Set default values in a JavaScript hash using values from an Array as keys

Question

Suppose I have a hash h and an array a like this:

h = {'data': {}};
arr = ['2017-01-01', '2017-02-01','2017-03-01', '2017-04-01', ....];

What is the most concise and efficient way to construct a default hash table as below in JavaScript:

// desired outcome for 'h'
h = {
 'data': {
    '2017-01-01': 0,
    '2017-02-01': 0,
    '2017-03-01': 0,
    '2017-04-01': 0,
    //...there can be more date values here
  }
}

I have implemented the solution below, but would like to know if there is more JavaScript-y (and hopefully more efficient) way to accomplish below:

arr.forEach(function(a) {
  h['data'][a] = 0;
});

Thanks in advance for your suggestion/answers!

arr.forEach(e => h.data[e] = 0 )

Taki
– Taki

2018-06-13 06:23:47 +00:00
Commented Jun 13, 2018 at 6:23 — Taki
– Taki, Commented Jun 13, 2018 at 6:23

Muhammad Usman · Accepted Answer · 2018-06-13 06:11:28Z

1

You can first convert your array into object with .reduce like

arr.reduce((o, key) => ({ ...o, [key]: 0}), {})

this will return you an object of form

 {
    '2017-01-01': 0,
    '2017-02-01': 0,
    '2017-03-01': 0,
    '2017-04-01': 0,
    //...there can be more date values here
  }

Now, you can simply assign this object to your h.data with Object.assign like

var h = {'data': {}};
var arr = ['2017-01-01', '2017-02-01','2017-03-01', '2017-04-01'];

h.data = Object.assign(arr.reduce((o, key) => ({ ...o, [key]: 0}), {}));

console.log(h)

ps: in case you don't know ... is called spread operator

answered Jun 13, 2018 at 6:11

Muhammad Usman

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3 Comments

user1330974 Over a year ago

Thank you for explaining step by step! In case you know, what would be more computationally efficient: my naive approach or your advanced one (assuming a size-able default array to set)? Or they are almost equal? I will wait for a few more hours to see who comments with more suggestion and then accept one of the answers. Hope you understand. :)

Muhammad Usman Over a year ago

what would be more computationally efficient , the one you're using should be more efficient. I just added another way to do it. :)

user1330974 Over a year ago

Thank you for the follow-up response. :) I'm glad that I learned a more advanced way to do it.

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