Analyzing and inserting elements into an array of objects

Question

Analyzing and inserting elements into an array of objects

The following code aims to:

1) iterate through the two arrays,

2) if an item exists in both arrays, add its value to the value of the matching item in the first array,

3) if the item is found in arr2 but not in arr1, add the item to arr1. The code functions correctly when both arrays are the same size, but encounters issues when dealing with arrays of different sizes.

Current Result:

[[42, "Bowling Ball"], [4, "Dirty Sock"], [2, "cat"], [6, "mugs"], [2, "Dirty Sock"], [3, "rags"]]

Expected Result:

[[42, "Bowling Ball"], [4, "Dirty Sock"], [2, "cat"], [3, "rags"], [3, "mugs"]]

Here is the code snippet:

function updateInventory(arr1, arr2) {
  for (var i = 0; i < arr1.length; i++) {
    for (var j = i; j < arr2.length; j++) {
      if (arr1[i][1] === arr2[j][1]) {
        arr1[i][0] += arr2[j][0];
      }
      if (arr1[i].indexOf(arr2[j][1]) === -1) {
        arr1.push(arr2[j]);
      }
      if (arr2.length > arr1.length) {
        arr1.push(arr2[arr2.length - 1]);
      }
      else
        break;
    }
  }
  return arr1;
}

var curInv = [
    [21, "Bowling Ball"],
    [2, "Dirty Sock"],
    [2, "cat"],
];

var newInv = [
    [21, "Bowling Ball"],
    [2, "Dirty Sock"],
    [3, "rags"],
    [3, "mugs"]
];

updateInventory(curInv, newInv);

What seems to be the issue in this code snippet?

javascript arrays multidimensional-array

Answer 1

Answer №1

To efficiently manage your inventory, consider utilizing a hash table to compare and update the currInv data.

let curInv = [[21, "Bowling Ball"], [2, "Dirty Sock"], [2, "cat"]],
    newInv = [[21, "Bowling Ball"], [2, "Dirty Sock"], [3, "rags"], [3, "mugs"]],
    inventory = Object.create(null);

curInv.forEach(function (a) {
    this[a[1]] = a;
}, inventory);

newInv.forEach(function (a) {
    if (!this[a[1]]) {
        this[a[1]] = [0, a[1]];
        curInv.push(this[a[1]]);
    }
    this[a[1]][0] += a[0];
}, inventory);

console.log(curInv);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

To efficiently manage your inventory, consider utilizing a hash table to compare and update the currInv data.

let curInv = [[21, "Bowling Ball"], [2, "Dirty Sock"], [2, "cat"]],
    newInv = [[21, "Bowling Ball"], [2, "Dirty Sock"], [3, "rags"], [3, "mugs"]],
    inventory = Object.create(null);

curInv.forEach(function (a) {
    this[a[1]] = a;
}, inventory);

newInv.forEach(function (a) {
    if (!this[a[1]]) {
        this[a[1]] = [0, a[1]];
        curInv.push(this[a[1]]);
    }
    this[a[1]][0] += a[0];
}, inventory);

console.log(curInv);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 3

Answer №2

Perhaps you are making things more complex than necessary. Here's a functional code snippet.

function mergeInventory(currentInventory, newInventory) {
  for (var i = 0; i < newInventory.length; i++) {
    var foundMatch = false;
    for (var j = 0; j < currentInventory.length; j++) {
      if (currentInventory[j][1] === newInventory[i][1]) {
        currentInventory[j][0] += newInventory[i][0];
        foundMatch = true;
      }
    }

    if (!foundMatch) {
      currentInventory.push(newInventory[i]);
    }
  }
  return currentInventory;
}


var currentInv = [
  [21, "Bowling Ball"],
  [2, "Dirty Sock"],
  [2, "cat"],
];

var additionalInv = [
  [21, "Bowling Ball"],
  [2, "Dirty Sock"],
  [3, "rags"],
  [3, "mugs"]
];

console.log(mergeInventory(currentInv, additionalInv));

Functionality Details

Iterate through all items in newInventory.
Check each item against the existing currentInventory.
If a match is found, increment the quantity in currentInventory by that in newInventory.
If no match is found, add the item from newInventory to currentInventory.

Answer 4

Perhaps you are making things more complex than necessary. Here's a functional code snippet.

function mergeInventory(currentInventory, newInventory) {
  for (var i = 0; i < newInventory.length; i++) {
    var foundMatch = false;
    for (var j = 0; j < currentInventory.length; j++) {
      if (currentInventory[j][1] === newInventory[i][1]) {
        currentInventory[j][0] += newInventory[i][0];
        foundMatch = true;
      }
    }

    if (!foundMatch) {
      currentInventory.push(newInventory[i]);
    }
  }
  return currentInventory;
}


var currentInv = [
  [21, "Bowling Ball"],
  [2, "Dirty Sock"],
  [2, "cat"],
];

var additionalInv = [
  [21, "Bowling Ball"],
  [2, "Dirty Sock"],
  [3, "rags"],
  [3, "mugs"]
];

console.log(mergeInventory(currentInv, additionalInv));

Functionality Details

Iterate through all items in newInventory.
Check each item against the existing currentInventory.
If a match is found, increment the quantity in currentInventory by that in newInventory.
If no match is found, add the item from newInventory to currentInventory.

Answer 5

Answer №3

When dealing with arrays, you have the option to either merge them or combine them to achieve the desired outcome automatically.

var array3 = curInv.concat(newInv);

If you are looking to identify the unique elements within the arrays

// Merges both arrays and finds the unique items var array3 = **

arrayUnique(array1.concat(array2));

function arrayUnique(array) {
   var a = array.concat(); 
   for(var i=0; i<a.length; ++i) { 
          for(var j=i+1; j<a.length; ++j) {
             if(a[i] === a[j]) a.splice(j--, 1); } 
     } 
   return a; 
}

**

Answer 6

When dealing with arrays, you have the option to either merge them or combine them to achieve the desired outcome automatically.

var array3 = curInv.concat(newInv);

If you are looking to identify the unique elements within the arrays

// Merges both arrays and finds the unique items var array3 = **

arrayUnique(array1.concat(array2));

function arrayUnique(array) {
   var a = array.concat(); 
   for(var i=0; i<a.length; ++i) { 
          for(var j=i+1; j<a.length; ++j) {
             if(a[i] === a[j]) a.splice(j--, 1); } 
     } 
   return a; 
}

**

Answer 7

Answer №4

In this solution, a concise implementation using a hash (implemented as a Map) is presented following a functional programming style:

function updateInventory(currentInventory, newInventory) {
    return Array.from(
        newInventory.reduce( (map, [quantity, item]) => map.set(item, (map.get(item) || 0) + quantity),
                   new Map(currentInventory.map ( ([quantity, item]) => [item, quantity] )) ), // swap pairs
        ([item, quantity]) => [quantity, item]) // swap back afterwards;
}

// Sample data
var currentInventory = [
    [21, "Bowling Ball"],
    [2, "Dirty Sock"],
    [2, "cat"],
];

var newInventory = [
    [21, "Bowling Ball"],
    [2, "Dirty Sock"],
    [3, "rags"],
    [3, "mugs"]
];

// calling the function
currentInventory = updateInventory(currentInventory, newInventory);
// Output the result
console.log(currentInventory);

.as-console-wrapper { max-height: 100% !important; top: 0; }

If the data had pairs where the item name comes first and quantity second, the code would have been much shorter. This is due to how Map objects are initialized with key-value pairs, not value-key pairs:

function updateInventory(currentInventory, newInventory) {
    return [...newInventory.reduce( (map, [quantity, item]) => map.set(item, (map.get(item) || 0) + quantity), new Map(currentInventory) )];
}

// Sample data
var currentInventory = [
    ["Bowling Ball", 21],
    ["Dirty Sock", 2],
    ["cat", 2],
];

var newInventory = [
    ["Bowling Ball", 21],
    ["Dirty Sock", 2],
    ["rags", 2],
    ["mugs", 2]
];

// calling the function
currentInventory = updateInventory(currentInventory, newInventory);
// Output the result
console.log(currentInventory);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Since this function returns the result without mutating the input arguments, it is named updateInventory.

Improvements for the original code

The outer loop should iterate over the second array for practicality in identifying missing elements in the first array.
The inner loop should start at index 0 rather than the current position of the outer loop.
When a match is found, there is no need to continue looping, so the inner loop should be exited.
The section of code looking for matches should be outside the inner loop and corrected to prevent duplicates.
Eliminate unnecessary break statements in the loops to prevent missing matches and duplicates.

Answer 8

In this solution, a concise implementation using a hash (implemented as a Map) is presented following a functional programming style:

function updateInventory(currentInventory, newInventory) {
    return Array.from(
        newInventory.reduce( (map, [quantity, item]) => map.set(item, (map.get(item) || 0) + quantity),
                   new Map(currentInventory.map ( ([quantity, item]) => [item, quantity] )) ), // swap pairs
        ([item, quantity]) => [quantity, item]) // swap back afterwards;
}

// Sample data
var currentInventory = [
    [21, "Bowling Ball"],
    [2, "Dirty Sock"],
    [2, "cat"],
];

var newInventory = [
    [21, "Bowling Ball"],
    [2, "Dirty Sock"],
    [3, "rags"],
    [3, "mugs"]
];

// calling the function
currentInventory = updateInventory(currentInventory, newInventory);
// Output the result
console.log(currentInventory);

.as-console-wrapper { max-height: 100% !important; top: 0; }

If the data had pairs where the item name comes first and quantity second, the code would have been much shorter. This is due to how Map objects are initialized with key-value pairs, not value-key pairs:

function updateInventory(currentInventory, newInventory) {
    return [...newInventory.reduce( (map, [quantity, item]) => map.set(item, (map.get(item) || 0) + quantity), new Map(currentInventory) )];
}

// Sample data
var currentInventory = [
    ["Bowling Ball", 21],
    ["Dirty Sock", 2],
    ["cat", 2],
];

var newInventory = [
    ["Bowling Ball", 21],
    ["Dirty Sock", 2],
    ["rags", 2],
    ["mugs", 2]
];

// calling the function
currentInventory = updateInventory(currentInventory, newInventory);
// Output the result
console.log(currentInventory);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Since this function returns the result without mutating the input arguments, it is named updateInventory.

Improvements for the original code

The outer loop should iterate over the second array for practicality in identifying missing elements in the first array.
The inner loop should start at index 0 rather than the current position of the outer loop.
When a match is found, there is no need to continue looping, so the inner loop should be exited.
The section of code looking for matches should be outside the inner loop and corrected to prevent duplicates.
Eliminate unnecessary break statements in the loops to prevent missing matches and duplicates.

Analyzing and inserting elements into an array of objects

Answer №1

Answer №2

Functionality Details

Answer №3

Answer №4

Improvements for the original code

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