Automatically adjusting maps according to internal criteria

Question

Automatically adjusting maps according to internal criteria

I have an example of a map that looks like this

const Map = new Map().set('123', [ [ 'foo', 'bar' ] ]).set('456', [ [ 'baz', 'qux' ], [ 'quux', 'corge' ] ]);
/* 
The structure of the Map is as follows:
    Map {
       '123' => [ [ 'foo', 'bar' ] ],
       '456' => [ [ 'baz', 'qux' ], [ 'quux', 'corge' ] ] 
   } 
*/

If I wanted to delete the array where the first nested element equals 'quux', how would I achieve that in order to obtain the following result?

Map {
    '123' => [ [ 'foo', 'bar' ] ],
    '456' => [ [ 'baz', 'qux' ] ] 
}

I can easily remove the item using the following code:

Map.set('456', (Map.get('456')).filter(array => array[0] !== 'quux'));

However, this method only works because I know which key ('456') contains the element with value 'quux'. I am uncertain of how to programmatically search through the Map, identify the corresponding key, and then remove the item. The keys and values within the Map may vary dynamically, but the structure remains consistent. The search term will remain static, for instance: 'quux', meaning that while the contents of the Map may change, I am essentially conducting a search and removal process.

javascript dictionary ecmascript-6

Answer 1

Answer №1

To find the desired value in a map, you can iterate through the map and filter out arrays based on certain conditions.

const myMap = new Map([['123', [['foo', 'bar']]], ['456', [['baz', 'qux'], ['quux', 'corge']]]]);

myMap.forEach((value, key, map) => {
    if (value.some(item => item[0] === 'quux')) {
        map.set(key, value.filter(item => item[0] !== 'quux'));
    }
});

console.log([...myMap]);

Answer 2

To find the desired value in a map, you can iterate through the map and filter out arrays based on certain conditions.

const myMap = new Map([['123', [['foo', 'bar']]], ['456', [['baz', 'qux'], ['quux', 'corge']]]]);

myMap.forEach((value, key, map) => {
    if (value.some(item => item[0] === 'quux')) {
        map.set(key, value.filter(item => item[0] !== 'quux'));
    }
});

console.log([...myMap]);

Answer 3

Answer №2

To iterate through the elements of a Map, you can utilize the findIndex method on each value v to check if it contains an array with 'quux' as its first element, and then remove that array using the splice function:

const map = new Map().set('123', [ [ 'foo', 'bar' ] ]).set('456', [ [ 'baz', 'qux' ], [ 'quux', 'corge' ] ]);

console.log("before", [...map]);

for (const v of map.values()) {
  const index = v.findIndex((a) => a[0] === "quux");
  
  if (index > -1) {
    v.splice(index, 1);
  }
}

console.log("after", [...map]);

For a non-destructive approach, you can create a new Map by extracting entries from the original one and filtering out unwanted arrays using the map and filter methods:

const before = new Map().set('123', [ [ 'foo', 'bar' ] ]).set('456', [ [ 'baz', 'qux' ], [ 'quux', 'corge' ] ]);

console.log("before", [...before]);

const after = new Map([...before].map(([k, v]) => {
  return [k, v.filter((a) => a[0] !== "quux")];
})

console.log("after", [...after]);

Please note: The main distinction between these two approaches is that the second one removes all arrays where 'quux' is the first element, while the first one only removes the first occurrence. You can modify them according to your specific requirements.

Answer 4

To iterate through the elements of a Map, you can utilize the findIndex method on each value v to check if it contains an array with 'quux' as its first element, and then remove that array using the splice function:

const map = new Map().set('123', [ [ 'foo', 'bar' ] ]).set('456', [ [ 'baz', 'qux' ], [ 'quux', 'corge' ] ]);

console.log("before", [...map]);

for (const v of map.values()) {
  const index = v.findIndex((a) => a[0] === "quux");
  
  if (index > -1) {
    v.splice(index, 1);
  }
}

console.log("after", [...map]);

For a non-destructive approach, you can create a new Map by extracting entries from the original one and filtering out unwanted arrays using the map and filter methods:

const before = new Map().set('123', [ [ 'foo', 'bar' ] ]).set('456', [ [ 'baz', 'qux' ], [ 'quux', 'corge' ] ]);

console.log("before", [...before]);

const after = new Map([...before].map(([k, v]) => {
  return [k, v.filter((a) => a[0] !== "quux")];
})

console.log("after", [...after]);

Please note: The main distinction between these two approaches is that the second one removes all arrays where 'quux' is the first element, while the first one only removes the first occurrence. You can modify them according to your specific requirements.

Answer 5

Answer №3

If you want to dynamically generate keys, consider using a for...of loop in your code:

Don't forget to open your devtools to view the new map as it may not display correctly in the code snippet.

const myMap = new Map().set('123', [
  ['foo', 'bar']
]).set('456', [
  ['baz', 'qux'],
  ['quux', 'corge']
]);

for (let keyVal of myMap) {
  myMap.set(keyVal[0], (myMap.get(keyVal[0])).filter(array => array[0] !== 'quux'));
}

console.log(myMap);

I hope this solution meets your requirements. Feel free to leave a comment if you need further assistance! ;)

Answer 6

If you want to dynamically generate keys, consider using a for...of loop in your code:

Don't forget to open your devtools to view the new map as it may not display correctly in the code snippet.

const myMap = new Map().set('123', [
  ['foo', 'bar']
]).set('456', [
  ['baz', 'qux'],
  ['quux', 'corge']
]);

for (let keyVal of myMap) {
  myMap.set(keyVal[0], (myMap.get(keyVal[0])).filter(array => array[0] !== 'quux'));
}

console.log(myMap);

I hope this solution meets your requirements. Feel free to leave a comment if you need further assistance! ;)

Answer 7

Answer №4

To iterate through the key-value pairs of a map, we can access the outer array containing the inner arrays in order to filter out the specific value we're searching for. By using the forEach function, we can determine the index of the inner array and then utilize the splice function to remove it from the outer array.

const dataMap = new Map().set('123', [ [ 'foo', 'bar' ] ]).set('456', [ [ 'baz', 'qux' ], [ 'quux', 'corge' ] ]);
dataMap.forEach((value, key) => {
   value.forEach((array, index) => {
      if (array.includes('quux')) {
         value.splice(index, 1);
      }
   });
});
console.log(dataMap);

Answer 8

To iterate through the key-value pairs of a map, we can access the outer array containing the inner arrays in order to filter out the specific value we're searching for. By using the forEach function, we can determine the index of the inner array and then utilize the splice function to remove it from the outer array.

const dataMap = new Map().set('123', [ [ 'foo', 'bar' ] ]).set('456', [ [ 'baz', 'qux' ], [ 'quux', 'corge' ] ]);
dataMap.forEach((value, key) => {
   value.forEach((array, index) => {
      if (array.includes('quux')) {
         value.splice(index, 1);
      }
   });
});
console.log(dataMap);

Answer 9

Answer №5

Should we always utilize Array.prototype.filter for better performance, or is it more efficient to first use Array.prototype.some before filtering the array?

In this scenario, the code filters all arrays without first checking for the presence of 'quux'.

const map = new Map().set('123', [ ['foo', 'bar' ] ]).set('456', [ [ 'baz', 'qux' ], [ 'quux', 'corge' ] ]);

map.forEach((val, key) => {
  val = val.filter(arr => arr[0] !== 'quux');
  map.set(key, val);
});

console.log(map);

Answer 10

Should we always utilize Array.prototype.filter for better performance, or is it more efficient to first use Array.prototype.some before filtering the array?

In this scenario, the code filters all arrays without first checking for the presence of 'quux'.

const map = new Map().set('123', [ ['foo', 'bar' ] ]).set('456', [ [ 'baz', 'qux' ], [ 'quux', 'corge' ] ]);

map.forEach((val, key) => {
  val = val.filter(arr => arr[0] !== 'quux');
  map.set(key, val);
});

console.log(map);

Automatically adjusting maps according to internal criteria

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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