Tips for filtering data using an array within an object containing arrays

Question

Tips for filtering data using an array within an object containing arrays

Below is the provided information:

userRoom = ['rm1'];

    data = [{
      name: 'building 1',
      building: [{
       room: 'rm1',
       name: 'Room 1'
      },{
       room: 'rm2',
       name: 'Room 2'
      }]
     },{
      name: 'building 2',
      building: [{
       room: 'rm3',
       name: 'Room 3'
      }]
     },{
      name: 'building 3',
      building: [{
       room: 'rm1',
       name: 'Room 1'
      }]
     }]

The objective here is to display the data that includes a building with room1 and apply a filter.

The expected result should look like this:

 [{
      name: 'building 1',
      building: [{
       room: 'rm1',
       name: 'Room 1'
      },{
       room: 'rm2',
       name: 'Room 2'
      }]
     },{
      name: 'building 3',
      building: [{
       room: 'rm1',
       name: 'Room 1'
      }]
     }]

I attempted to achieve this by utilizing the following code snippet:

data.map(x => x['building'].filter(y=> userRoom.includes(y.room));

However, I encountered issues with its functionality.

javascript typescript

Answer 1

Answer №1

In my earlier comment, I suggested utilizing the filter, followed by some, and then includes for a more organized solution:

userRoom = ['rm1'];

data = [{
  name: 'building 1',
  building: [
    { room: 'rm1', name: 'Room 1' },
    { room: 'rm2', name: 'Room 2' }
  ]
}, {
  name: 'building 2',
  building: [
    { room: 'rm3', name: 'Room 3' }
  ]
}, {
  name: 'building 3',
  building: [
    { room: 'rm1', name: 'Room 1' }
  ]
}];

let filtered = data.filter(i => 
  i.building.some(j => 
    userRoom.includes(j.room)));

console.log(filtered);

This approach works effectively due to:

The need to selectively filter the entries in the data array based on specific criteria.
For a building to pass the filtering process, at least one room within it must match certain conditions as determined by the some function.
The condition being that the room's identifier must be part of the userRoom array, hence the usage of includes.

Answer 2

In my earlier comment, I suggested utilizing the filter, followed by some, and then includes for a more organized solution:

userRoom = ['rm1'];

data = [{
  name: 'building 1',
  building: [
    { room: 'rm1', name: 'Room 1' },
    { room: 'rm2', name: 'Room 2' }
  ]
}, {
  name: 'building 2',
  building: [
    { room: 'rm3', name: 'Room 3' }
  ]
}, {
  name: 'building 3',
  building: [
    { room: 'rm1', name: 'Room 1' }
  ]
}];

let filtered = data.filter(i => 
  i.building.some(j => 
    userRoom.includes(j.room)));

console.log(filtered);

This approach works effectively due to:

The need to selectively filter the entries in the data array based on specific criteria.
For a building to pass the filtering process, at least one room within it must match certain conditions as determined by the some function.
The condition being that the room's identifier must be part of the userRoom array, hence the usage of includes.

Answer 3

Answer №2

Using Array.map won't work in this scenario, as it always returns a node for each element in the input array. Consider using a different method of iteration.

Instead, you can achieve the desired outcome by combining Array.filter with Array.some.

const userRoom = ['rm1'];
const data = [{"name":"building 1","building":[{"room":"rm1","name":"Room 1"},{"room":"rm2","name":"Room 2"}]},{"name":"building 2","building":[{"room":"rm3","name":"Room 3"}]},{"name":"building 3","building":[{"room":"rm1","name":"Room 1"}]}];
const output = data.filter((curr) => curr['building'].some(y=> userRoom.includes(y.room)));
console.log(output);

To solve this problem, apply a filter on the data array based on specific conditions.

Your condition is valid, so utilize Array.some to filter the data accordingly.

Answer 4

Using Array.map won't work in this scenario, as it always returns a node for each element in the input array. Consider using a different method of iteration.

Instead, you can achieve the desired outcome by combining Array.filter with Array.some.

const userRoom = ['rm1'];
const data = [{"name":"building 1","building":[{"room":"rm1","name":"Room 1"},{"room":"rm2","name":"Room 2"}]},{"name":"building 2","building":[{"room":"rm3","name":"Room 3"}]},{"name":"building 3","building":[{"room":"rm1","name":"Room 1"}]}];
const output = data.filter((curr) => curr['building'].some(y=> userRoom.includes(y.room)));
console.log(output);

To solve this problem, apply a filter on the data array based on specific conditions.

Your condition is valid, so utilize Array.some to filter the data accordingly.

Answer 5

Answer №3

Implementing some, filter, and includes

const data = [{"name":"building 1","building":[{"room":"rm1","name":"Room 1"},{"room":"rm2","name":"Room 2"}]},{"name":"building 2","building":[{"room":"rm3","name":"Room 3"}]},{"name":"building 3","building":[{"room":"rm1","name":"Room 1"}]}];

const anyUserRoomInBuilding = 
(building, userRoom) => building.some(({room}) => userRoom.includes(room));

const getBuildingsByUserRoom = 
(data, userRoom) => data.filter(({building}) => anyUserRoomInBuilding(building, userRoom))

console.log(getBuildingsByUserRoom(data, ['rm1']));

// console.log(getBuildingsByUserRoom(data, ['rm3', 'rm1']));

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

Answer 6

Implementing some, filter, and includes

const data = [{"name":"building 1","building":[{"room":"rm1","name":"Room 1"},{"room":"rm2","name":"Room 2"}]},{"name":"building 2","building":[{"room":"rm3","name":"Room 3"}]},{"name":"building 3","building":[{"room":"rm1","name":"Room 1"}]}];

const anyUserRoomInBuilding = 
(building, userRoom) => building.some(({room}) => userRoom.includes(room));

const getBuildingsByUserRoom = 
(data, userRoom) => data.filter(({building}) => anyUserRoomInBuilding(building, userRoom))

console.log(getBuildingsByUserRoom(data, ['rm1']));

// console.log(getBuildingsByUserRoom(data, ['rm3', 'rm1']));

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

Answer 7

Answer №4

const filteredData = data.filter(item => {
  return item.building.filter(buildingItem => {
    return userRooms.includes(buildingItem.room);
  }).length > 0;
});

userRooms = ['rm1'];

data = [{
  name: 'building 1',
  building: [{
    room: 'rm1',
    name: 'Room 1'
  }, {
    room: 'rm2',
    name: 'Room 2'
  }]
}, {
  name: 'building 2',
  building: [{
    room: 'rm3',
    name: 'Room 3'
  }]
}, {
  name: 'building 3',
  building: [{
    room: 'rm1',
    name: 'Room 1'
  }]
}]

const filteredData = data.filter(item => {
  return item.building.filter(buildingItem => {
    return userRooms.includes(buildingItem.room);
  }).length > 0;
});

console.log(filteredData);

Answer 8

const filteredData = data.filter(item => {
  return item.building.filter(buildingItem => {
    return userRooms.includes(buildingItem.room);
  }).length > 0;
});

userRooms = ['rm1'];

data = [{
  name: 'building 1',
  building: [{
    room: 'rm1',
    name: 'Room 1'
  }, {
    room: 'rm2',
    name: 'Room 2'
  }]
}, {
  name: 'building 2',
  building: [{
    room: 'rm3',
    name: 'Room 3'
  }]
}, {
  name: 'building 3',
  building: [{
    room: 'rm1',
    name: 'Room 1'
  }]
}]

const filteredData = data.filter(item => {
  return item.building.filter(buildingItem => {
    return userRooms.includes(buildingItem.room);
  }).length > 0;
});

console.log(filteredData);

Tips for filtering data using an array within an object containing arrays

Answer №1

Answer №2

Answer №3

Answer №4

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