How to tally duplicate values in an array using Javascript

Question

How to tally duplicate values in an array using Javascript

I'm honing my algorithm skills and challenging myself to write my own code rather than relying on existing solutions. It's a fun way for me to gauge my understanding.

INPUT:

arr1 = ['asd','ew','lol','asd']

EXPECTED OUTPUT:

{ asd: 2, ew: 1, lol: 1 }

This is my implementation:

arr1 = ['asd', 'ew', 'lol', 'asd']
      arr2 = []
      results = {}

function checkIfExists(word) {
  if (arr2.length != 0) {
    for (i = 0; i < arr2.length; i++) {
      if (arr2[i] == word) {
        results[word] += 1
      } else {
        arr2.push(word)
        results[word] = 1
      }
    }
  } else {
    arr2.push(word)
    results[word] = 1
  }

for (i = 0; i < arr1.length; i++) {
  checkIfExists(arr1[i])
}
console.log(results)

ACTUAL OUTPUT:

{ asd: 2, ew: 2 }

javascript algorithm

Answer 1

Answer №1

Make sure not to use i as a global variable, as it can lead to issues when used in multiple loops. Another mistake is in the increment algorithm where there are more counts added to the results array than necessary. Here's an improved approach:

arr1 = ['cat','dog','fish','cat']
arr2 = [] 
results = {}

function checkIfExists(word){
    if (arr2.length != 0){
        var exists = false;
        for (var j = 0; j < arr2.length; j++){
            if(arr2[j] == word){
                results[word] += 1
                exists = true;
                break;          
            }
        }
        if(!exists) {
            arr2.push(word)
            results[word] = 1
        }
    }else{
        arr2.push(word)
        results[word] = 1 
    }
}

for (var k = 0; k < arr1.length; k++) {
    checkIfExists(arr1[k])
}
console.log(results)

Answer 2

Make sure not to use i as a global variable, as it can lead to issues when used in multiple loops. Another mistake is in the increment algorithm where there are more counts added to the results array than necessary. Here's an improved approach:

arr1 = ['cat','dog','fish','cat']
arr2 = [] 
results = {}

function checkIfExists(word){
    if (arr2.length != 0){
        var exists = false;
        for (var j = 0; j < arr2.length; j++){
            if(arr2[j] == word){
                results[word] += 1
                exists = true;
                break;          
            }
        }
        if(!exists) {
            arr2.push(word)
            results[word] = 1
        }
    }else{
        arr2.push(word)
        results[word] = 1 
    }
}

for (var k = 0; k < arr1.length; k++) {
    checkIfExists(arr1[k])
}
console.log(results)

Answer 3

Answer №2

Upon initial inspection, I have pinpointed three questions that you may have intended to pose.

What issues are present in my code?
- Summarizing the situation succinctly is best done by referencing Felix Klings comment:
  
  The main problem is not utilizing var to declare i.
- The loop, seemingly designed to find an element within arr2, actually leads to multiple modifications of arr2... quite prominently!
Why is my code generating inaccurate results? Although your code correctly produces output based on the applied logic, it seems probable that the intended logic does not align with what you had in mind. This scenario is quite common.
How can I rectify my code?
- Initiate improvements by altering your loop patterns from for (i = ...) to for (var i = ...).
- Consider the specific objective of the said loop. If its purpose entails locating an item within arr2, then there should be no necessity for altering arr2 during this process. It's possible that the loop itself might be unnecessary;
  1. You could substitute the entire loop with Array.prototype.indexOf or Array.prototype.includes!
  2. You could potentially replace the complete function with
```
function checkIfExists(word) { results[word] = arr1.filter(function(w) { return w === word; }).length; }
```
    !
  3. Enhancing your comprehension in designing functions would be beneficial, so exploring certain aspects in this code might prove helpful:

var arr1 = ['asd','ew','lol','asd'];
var result = arr1.reduce(function(result, w) { result[w] = result[w] || 0;
                                               result[w]++;
                                               return result;              }, {}));
console.log(result);

Answer 4

Upon initial inspection, I have pinpointed three questions that you may have intended to pose.

What issues are present in my code?
- Summarizing the situation succinctly is best done by referencing Felix Klings comment:
  
  The main problem is not utilizing var to declare i.
- The loop, seemingly designed to find an element within arr2, actually leads to multiple modifications of arr2... quite prominently!
Why is my code generating inaccurate results? Although your code correctly produces output based on the applied logic, it seems probable that the intended logic does not align with what you had in mind. This scenario is quite common.
How can I rectify my code?
- Initiate improvements by altering your loop patterns from for (i = ...) to for (var i = ...).
- Consider the specific objective of the said loop. If its purpose entails locating an item within arr2, then there should be no necessity for altering arr2 during this process. It's possible that the loop itself might be unnecessary;
  1. You could substitute the entire loop with Array.prototype.indexOf or Array.prototype.includes!
  2. You could potentially replace the complete function with
```
function checkIfExists(word) { results[word] = arr1.filter(function(w) { return w === word; }).length; }
```
    !
  3. Enhancing your comprehension in designing functions would be beneficial, so exploring certain aspects in this code might prove helpful:

var arr1 = ['asd','ew','lol','asd'];
var result = arr1.reduce(function(result, w) { result[w] = result[w] || 0;
                                               result[w]++;
                                               return result;              }, {}));
console.log(result);

How to tally duplicate values in an array using Javascript

Answer №1

Answer №2

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