Dynamically restructuring an array of objects each containing two property value pairs

Question

I'm drawing data from a database and ending up with an array of objects each containing waiter names and the the days they have worked. I want to dynamically restructure the array by condensing it to a single object with two property value pairs per waiter. One property for the name and another for the days working.

I am really curious to see your feedback. I started off by extracting the names and removing duplicates. Leaving me with an array like this: let waiterNames = ['John','Mark','Jess'].

Then I was attempting a double for of loop and somehow creating a new array of objects, but I'm stuck.

for (entry in waiterInfo) {
    for (waiter in waiterNames) {
        if (entry.waiters == waiter) {
        ???
        }
    }
}

waiterInfo is what I start with, newInfo is what I want to achieve. I need this to happen dynamically because the data drawn from the database is unpredictable.

let waiterInfo = [{ waiters: 'John', weekdays: 'Monday' },
{ waiters: 'John', weekdays: 'Tuesday' },
{ waiters: 'John', weekdays: 'Wednesday' },
{ waiters: 'Mark', weekdays: 'Monday' },
{ waiters: 'Mark', weekdays: 'Tuesday' },
{ waiters: 'Jess', weekdays: 'Monday' },
{ waiters: 'Jess', weekdays: 'Tuesday' },
{ waiters: 'Jess', weekdays: 'Wednesday' },
{ waiters: 'Jess', weekdays: 'Thursday' }]

let newInfo = [{ waiters: 'John', weekdays: 'Monday, Tuesday, Wednesday'},
{ waiters: 'Mark', weekdays: 'Monday, Tuesday' },
{ waiters: 'Jess', weekdays: 'Monday, Tuesday, Wednesday, Thursday' }]

Hi, thank you four reply. The array is shown above - waiterInfo. I want to change it dynamically to look like newInfo. — Warwick
– Warwick, Commented Oct 28, 2019 at 10:59
Which DB do you use? Perhaps it would be easier to do this thing directly in the db. — georg
– georg, Commented Oct 28, 2019 at 11:33
This is a valid point - but I was curious to see how to solve this problem anyhow. I'm using postgreSQL. My table has four columns: id, waiter_username, weekdays_working and waiters_id which is unique for each waiter. I was attempting to write a query that would remove the duplicate waiters names, but was unsuccessful. This is the query I used which returned the waiterInfo array above -> select waiter_username as waiters, weekdays_working as weekdays from waiters order by waiter_username; — Warwick
– Warwick, Commented Oct 28, 2019 at 12:02
@Warwick: select waiter_username, array_agg(weekdays_working) as weekdays from waiters group by waiter_username should do the trick. — georg
– georg, Commented Oct 28, 2019 at 13:55

Kobe · Accepted Answer · 2019-10-28 11:20:18Z

2

You can use a map alongside reduce:

let waiterInfo = [{ waiters: 'John', weekdays: 'Monday' },
  { waiters: 'John', weekdays: 'Tuesday' },
  { waiters: 'John', weekdays: 'Wednesday' },
  { waiters: 'Mark', weekdays: 'Monday' },
  { waiters: 'Mark', weekdays: 'Tuesday' },
  { waiters: 'Jess', weekdays: 'Monday' },
  { waiters: 'Jess', weekdays: 'Tuesday' },
  { waiters: 'Jess', weekdays: 'Wednesday' },
  { waiters: 'Jess', weekdays: 'Thursday' }
]


const map = new Map()

const newInfo = waiterInfo.reduce((a, o) => {
  const i = map.get(o.waiters)
  if(i !== undefined) {
    a[i].weekdays = [a[i].weekdays, o.weekdays].join(', ')
  } else {
    map.set(o.waiters, a.push(o) - 1)
  }
  return a
}, [])

console.log(newInfo)

This code's time complexity is O(n), since we are using a map.

(Also, this is one of the only times I've seen push's return value be useful 😜)

edited Oct 28, 2019 at 11:20

answered Oct 28, 2019 at 11:00

Kobe

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3 Comments

Warwick Over a year ago

Incredible - you guys are very good. I am very inspired by your feedback!

Kobe Over a year ago

@Warwick No problem, glad to help. If you need more help, be it understanding how this code works or anything else, feel free to ask.

Warwick Over a year ago

I'm going on lunch now.. will attempt implementing and understanding the code into my program afterwards. Will post a question here if I don't understand - but think I can figure it out with all the information I've obtained in this thread. Thank you. =)

Yosvel Quintero · Accepted Answer · 2019-10-28 11:34:32Z

1

You can do:

const waiterInfo = [{ waiters: 'John', weekdays: 'Monday' },{ waiters: 'John', weekdays: 'Tuesday' },{ waiters: 'John', weekdays: 'Wednesday' },{ waiters: 'Mark', weekdays: 'Monday' },{ waiters: 'Mark', weekdays: 'Tuesday' },{ waiters: 'Jess', weekdays: 'Monday' },{ waiters: 'Jess', weekdays: 'Tuesday' },{ waiters: 'Jess', weekdays: 'Wednesday' },{ waiters: 'Jess', weekdays: 'Thursday' },]

const result = Object
  .values(
    waiterInfo.reduce((a, { waiters, weekdays }) => {
      a[waiters] = a[waiters] || { waiters, weekdays: [] }
      a[waiters].weekdays.push(weekdays)
      return a
    }, {})
  )
  .map(({ waiters, weekdays }) => ({ waiters, weekdays: weekdays.join(', ') }))

console.log(result)

edited Oct 28, 2019 at 11:34

answered Oct 28, 2019 at 11:00

Yosvel Quintero

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Comments

T.J. Crowder · Accepted Answer · 2019-10-28 11:01:07Z

The easiest way to do this is to use a temporary Map to keep track of the objects for the waiters you've seen before, so you can add to their list of days.

const waiterMap = new Map();
const newInfo = [];
for (const {waiters, weekdays} of waiterInfo) {
    let entry = waiterMap.get(waiters);
    if (!entry) {
        // New waiter, create the object and put it in the map and result array
        const copy = {waiters, weekdays};
        waiterMap.set(waiters, copy);
        newInfo.push(copy);
    } else {
        // Existing entry, just add to it
        entry.weekdays += `, ${weekdays}`;
    }
}

Live Example:

let waiterInfo = [{ waiters: 'John', weekdays: 'Monday' },
{ waiters: 'John', weekdays: 'Tuesday' },
{ waiters: 'John', weekdays: 'Wednesday' },
{ waiters: 'Mark', weekdays: 'Monday' },
{ waiters: 'Mark', weekdays: 'Tuesday' },
{ waiters: 'Jess', weekdays: 'Monday' },
{ waiters: 'Jess', weekdays: 'Tuesday' },
{ waiters: 'Jess', weekdays: 'Wednesday' },
{ waiters: 'Jess', weekdays: 'Thursday' }];

const waiterMap = new Map();
const newInfo = [];
for (const {waiters, weekdays} of waiterInfo) {
    let entry = waiterMap.get(waiters);
    if (!entry) {
        // New waiter, create the object and put it in the map and result array
        const copy = {waiters, weekdays};
        waiterMap.set(waiters, copy);
        newInfo.push(copy);
    } else {
        // Existing entry, just add to it
        entry.weekdays += `, ${weekdays}`;
    }
}

console.log(newInfo);

That avoids modifying the existing data, but you could reuse the original object if you don't care about waiterInfo once you're done.

T.J. Crowder magnificent! I am very impressed by your speed.

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