JavaScript: Convert array of objects into hashmap in Generic

Question

I have a set of values in an array where each value has an ID and LABEL.

const data = [
{ID: 0, LABEL: 'turbo'},
{ID: 1, LABEL: 'classic'},
{ID: 7, LABEL: 'unknown'}
];

const hashMap = data.reduce((result, item) => {
  return { ...result, [ item.ID ] : item.LABEL };
}, {});

const hashMapJson = JSON.stringify(hashMap);

console.log('hashMap', hashMap);
console.log('hashMapJson', hashMapJson);

But if I have the following json object where key is slightly different, it does not handle of course. I wonder how I could able to make the above solution in more generic to tackle different inputs.

const data = [
{K_ID: 0, K_LABEL: 'turbo'},
{L_ID: 1, K_LABEL: 'classic'},
{S_ID: 7, K_LABEL: 'unknown'}
];

do you have same pattern with ID and LABEL? please add the wanted result as well. — Nina Scholz
– Nina Scholz, Commented Jun 25, 2019 at 17:48
You could use Array#map to change all your objects to the first form: const normalizedData = data.map( obj => ({ [obj[Object.keys( obj ).find( key => /_ID$/.test(key) )]]: obj[Object.keys( obj ).find( key => /_LABEL$/.test(key) )] }) );, then convert it to a map the same way you already do: const hashMap = nomalizedData.reduce( ... — Paul
– Paul, Commented Jun 25, 2019 at 17:51

Paul · Accepted Answer · 2019-06-25 17:58:38Z

2

You could change your hashMap function to something like the following :

const data = [{K_ID: 0, K_LABEL: 'turbo'},{L_ID: 1, K_LABEL: 'classic'},{S_ID: 7, K_LABEL: 'unknown'}];

const hashMap = data.reduce((result, item) => {
    let id
    let label

    Object.keys(item).forEach((key) => {
        if (key.includes('ID')) {
            id = key
        } else if (key.includes('LABEL')) {
            label = key
        }
    })

    return { ...result, [ item[id] ] : item[label] };
}, {});

console.log( hashMap );

This is assuming your data set you are passing (data) is similar the the one in the example you provided. As long as a key contains ID, it'll make that the id key and as long as a key contains LABEL it will make that the key for the LABEL

edited Jun 25, 2019 at 17:58

Paul

142k28 gold badges285 silver badges272 bronze badges

answered Jun 25, 2019 at 17:54

Jonathan Beadle

4182 silver badges8 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Paul Over a year ago

I think this is the best answer, but you might want to include the underscore in your searches to handle some weird cases like {LABEL_ID: 121, LABEL_LABEL: 'label'}

BJRINT · Accepted Answer · 2019-06-25 17:57:19Z

2

It really depends of your data model, but this should work...

const data = [
{K_ID: 0, K_LABEL: 'turbo'},
{L_ID: 1, K_LABEL: 'classic'},
{S_ID: 7, K_LABEL: 'unknown'}
]

const hashMap = data.reduce((ac, cv) => {
  const keys = Object.keys(cv)
  let id, label
  keys.forEach(key => {
	if(key.endsWith('_LABEL'))
		label = cv[key]
	else if(key.endsWith('_ID'))
		id = cv[key]
  })
  
  if(id !== undefined && label !== undefined)
	ac[id] = label
  
  return ac
}, {})

console.log(hashMap)

answered Jun 25, 2019 at 17:57

BJRINT

6673 silver badges9 bronze badges

Comments

Nina Scholz · Accepted Answer · 2019-06-25 17:58:19Z

You could take a function for finding a similar key of the object.

const 
    getSimilarKey = (object, key) => Object.keys(object).find(k => k.toLowerCase().includes(key.toLowerCase())),
    data = [{ K_ID: 0, K_LABEL: 'turbo' }, { L_ID: 1, K_LABEL: 'classic' }, { S_ID: 7, K_LABEL: 'unknown' }],
    key = 'id', 
    value = 'label',
    result = Object.assign(
        {},
        ...data.map(o => ({ [o[getSimilarKey(o, key)]]: o[getSimilarKey(o, value)] }))
    );

console.log(result);

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

With upcoming Object.fromEntries

const 
    getSimilarKey = (object, key) => Object.keys(object).find(k => k.toLowerCase().includes(key.toLowerCase())),
    data = [{ K_ID: 0, K_LABEL: 'turbo' }, { L_ID: 1, K_LABEL: 'classic' }, { S_ID: 7, K_LABEL: 'unknown' }],
    key = 'id', 
    value = 'label',
    result = Object.fromEntries(data.map(o => ({ [o[getSimilarKey(o, key)]]: o[getSimilarKey(o, value)] })));

console.log(result);

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

Nikhil Aggarwal · Accepted Answer · 2019-06-25 18:00:25Z

2

You can try following using String.endsWith

const data = [{K_ID: 0, K_LABEL: 'turbo'},{L_ID: 1, K_LABEL: 'classic'},{S_ID: 7, K_LABEL: 'unknown'}];

const hashMap = data.reduce((result, item) => {
  const temp = Object.keys(item);
  const key = temp[0].endsWith("_ID") ? 0 : 1; // check for 1st key ending with _ID
  const value = +!key; // if key is 1, value be 0 and other way around
  result[item[temp[key]]] = item[temp[value]];
  return result;
}, {});

const hashMapJson = JSON.stringify(hashMap);
console.log(hashMapJson);

edited Jun 25, 2019 at 18:00

answered Jun 25, 2019 at 17:51

Nikhil Aggarwal

28.5k4 gold badges46 silver badges61 bronze badges

5 Comments

Paul Over a year ago

This assumes the objects all only have two keys (probably a valid assumption).

Nikhil Aggarwal Over a year ago

@Paulpro - temp[0].indexOf("_ID") === temp[0].length - 3 will work for AA_ID as well

Nikhil Aggarwal Over a year ago

@Paulpro - I have though updated to endsWith which is what I wanted to achieve with indexOf and length

Pietro Nadalini Over a year ago

why are you using "_ID"? it won't work on the OP's base example unless its "ID"

Nikhil Aggarwal Over a year ago

In the comments OP has clarified the pattern will be SOMETHING_ID.

junvar · Accepted Answer · 2019-06-25 19:14:40Z

1

You can make create an extractIdLabel function that tries to guess the keys for ID and Label respectively. You can use regex (via string.prototype.match) or other tools to make it as robust as required.

In the example below,

1) it looks for the keys 'id' and 'label'. e.g. {id: 0, label: 'zero'}

2) If one or both are not found, then it looks for keys containing the strings 'id' and 'label'. e.g. {x_id: 0, y_label_z: 'zero'}

3) If still one of them is not found, but the other is found, and the item has 2 keys, it uses the unused key for the not-found one. e.g. {id: 0, obviously_not_id_so_must_be_L_a_b_e_l: 'zero'}

const data = [
  {K_ID: -1, K_LABEL: 'turbo'},
  {L_ID: 1, K_LABEL: 'classic'},
  {S_ID: 7, K_LABEL: 'unknown'},
  {ID: 10, LABEL: 'ten', garbage: 'garbage', bad_label: 'bad_label'},
  {KEY: 11, X_LABEL: 'eleven'},
  {Y_ID: 12, VALUE: 'twelve'},
  {id: 0, obviously_not_id_so_must_be_L_a_b_e_l: 'zero'}
];

let extractIdLabel = item => {
  let keys = Object.keys(item);
  let id = keys.find(key => key.toLowerCase() === 'id') ||
      keys.find(key => key.toLowerCase().includes('id'));
  let label = keys.find(key => key.toLowerCase() === 'label') ||
      keys.find(key => key.toLowerCase().includes('label'));
  if (keys.length === 2 && !id ^ !label)
    if (!id)
      id = keys.find(key => key !== label);
    else
      label = keys.find(key => key !== id);
  return [item[id], item[label]];
};

const hashMap = data.reduce((result, item) => {
  let [id, label] = extractIdLabel(item);
  return {...result, [id]: label};
}, {});

const hashMapJson = JSON.stringify(hashMap);

console.log('hashMap', hashMap);
console.log('hashMapJson', hashMapJson);

edited Jun 25, 2019 at 19:14

answered Jun 25, 2019 at 17:57

junvar

11.6k2 gold badges35 silver badges52 bronze badges

8 Comments

Paul Over a year ago

I'm not sure why you used an XOR operator, but no matter what operator you use !id is not useful, because it gives the same value for id === 0 as it does for no id being found.

Paul Over a year ago

I think you can just do something like this:

if ( id !== undefined && label === undefined && keys.length === 2 ) label = keys.find(key => key !== id); if ( label !== undefined && id === undefined && keys.length === 2 ) id = keys.find(key => key !== label);

junvar Over a year ago

!id ^ !label is a succinct way of checking if one of them was found, and the other is still undefined. Your suggestion isn't wrong, but it's like a paragraph when a few words would have done just fine.

Paul Over a year ago

I disagree, my suggestion is less lines, less nesting, and clearer code IMO. Your code is the equivalent of "if exactly one of these is set, then if the first one is not set, do X, otherwise do Y" which has an extra condition and is harder to understand than necessary (which "if" is the "otherwise" applying to?) Mine is "if the first one is not set and the second one is, do X. if the second one is set and the first one is not, do Y." and, in my opinion, it is much clearer what is happening in the latter. Also, my code works.

Paul Over a year ago

You could make yours work too with something like id === undefined ^ label === undefined and also changing the next if statement to if ( id === undefined )

|

Collectives™ on Stack Overflow

JavaScript: Convert array of objects into hashmap in Generic

5 Answers 5

1 Comment

Comments

Comments

5 Comments

8 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

1 Comment

Comments

Comments

5 Comments

8 Comments

Your Answer

Sign up or log in

Post as a guest

Related