Map javascript array to array of height gaps

Question

I have a decently tricky javascript data manipulation problem that I am struggling with. I have the following array of data:

var myArray = [8, 13, 11, 17, 5, 13, 13];

And I would like to map this array into an array of equal length, where each value in the new array is equal to the difference between the array's previous value at that spot, and the next lowest value in the previous array. This is wordy, but for example, I want the output to be:

var newArray = [3, 2, 3, 4, 5, 0, 0];

3 in the first element is (8 - 5), where 8 is the array's previous value in the first slot, and 5 is the next lowest value in the previous array. 2 in the second element is (13 - 11), ... etc.

I'm not particularly caring how to handle ties (for example there are 2 values of 13 in the original array), so long as exactly one instance of 13 is replaced with (13 - next lowest value), and all other instances of 13 are replaced with zero (13 - 13 essentially).

Thanks!

I've spent time pondering the Q, but no I do not have code attempting a solution. This has the feeling of a Q where, once I have an approach on how to calculate it, the code is written in <1 minute. — Canovice
– Canovice, Commented Jun 4, 2018 at 19:58
I'm actively working on it and will certainly post if i come up with a solution before an answer is posted — Canovice
– Canovice, Commented Jun 4, 2018 at 19:59
@VishnuKyatannawar there will never be negative values in my dataset — Canovice
– Canovice, Commented Jun 4, 2018 at 19:59

charlietfl · Accepted Answer · 2018-06-04 20:57:22Z

3

Using a reduce() within map() and checking if number is first instance or not using indexOf()

var arr = [8, 13, 11, 17, 5, 13, 13];

var res = arr.map((n, i) => {
  if (i !== arr.indexOf(n)) return 0;
  return n - arr.reduce((a, c) => c < n && c > a ? c : a, 0);
})

console.log(JSON.stringify(res))/// [3, 2, 3, 4, 5, 0, 0]

Alternate approach that creates descending sorted uniques array and then looks for number in next index within that array

var arr = [8, 13, 11, 17, 5, 13, 13];
var sortedUniques  = [...new Set(arr)].sort((a,b) => b-a);

var res = arr.map((n, i) => {
  if (i !== arr.indexOf(n)) return 0;
  var sortedIdx = sortedUniques.indexOf(n);
  return n - ( sortedUniques[sortedIdx+1] || 0 );
})

console.log(JSON.stringify(res))/// [3, 2, 3, 4, 5, 0, 0]

edited Jun 4, 2018 at 20:57

answered Jun 4, 2018 at 20:14

charlietfl

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5 Comments

charlietfl Over a year ago

@VishnuKyatannawar why not? That seems like only an opinion. Was easy enough for me to read while writing it

Luis felipe De jesus Munoz Over a year ago

Code is very intuitive. I may do a challenge in codegolf with this question and take this answer as an example

Vishnu Kyatannawar Over a year ago

@charlietfl yup its just my opinion. Not used to writing code like this. This makes sense to me though.

charlietfl Over a year ago

@LuisfelipeDejesusMunoz think I have a more efficient idea. working on it

charlietfl Over a year ago

@LuisfelipeDejesusMunoz added what may be slightly more performant approach

wiesion · Accepted Answer · 2018-06-04 21:15:07Z

1

Big thanks to @charlietfl who helped me to understand the part about the duplicates :) In general i came up with an almost identical solution, save for using Array.filter and Math.max instead of minimizing that to a single Array.reduce statement.

let myArray = [8, 13, 11, 17, 5, 13, 13],
    cmpArray = myArray.concat([0]);

console.log(
  myArray.map((value, index) => {
    if(index !== myArray.indexOf(value)) return 0;
    return value - Math.max(...cmpArray.filter(number => number < value));
  })
);

edited Jun 4, 2018 at 21:15

answered Jun 4, 2018 at 20:41

wiesion

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4 Comments

wiesion Over a year ago

i see, i am having a hard time deciphering the requirements

charlietfl Over a year ago

if its a dup just return zero ... "all other instances of 13 are replaced with zero"

wiesion Over a year ago

can i replace all 13 with 0, even if there would be 3 times 13 in it?

charlietfl Over a year ago

First one found is supposed to be evaluated. That's why I used indexOf() to check

Nina Scholz · Accepted Answer · 2018-06-05 06:04:36Z

1

You could collect the next values in an array and all array in an object and pop the values until all values are taken.

var array = [8, 13, 11, 17, 5, 13, 13],
    temp = array
        .slice()
        .sort((a, b) => b - a)
        .reduce(
            (r, v, i, a) => ((r[v] = r[v] || []).push(v - (a[i + 1] || 0)), r),
            Object.create(null)
        ),
    result = array.map(v => temp[v].pop());

console.log(result);

edited Jun 5, 2018 at 6:04

answered Jun 4, 2018 at 21:51

Nina Scholz

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Comments

Canovice · Accepted Answer · 2018-06-04 20:12:59Z

0

Something along these lines is getting close for me.

let newArray = [];
myArray.forEach(function(d){
  let temp = myArray.filter(val => val < d)
  let maxRemVal = d3.max(temp)
  maxRemVal = isNaN(maxRemVal) ? 0 : maxRemVal;
  newArray.push(d - maxRemVal)
});

answered Jun 4, 2018 at 20:12

Canovice

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