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Sort array of years (including BCE)

There is an array of objects containing a chemical element with it's discovery year:

[
   {"name": "hydrogen", "discovered": "1766"},
   {"name": "boron", "discovered": "1808"},
   {"name": "copper", "discovered": "9000 BCE"},
   {"name": "argon", "discovered": "1894"},
   {"name": "iron", "discovered": "before 5000 BCE"},
   {"name": "phosphorus", "discovered": "1669"}
]

What I want is to sort the array in terms of the year discovered, so the oldest (copper in this case) should be the first item and the recent(argon) should be the last item.

//It should look like this in the end:
[
   {"name": "copper", "discovered": "9000 BCE"},
   {"name": "iron", "discovered": "before 5000 BCE"},
   {"name": "phosphorus", "discovered": "1669"},
   {"name": "hydrogen", "discovered": "1766"},
   {"name": "boron", "discovered": "1808"},
   {"name": "argon", "discovered": "1894"}
]

The problem I encountered was the few keywords like "before", "BCE", "CE", that I don't know how to handle.

Is there any solution to handle this problem?

about 4 years ago · Juan Pablo Isaza
2 Respuestas
Responde la pregunta

0

9000 BCE is same as year -9000.

You can simply treat those BCE (keyword before might be ignored, cause it contains BCE anyway) data strings as -year in sort function, as follows:

const data = [{name:"hydrogen",discovered:"1766"},{name:"boron",discovered:"1808"},{name:"copper",discovered:"9000 BCE"},{name:"argon",discovered:"1894"},{name:"iron",discovered:"before 5000 BCE"},{name:"phosphorus",discovered:"1669"}];

const res = data.sort((a, b) => {
  const numA = (a.discovered.includes("BCE") ? -1 : 1)
              * a.discovered.replace(/\D/g, '')
  const numB = (b.discovered.includes("BCE") ? -1 : 1)
              * b.discovered.replace(/\D/g, '')
  return numA - numB
})

console.log(res)
.as-console-wrapper { max-height: 100% !important; top: 0; } /* ignore this */

about 4 years ago · Juan Pablo Isaza Denunciar

0

Alright, so first you need to create a function that will convert the dates to a common format. In this example I will convert it to a number relative to 0 (year 0):

const beforeRegExp = new RegExp('(before )\\d*( BCE)');
const bceRegExp = new RegExp('\\d*( BCE)');

function toCommonDateFormat(date) {
  if (beforeRegExp.test(date)) {
    return date.replace("before ", "").replace(" BCE", "") * -1
  } else if (bceRegExp.test(date)) {
    return date.replace(" BCE", "") * -1
  } else {
    return date * 1;
  }
}

Now we can use this function in a sort function:

function compareDates(firstDate, secondDate) {
  return toCommonDateFormat(firstDate) - toCommonDateFormat(secondDate);
}

And now, to tie it all together:

const list = [{
    "name": "hydrogen",
    "discovered": "1766"
  },
  {
    "name": "boron",
    "discovered": "1808"
  },
  {
    "name": "copper",
    "discovered": "9000 BCE"
  },
  {
    "name": "argon",
    "discovered": "1894"
  },
  {
    "name": "iron",
    "discovered": "before 5000 BCE"
  },
  {
    "name": "phosphorus",
    "discovered": "1669"
  }
]

const beforeRegExp = new RegExp('(before )\\d*( BCE)');
const bceRegExp = new RegExp('\\d*( BCE)');

function toCommonDateFormat(date) {
  if (beforeRegExp.test(date)) {
    return date.replace("before ", "").replace(" BCE", "") * -1
  } else if (bceRegExp.test(date)) {
    return date.replace(" BCE", "") * -1
  } else {
    return date * 1;
  }
}

function compareDates(firstDate, secondDate) {
  return toCommonDateFormat(firstDate) - toCommonDateFormat(secondDate);
}

const orderedList = list.sort(function(firstItem, secondItem) {
  return compareDates(firstItem.discovered, secondItem.discovered);
});

console.log(orderedList);

about 4 years ago · Juan Pablo Isaza Denunciar
Responde la pregunta
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