How to find objects in array where inner array includes other objects by using lodash

Question

I have the following arrays:

const tasks = [
  {id: 0, name: 'a', tags: [{id: 0, name: 'q'}, {id: 7, name: 'i'}]},
  {id: 1, name: 'b', tags: [{id: 2, name: 'e'}, {id: 4, name: 't'}, {id: 2, name: 'e'}, {id: 0, name: 'q'}]},
  {id: 2, name: 'c', tags: []},
  {id: 3, name: 'd', tags: [{id: 7, name: 'i'}, {id: 3, name: 'r'}, {id: 0, name: 'q'}]},
  {id: 6, name: 'g', tags: [{id: 7, name: 'i'}, {id: 4, name: 't'}]},
]

const tags = [
  {id: 0, name: 'q'},
  {id: 1, name: 'w'},
  {id: 2, name: 'e'},
  {id: 3, name: 'r'},
  {id: 4, name: 't'},
  {id: 7, name: 'i'},
  {id: 11, name: 's'}
]

let selectedTags = [0, 5]

selectedTags is an Array of indexes of tags Array. Now I need to find all objects in tasks Array, where property tags includes ALL of the selected tags. So in this case the output should be:

let result = [
  {id: 0, name: 'a', tags: [{id: 0, name: 'q'}, {id: 7, name: 'i'}]},
  {id: 3, name: 'd', tags: [{id: 7, name: 'i'}, {id: 3, name: 'r'}, {id: 0, name: 'q'}]}
]

I tried to do something like this:

let result= []
_.forEach(selectedTags, index => {
  const tagId = tags[index]._id
  result = _.filter(tasks, task => _.find(task.tags, ['_id', tagId]))
})

But the result array included objects where tags array had only one of the selected tags (always the same tag).

Any ideas, please?

It looks similar to stackoverflow.com/questions/60468021/… — Ele
– Ele, Commented Mar 2, 2020 at 13:23
result gets overwritten in the forEach loop (that would explain why its always the same/last tag that works) — Roland Starke
– Roland Starke, Commented Mar 2, 2020 at 13:23
Your test data doesn't seem to have the results you say it should, your selected tags is [0, 5] but your expected results doesn't have 5, it has 7, is this a typo.? — Keith
– Keith, Commented Mar 2, 2020 at 13:27
@Keith I believe it is the object at index 5 from the tags array — Nick Parsons
– Nick Parsons, Commented Mar 2, 2020 at 13:28
@Ele Yes, it is similar. The difference is here I need to find objects with ALL selected tags, while in the other question it was ANY of the selected. — Igal
– Igal, Commented Mar 2, 2020 at 13:30

Keith · Accepted Answer · 2020-03-02 14:20:09Z

1

One way is to map your selectedTags into the equivalent tags array.

Next use every to make sure every item in the selectedTags exists in the tasks.tags array using some..

Example below..

const tasks = JSON.parse('[{"id":0,"name":"a","tags":[{"id":0,"name":"q"},{"id":7,"name":"i"}]},{"id":1,"name":"b","tags":[{"id":2,"name":"e"},{"id":4,"name":"t"},{"id":2,"name":"e"},{"id":0,"name":"q"}]},{"id":2,"name":"c","tags":[]},{"id":3,"name":"d","tags":[{"id":7,"name":"i"},{"id":3,"name":"r"},{"id":0,"name":"q"}]},{"id":6,"name":"g","tags":[{"id":7,"name":"i"},{"id":4,"name":"t"}]}]');
const tags = JSON.parse('[{"id":0,"name":"q"},{"id":1,"name":"w"},{"id":2,"name":"e"},{"id":3,"name":"r"},{"id":4,"name":"t"},{"id":7,"name":"i"},{"id":11,"name":"s"}]');
let selectedTags = [0, 5];


const result = tasks.filter(f =>
  selectedTags.map(m => tags[m]).every(
    e => f.tags.some(s => s.id === e.id)));

console.log(result);

ps. The map could be done separately if performance was a concern.

edited Mar 2, 2020 at 14:20

answered Mar 2, 2020 at 14:02

Keith

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1 Comment

Igal Over a year ago

Thank you very much! Worked flawlessly! I just rewrote it the lodash way:

let result = _.filter(tasks, task => _.every(_.map(selectedTags, i => tags[i]), sTag => _.some(task.tags, tag => tag.id === sTag.id)))

StepUp · Accepted Answer · 2020-03-02 14:36:21Z

It is possible to use reduce method and check on every iteration whether an element contains desired tags:

const tagsToFind = selectedTags.map(s => tags[s]);

const result = tasks.reduce((a, c) => {
    if (tagsToFind.every(t => c.tags.some(s => s.name == t.name)))
        a.push(c);
    return a;
},[])

An example:

const tasks = [
    {id: 0, name: 'a', tags: [{id: 0, name: 'q'}, {id: 7, name: 'i'}]},
    {id: 1, name: 'b', tags: [{id: 2, name: 'e'}, {id: 4, name: 't'}, 
        {id: 2, name: 'e'}, {id: 0, name: 'q'}]},
    {id: 2, name: 'c', tags: []},
    {id: 3, name: 'd', tags: [{id: 7, name: 'i'}, {id: 3, name: 'r'}, 
        {id: 0, name: 'q'}]},
    {id: 6, name: 'g', tags: [{id: 7, name: 'i'}, {id: 4, name: 't'}]},
];

const tags = [
    {id: 0, name: 'q'},
    {id: 1, name: 'w'},
    {id: 2, name: 'e'},
    {id: 3, name: 'r'},
    {id: 4, name: 't'},
    {id: 7, name: 'i'},
    {id: 11, name: 's'}
];

let selectedTags = [0, 5];
const tagsToFind = selectedTags.map(s => tags[s]);

const result = tasks.reduce((a, c) => {
    if (tagsToFind.every(t => c.tags.some(s => s.name == t.name)))
        a.push(c);
    return a;
},[])

console.log(result);

Siva Kondapi Venkata · Accepted Answer · 2020-03-02 15:36:44Z

0

First build names from selectedTags and use filter on tasks should work.

const filter = (tasks, tags, selectedTags) => {
  const names = selectedTags.map(tag => tags[tag].name);
  return tasks.filter(task =>
    names.every(name => task.tags.map(x => x.name).includes(name))
  );
};

const tasks = [
  {
    id: 0,
    name: "a",
    tags: [
      { id: 0, name: "q" },
      { id: 7, name: "i" }
    ]
  },
  {
    id: 1,
    name: "b",
    tags: [
      { id: 2, name: "e" },
      { id: 4, name: "t" },
      { id: 2, name: "e" },
      { id: 0, name: "q" }
    ]
  },
  { id: 2, name: "c", tags: [] },
  {
    id: 3,
    name: "d",
    tags: [
      { id: 7, name: "i" },
      { id: 3, name: "r" },
      { id: 0, name: "q" }
    ]
  },
  {
    id: 6,
    name: "g",
    tags: [
      { id: 7, name: "i" },
      { id: 4, name: "t" }
    ]
  }
];

const tags = [
  { id: 0, name: "q" },
  { id: 1, name: "w" },
  { id: 2, name: "e" },
  { id: 3, name: "r" },
  { id: 4, name: "t" },
  { id: 7, name: "i" },
  { id: 11, name: "s" }
];

let selectedTags = [0, 5];

console.log(filter(tasks, tags, selectedTags));

edited Mar 2, 2020 at 15:36

answered Mar 2, 2020 at 13:39

Siva Kondapi Venkata

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3 Comments

Igal Over a year ago

Thanks for your help! However the returned object also includes object with ID 1, which tags array don't have all the selected tags.

Igal Over a year ago

Yes, but it doesn't have tag 'i'.

Siva Kondapi Venkata Over a year ago

Ah my bad. [0, 5] didn't realize that these indexes. @Igal, Updated the answer with fix.

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