How to get an array of unique values of a specific key provided I only have an array of id's of that object...

I have a store set up for a list of candidate objects like this

const candidatesStore = [

   { id: 1, status: active}

   {id:2, status: inactive}

   {id: 3 status: failed}

   {id: 4 status:inactive}

   {id: 5 status:failed}



    .

    ... and so on ]

I only have the list of id's in an array, ex: requiredArray = [2,3,4]

How to get a list of unique statuses by just using requiredArray using ES6 in the most efficient way

The result should be [inactive, failed]
I preferably want to use reduce to get unique value, and find to get the list of candidates from candidatesStore

edited Nov 13 '18 at 22:23

asked Nov 12 '18 at 19:53

Anand Dharne

What's going on with these candidate1 = assignments in your array literal?
– Bergi
Nov 12 '18 at 19:59

How about Array.from(new Set(candidatesStore.filter(cand => requiredArray.includes(cand.id)).map(cand => cand.status)))? I don't see how reduce or find could be useful here.
– Bergi
Nov 12 '18 at 20:00

add a comment |

I have a store set up for a list of candidate objects like this

const candidatesStore = [

   { id: 1, status: active}

   {id:2, status: inactive}

   {id: 3 status: failed}

   {id: 4 status:inactive}

   {id: 5 status:failed}



    .

    ... and so on ]

I only have the list of id's in an array, ex: requiredArray = [2,3,4]

How to get a list of unique statuses by just using requiredArray using ES6 in the most efficient way

The result should be [inactive, failed]
I preferably want to use reduce to get unique value, and find to get the list of candidates from candidatesStore

edited Nov 13 '18 at 22:23

asked Nov 12 '18 at 19:53

Anand Dharne

What's going on with these candidate1 = assignments in your array literal?
– Bergi
Nov 12 '18 at 19:59

How about Array.from(new Set(candidatesStore.filter(cand => requiredArray.includes(cand.id)).map(cand => cand.status)))? I don't see how reduce or find could be useful here.
– Bergi
Nov 12 '18 at 20:00

add a comment |

I have a store set up for a list of candidate objects like this

const candidatesStore = [

   { id: 1, status: active}

   {id:2, status: inactive}

   {id: 3 status: failed}

   {id: 4 status:inactive}

   {id: 5 status:failed}



    .

    ... and so on ]

I only have the list of id's in an array, ex: requiredArray = [2,3,4]

How to get a list of unique statuses by just using requiredArray using ES6 in the most efficient way

The result should be [inactive, failed]
I preferably want to use reduce to get unique value, and find to get the list of candidates from candidatesStore

edited Nov 13 '18 at 22:23

asked Nov 12 '18 at 19:53

Anand Dharne

I have a store set up for a list of candidate objects like this

const candidatesStore = [

   { id: 1, status: active}

   {id:2, status: inactive}

   {id: 3 status: failed}

   {id: 4 status:inactive}

   {id: 5 status:failed}



    .

    ... and so on ]

I only have the list of id's in an array, ex: requiredArray = [2,3,4]

How to get a list of unique statuses by just using requiredArray using ES6 in the most efficient way

The result should be [inactive, failed]
I preferably want to use reduce to get unique value, and find to get the list of candidates from candidatesStore

javascript ecmascript-6

edited Nov 13 '18 at 22:23

asked Nov 12 '18 at 19:53

Anand Dharne

edited Nov 13 '18 at 22:23

asked Nov 12 '18 at 19:53

Anand Dharne

edited Nov 13 '18 at 22:23

asked Nov 12 '18 at 19:53

Anand Dharne

asked Nov 12 '18 at 19:53

Anand Dharne

asked Nov 12 '18 at 19:53

Anand Dharne

What's going on with these candidate1 = assignments in your array literal?
– Bergi
Nov 12 '18 at 19:59

How about Array.from(new Set(candidatesStore.filter(cand => requiredArray.includes(cand.id)).map(cand => cand.status)))? I don't see how reduce or find could be useful here.
– Bergi
Nov 12 '18 at 20:00

add a comment |

What's going on with these candidate1 = assignments in your array literal?
– Bergi
Nov 12 '18 at 19:59

How about Array.from(new Set(candidatesStore.filter(cand => requiredArray.includes(cand.id)).map(cand => cand.status)))? I don't see how reduce or find could be useful here.
– Bergi
Nov 12 '18 at 20:00

What's going on with these candidate1 = assignments in your array literal?
– Bergi
Nov 12 '18 at 19:59

How about Array.from(new Set(candidatesStore.filter(cand => requiredArray.includes(cand.id)).map(cand => cand.status)))? I don't see how reduce or find could be useful here.
– Bergi
Nov 12 '18 at 20:00

add a comment |

2 Answers
2

active

oldest

votes

You can use "reduce" as well as "filter" for this.

Note - I have to change your input object little bit as "candidate1 = ..." was not making much sense here.

const candidates = [

    {id: 1, status: 'active'}

    ,{id: 2, status: 'inactive'}

    ,{id: 3, status: 'failed'}

    ,{id: 4, status: 'inactive'}

    ,{id: 5, status: 'failed'}

]



const requiredArray = [2,3,4]



let result = [...candidates.filter(({ id }) =>  requiredArray.includes(id))

                           .reduce((s, d) => s.add(d.status) , new Set)

             ]



console.log(result)

edited Nov 13 '18 at 6:38

answered Nov 12 '18 at 20:01

Nitish Narang

2,948815

FWIW, you can also skip the .map step and do .reduce((s, d) => s.add(d.status) , new Set)
– Felix Kling
Nov 13 '18 at 1:44

Yeah Correct. Thank you for pointing Sir!! BTW been long time follower of your detailed answers, which has helped me understand JS better. Want to take this chance to Thank you for spreading knowledge!! 😊
– Nitish Narang
Nov 13 '18 at 3:57

1

I'm glad to hear that and you are very welcome :)
– Felix Kling
Nov 13 '18 at 17:13

add a comment |

Try this.

const candidates = [

    {id: 1, status: "active"},

    {id: 2, status: "inactive"},

    {id: 3, status: "failed"},

    {id: 4, status: "inactive"},

    {id: 5, status: "failed"},

];

let requiredArray = [2,3,4];

let newArray = candidates.filter(x=>requiredArray.includes(x.id))

console.log(newArray);

answered Nov 12 '18 at 20:04

eag845

878611

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

You can use "reduce" as well as "filter" for this.

Note - I have to change your input object little bit as "candidate1 = ..." was not making much sense here.

const candidates = [

    {id: 1, status: 'active'}

    ,{id: 2, status: 'inactive'}

    ,{id: 3, status: 'failed'}

    ,{id: 4, status: 'inactive'}

    ,{id: 5, status: 'failed'}

]



const requiredArray = [2,3,4]



let result = [...candidates.filter(({ id }) =>  requiredArray.includes(id))

                           .reduce((s, d) => s.add(d.status) , new Set)

             ]



console.log(result)

edited Nov 13 '18 at 6:38

answered Nov 12 '18 at 20:01

Nitish Narang

2,948815

FWIW, you can also skip the .map step and do .reduce((s, d) => s.add(d.status) , new Set)
– Felix Kling
Nov 13 '18 at 1:44

Yeah Correct. Thank you for pointing Sir!! BTW been long time follower of your detailed answers, which has helped me understand JS better. Want to take this chance to Thank you for spreading knowledge!! 😊
– Nitish Narang
Nov 13 '18 at 3:57

1

I'm glad to hear that and you are very welcome :)
– Felix Kling
Nov 13 '18 at 17:13

add a comment |

You can use "reduce" as well as "filter" for this.

Note - I have to change your input object little bit as "candidate1 = ..." was not making much sense here.

const candidates = [

    {id: 1, status: 'active'}

    ,{id: 2, status: 'inactive'}

    ,{id: 3, status: 'failed'}

    ,{id: 4, status: 'inactive'}

    ,{id: 5, status: 'failed'}

]



const requiredArray = [2,3,4]



let result = [...candidates.filter(({ id }) =>  requiredArray.includes(id))

                           .reduce((s, d) => s.add(d.status) , new Set)

             ]



console.log(result)

edited Nov 13 '18 at 6:38

answered Nov 12 '18 at 20:01

Nitish Narang

2,948815

FWIW, you can also skip the .map step and do .reduce((s, d) => s.add(d.status) , new Set)
– Felix Kling
Nov 13 '18 at 1:44

Yeah Correct. Thank you for pointing Sir!! BTW been long time follower of your detailed answers, which has helped me understand JS better. Want to take this chance to Thank you for spreading knowledge!! 😊
– Nitish Narang
Nov 13 '18 at 3:57

1

I'm glad to hear that and you are very welcome :)
– Felix Kling
Nov 13 '18 at 17:13

add a comment |

You can use "reduce" as well as "filter" for this.

Note - I have to change your input object little bit as "candidate1 = ..." was not making much sense here.

const candidates = [

    {id: 1, status: 'active'}

    ,{id: 2, status: 'inactive'}

    ,{id: 3, status: 'failed'}

    ,{id: 4, status: 'inactive'}

    ,{id: 5, status: 'failed'}

]



const requiredArray = [2,3,4]



let result = [...candidates.filter(({ id }) =>  requiredArray.includes(id))

                           .reduce((s, d) => s.add(d.status) , new Set)

             ]



console.log(result)

edited Nov 13 '18 at 6:38

answered Nov 12 '18 at 20:01

Nitish Narang

2,948815

You can use "reduce" as well as "filter" for this.

Note - I have to change your input object little bit as "candidate1 = ..." was not making much sense here.

const candidates = [

    {id: 1, status: 'active'}

    ,{id: 2, status: 'inactive'}

    ,{id: 3, status: 'failed'}

    ,{id: 4, status: 'inactive'}

    ,{id: 5, status: 'failed'}

]



const requiredArray = [2,3,4]



let result = [...candidates.filter(({ id }) =>  requiredArray.includes(id))

                           .reduce((s, d) => s.add(d.status) , new Set)

             ]



console.log(result)

const candidates = [

    {id: 1, status: 'active'}

    ,{id: 2, status: 'inactive'}

    ,{id: 3, status: 'failed'}

    ,{id: 4, status: 'inactive'}

    ,{id: 5, status: 'failed'}

]



const requiredArray = [2,3,4]



let result = [...candidates.filter(({ id }) =>  requiredArray.includes(id))

                           .reduce((s, d) => s.add(d.status) , new Set)

             ]



console.log(result)

const candidates = [

    {id: 1, status: 'active'}

    ,{id: 2, status: 'inactive'}

    ,{id: 3, status: 'failed'}

    ,{id: 4, status: 'inactive'}

    ,{id: 5, status: 'failed'}

]



const requiredArray = [2,3,4]



let result = [...candidates.filter(({ id }) =>  requiredArray.includes(id))

                           .reduce((s, d) => s.add(d.status) , new Set)

             ]



console.log(result)

edited Nov 13 '18 at 6:38

answered Nov 12 '18 at 20:01

Nitish Narang

2,948815

edited Nov 13 '18 at 6:38

answered Nov 12 '18 at 20:01

Nitish Narang

2,948815

answered Nov 12 '18 at 20:01

Nitish Narang

2,948815

answered Nov 12 '18 at 20:01

Nitish Narang

2,948815

FWIW, you can also skip the .map step and do .reduce((s, d) => s.add(d.status) , new Set)
– Felix Kling
Nov 13 '18 at 1:44

Yeah Correct. Thank you for pointing Sir!! BTW been long time follower of your detailed answers, which has helped me understand JS better. Want to take this chance to Thank you for spreading knowledge!! 😊
– Nitish Narang
Nov 13 '18 at 3:57

1

I'm glad to hear that and you are very welcome :)
– Felix Kling
Nov 13 '18 at 17:13

add a comment |

FWIW, you can also skip the .map step and do .reduce((s, d) => s.add(d.status) , new Set)
– Felix Kling
Nov 13 '18 at 1:44

Yeah Correct. Thank you for pointing Sir!! BTW been long time follower of your detailed answers, which has helped me understand JS better. Want to take this chance to Thank you for spreading knowledge!! 😊
– Nitish Narang
Nov 13 '18 at 3:57

1

I'm glad to hear that and you are very welcome :)
– Felix Kling
Nov 13 '18 at 17:13

FWIW, you can also skip the .map step and do .reduce((s, d) => s.add(d.status) , new Set)
– Felix Kling
Nov 13 '18 at 1:44

Yeah Correct. Thank you for pointing Sir!! BTW been long time follower of your detailed answers, which has helped me understand JS better. Want to take this chance to Thank you for spreading knowledge!! 😊
– Nitish Narang
Nov 13 '18 at 3:57

I'm glad to hear that and you are very welcome :)
– Felix Kling
Nov 13 '18 at 17:13

add a comment |

Try this.

const candidates = [

    {id: 1, status: "active"},

    {id: 2, status: "inactive"},

    {id: 3, status: "failed"},

    {id: 4, status: "inactive"},

    {id: 5, status: "failed"},

];

let requiredArray = [2,3,4];

let newArray = candidates.filter(x=>requiredArray.includes(x.id))

console.log(newArray);

answered Nov 12 '18 at 20:04

eag845

878611

add a comment |

Try this.

const candidates = [

    {id: 1, status: "active"},

    {id: 2, status: "inactive"},

    {id: 3, status: "failed"},

    {id: 4, status: "inactive"},

    {id: 5, status: "failed"},

];

let requiredArray = [2,3,4];

let newArray = candidates.filter(x=>requiredArray.includes(x.id))

console.log(newArray);

answered Nov 12 '18 at 20:04

eag845

878611

add a comment |

Try this.

const candidates = [

    {id: 1, status: "active"},

    {id: 2, status: "inactive"},

    {id: 3, status: "failed"},

    {id: 4, status: "inactive"},

    {id: 5, status: "failed"},

];

let requiredArray = [2,3,4];

let newArray = candidates.filter(x=>requiredArray.includes(x.id))

console.log(newArray);

answered Nov 12 '18 at 20:04

eag845

878611

Try this.

const candidates = [

    {id: 1, status: "active"},

    {id: 2, status: "inactive"},

    {id: 3, status: "failed"},

    {id: 4, status: "inactive"},

    {id: 5, status: "failed"},

];

let requiredArray = [2,3,4];

let newArray = candidates.filter(x=>requiredArray.includes(x.id))

console.log(newArray);

const candidates = [

    {id: 1, status: "active"},

    {id: 2, status: "inactive"},

    {id: 3, status: "failed"},

    {id: 4, status: "inactive"},

    {id: 5, status: "failed"},

];

let requiredArray = [2,3,4];

let newArray = candidates.filter(x=>requiredArray.includes(x.id))

console.log(newArray);

const candidates = [

    {id: 1, status: "active"},

    {id: 2, status: "inactive"},

    {id: 3, status: "failed"},

    {id: 4, status: "inactive"},

    {id: 5, status: "failed"},

];

let requiredArray = [2,3,4];

let newArray = candidates.filter(x=>requiredArray.includes(x.id))

console.log(newArray);

answered Nov 12 '18 at 20:04

eag845

878611

answered Nov 12 '18 at 20:04

eag845

878611

answered Nov 12 '18 at 20:04

eag845

878611

answered Nov 12 '18 at 20:04

eag845

878611

add a comment |

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