from an array in the most efficient manner

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

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