Looking to compare two elements within different arrays? The method outlined below is specifically designed for comparing to individual values rather than entire arrays

Question

Looking to compare two elements within different arrays? The method outlined below is specifically designed for comparing to individual values rather than entire arrays

Is this the right approach? How can we iterate through each array to compare values? Should these data structures be modified or transformed first?

Below is the data that needs to be compared. The objective is to match userID with DocumentID.

const videos = 
  [ { userID: '5lyU0TCyqRcTD3y7Rs2FGV8h2Sd2', name: 'Wedge Antilles',  faction: 'Rebels' } 
  , { userID: '8',                            name: 'Ciena Ree',       faction: 'Empire' } 
  , { userID: '40',                           name: 'Iden Versio',     faction: 'Empire' } 
  , { userID: '66',                           name: 'Thane Kyrell',    faction: 'Rebels' } 
  ] 
 
const blocked = 
  [ { id:  2, name: 'Wedge Antilles', documentID: '5lyU0TCyqRcTD3y7Rs2FGV8h2Sd2' } 
  , { id:  8, name: 'Ciena Ree',      documentID: 'Empire'                       } 
  , { id: 40, name: 'Iden Versio',    documentID: 'Empire'                       } 
  , { id: 66, name: 'Thane Kyrell',   documentID: 'Rebels'                       } 
  ] 
var result = videos.filter(function(video) { 
  return blocked.some(blockedItem => video.userID === blockedItem.documentID)
})

console.log(result)

javascript arrays filter

Answer 1

Answer №1

There are various approaches you can take to achieve this task. One method involves utilizing javascript functions such as map and includes. Here is an example implementation:

var blockedIds = blocked.map(function(blockedItem, index) {
  return blockedItem.documentID;
});
var result = videos
  .filter(function(videos, index) {
    return blockedIds.includes(videos["userID"]);
  });

It's important to note that this approach has a time complexity of O(nxm) (where n represents the size of the first array and m denotes the size of the second array).

Answer 2

There are various approaches you can take to achieve this task. One method involves utilizing javascript functions such as map and includes. Here is an example implementation:

var blockedIds = blocked.map(function(blockedItem, index) {
  return blockedItem.documentID;
});
var result = videos
  .filter(function(videos, index) {
    return blockedIds.includes(videos["userID"]);
  });

It's important to note that this approach has a time complexity of O(nxm) (where n represents the size of the first array and m denotes the size of the second array).

Answer 3

Answer №2

To improve efficiency, I suggest converting the blocked items into an object or map format and then using that to filter through.

const blockedMap = Object.fromEntries(Object.values(blocked).map(item => [item.documentID, item]))

// apply filtering
var filteredVideos = videos.filter((video) => blockedMap[video.userID])

Answer 4

To improve efficiency, I suggest converting the blocked items into an object or map format and then using that to filter through.

const blockedMap = Object.fromEntries(Object.values(blocked).map(item => [item.documentID, item]))

// apply filtering
var filteredVideos = videos.filter((video) => blockedMap[video.userID])

Answer 5

Answer №3

To achieve this, follow the steps below:

Retrieve the documentID by using the Array.map() method in a separate array, only comparing documentID with the userID.
Utilize the Array.filter() method to compare the videos array with the documentIDArray and filter out the matching results.

Example :

const videos = [{
  userID: '5lyU0TCyqRcTD3y7Rs2FGV8h2Sd2',
  name: 'Wedge Antilles',
  faction: 'Rebels'
}, {
  userID: '8',
  name: 'Ciena Ree',
  faction: 'Empire'
}, {
  userID: '40',
  name: 'Iden Versio',
  faction: 'Empire'
}, {
  userID: '66',
  name: 'Thane Kyrell',
  faction: 'Rebels'
}]; 

const blocked = [{
  id: 2,
  name: 'Wedge Antilles',
  documentID: '5lyU0TCyqRcTD3y7Rs2FGV8h2Sd2'
}, {
  id: 8,
  name: 'Ciena Ree',
  documentID: 'Empire'
}, {
  id: 40,
  name: 'Iden Versio',
  documentID: 'Empire'
}, {
  id: 66,
  name: 'Thane Kyrell',
  documentID: 'Rebels'
}];

// Retrieve documentID using Array.map() method in a separate array for comparison.
const documentIDArray = blocked.map((obj) => obj.documentID);

// Use Array.filter() method to compare videos array with documentIDArray to filter out matched results.
var result = videos.filter(video => documentIDArray.includes(video['userID']));

console.log(result);

Answer 6

To achieve this, follow the steps below:

Retrieve the documentID by using the Array.map() method in a separate array, only comparing documentID with the userID.
Utilize the Array.filter() method to compare the videos array with the documentIDArray and filter out the matching results.

Example :

const videos = [{
  userID: '5lyU0TCyqRcTD3y7Rs2FGV8h2Sd2',
  name: 'Wedge Antilles',
  faction: 'Rebels'
}, {
  userID: '8',
  name: 'Ciena Ree',
  faction: 'Empire'
}, {
  userID: '40',
  name: 'Iden Versio',
  faction: 'Empire'
}, {
  userID: '66',
  name: 'Thane Kyrell',
  faction: 'Rebels'
}]; 

const blocked = [{
  id: 2,
  name: 'Wedge Antilles',
  documentID: '5lyU0TCyqRcTD3y7Rs2FGV8h2Sd2'
}, {
  id: 8,
  name: 'Ciena Ree',
  documentID: 'Empire'
}, {
  id: 40,
  name: 'Iden Versio',
  documentID: 'Empire'
}, {
  id: 66,
  name: 'Thane Kyrell',
  documentID: 'Rebels'
}];

// Retrieve documentID using Array.map() method in a separate array for comparison.
const documentIDArray = blocked.map((obj) => obj.documentID);

// Use Array.filter() method to compare videos array with documentIDArray to filter out matched results.
var result = videos.filter(video => documentIDArray.includes(video['userID']));

console.log(result);

Looking to compare two elements within different arrays? The method outlined below is specifically designed for comparing to individual values rather than entire arrays

Answer №1

Answer №2

Answer №3

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