How to create a object using two single arrays?

Question

I have two arrays. Those are

let x = [1,2,0,4,7,8,9,1,1,1];
let y = [10,50,80,70,906,80,70,80,15,11]

Important : x = y, That means 1 = 10, 2 = 50, 0 = 80 etc...

I want to find minimum and maximum value of x array and want to get y values related to that data, as well as 0 values (min value should greater than 0 value)

expected output is:

let res = { 
  min: {
    minVal: 1,
    minvalues: [10, 80, 15, 11]
  },
  max:  {
    minVal: 9,
    minvalues: [70]
  } 
  empty: [80]
}

Here is what I tried. this way problem is I couldn't get only value from filter method it gives object

const res = x.map((key, index) => {
  return {[key]: y[index]};
}, {});

let info = {
  min : {},
  max : {},
  empty : {}
}

let min = Math.min.apply(this, x.filter(Number));
let max = Math.max(...x);
  
info.min['min'] = min
info.min['minVal'] = res.filter((el, idx) => el[min])`

and where is labelsAndHistoData defined?

Chris G
– Chris G

2023-02-07 17:08:10 +00:00
Commented Feb 7, 2023 at 17:08 — Chris G
– Chris G, Commented Feb 7, 2023 at 17:08

Nina Scholz · Accepted Answer · 2023-02-07 18:02:36Z

1

You could group and collect min and max values, then destrucure the result and build a new object.

const
    x = [ 1,  2,  0 , 4,   7,  8,  9,  1 , 1,  1],
    y = [10, 50, 80, 70, 906, 80, 70, 80, 15, 11],
    { minVal, maxVal, 0: empty, [minVal]: minvalues, [maxVal]: maxvalues } = x.reduce((r, value, i) => {
        (r[value] ??= []).push(y[i]);
        if (value && r.minVal > value) r.minVal = value;
        if (value && r.maxVal < value) r.maxVal = value;
        return r;
    }, { minVal: Number.MAX_VALUE, maxVal: -Number.MAX_VALUE })
    result = { min: { minVal, minvalues }, max: { maxVal, maxvalues }, empty };

console.log(result);

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answered Feb 7, 2023 at 18:02

Nina Scholz

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Comments

Unmitigated · Accepted Answer · 2023-02-07 17:08:43Z

0

You can store the indexes of the minimum x values, then use those indexes to get the corresponding y values with Array#map.

let x = [1,2,0,4,7,8,9,1,1,1];
let y = [10,50,80,70,906,80,70,80,15,11];
let minIdx = [0], maxIdx = [0], empty = [];
for (let i = 1; i < x.length; i++) {
  if (x[i] === 0) {
    empty.push(y[i]);
    continue;
  }
  if (x[i] === x[minIdx[0]]) minIdx.push(i);
  else if (x[i] < x[minIdx[0]]) minIdx = [i];
  if (x[i] === x[maxIdx[0]]) maxIdx.push(i);
  else if (x[i] > x[maxIdx[0]]) maxIdx = [i];
}
let res = {
  min : {
    minVal: x[minIdx[0]],
    minvalues: minIdx.map(i => y[i])
  },
  max: {
    maxVal: x[maxIdx[0]],
    maxvalues: maxIdx.map(i => y[i])
  },
  empty
};
console.log(res);

answered Feb 7, 2023 at 17:08

Unmitigated

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Comments

NetByMatt · Accepted Answer · 2023-02-07 17:12:19Z

0

It looks like you're trying to do everything in one line. Let's examine finding the desired min values as separate steps.

// first remove zero values from the x array (since those appear to get their own special handing in empty
// This returns all values that are not zero in the x array
const cleanX = x.filter(val=>val !== 0);

// now find the minimum of the x array
// the ... operator takes and array and turns it into a list of parameters thus [1,2,3] becomes Math.min(1,2,3)
const xMin = Math.min(...cleanX);

// then get the values associated with the minimum X
// this filter returns only values that match xMin by using the index (position in array) to match the x and y arrays
// we have to run y.filter, to return the values from the y array
const xMinValues = y.filter((yValue, index) => x[index] === xMin);

// a portion of your result would then look like this
const res = {
  min: {
    minVal: xMin,
    minvalues: xMinValues
  },

A similar method can be used for max, and empty. But empty will only look for the fixed value of 0.

answered Feb 7, 2023 at 17:12

NetByMatt

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1 Comment

kevin Over a year ago

Thanks for your answer. I could get more idea from this

Eliseo · Accepted Answer · 2023-02-07 17:42:50Z

0

let x = [1, 2, 0, 4, 7, 8, 9, 1, 1, 1];
let y = [10, 50, 80, 70, 906, 80, 70, 80, 15, 11];

//group by "x" and sort
const group = x
  .reduce((a: any, b: number, index: number) => {
    const el = a.find((x) => x.value == b);
    if (!el) a.push({ value: b, data: [y[index]] });
    else el.data.push(y[index]);
    return a;
  }, [])
  .sort((a, b) => a.value - b.value);

//create an object result
const result = {
  min: {
    minVal: group[0].value == 0 ? group[1].value : group[0].value,
    minValues: group[0].value == 0 ? group[1].data : group[1].data,
  },
  max: {
    minVal: group[group.length - 1].value,
    minValues: group[group.length - 1].data,
  },
  empty: group[0].value == 0 ? group[0].data : [],
};

answered Feb 7, 2023 at 17:42

Eliseo

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1 Comment

kevin Over a year ago

Thanks for your answer. I could get more idea from this

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