JavaScript - Calculating values based on an array

Question

I have an array, pictured below, which contains a number of values which I would like to do some calculations on a then push to a separate array.

This is is an example of the array and data I an working with:

I would like to push this data into an array called formatted, This array must contain the following information:

name - taken from the name form the original array

min - the lowest value from the original array

max - the largest value from the original array

Q1 - this is the lower quartile which is calculated from the original array by adding the p30 and p40 values and dividing by 2

Q4 - this is is the upper quartile value which is calculated by adding the p70 and p80 values together and dividing by 2

median - this is calculated by adding the p50 and p60 values together and dividing by 2

This is an example of how I tried to solve this but as seen below I had a error on every value:

 var formated = [];
        for(var i=0; i < final.length; i++) {    
            formated.push({
                name: final[i].name,
                min: final[i][0],
                max: final[i].length -1,
                Q1: (final[1] + final[2] / 2),
                Q2: (final[5] + final[6] / 2),
                meidan: (final[3] + final[4] / 2)
            });
        }

tex · Accepted Answer · 2016-06-15 15:33:20Z

2

Assuming all of the objects always have the same p-values, this is what you'd need:

 var formated = [];
        for(var i=0; i < final.length; i++) {    
            formated.push({
                name: final[i].name,
                min: final[i].p20,
                max: final[i].p90,
                Q1: ((final[i].p30 + final[i].p40) / 2),
                Q2: ((final[i].p70 + final[i].p80) / 2),
                median: ((final[i].p50 + final[i].p60) / 2)
            });
        }

A more elegant version of the above, using Array.prototype.map():

 var formated = final.map(function(item){
     return {
       name: item.name,
       min: item.p20,
       max: item.p90,
       Q1: ((item.p30 + item.p40) / 2),
       Q2: ((item.p70 + item.p80) / 2),
       median: ((item.p50 + item.p60) / 2)
     };
   });

edited Jun 15, 2016 at 15:33

answered Jun 15, 2016 at 15:23

tex

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6 Comments

tex Over a year ago

@JoseHermosillaRodrigo: Already fixed! A simple mistake in translating the original code to code based on map(). Thanks for pointing it out, though.

Jose Hermosilla Rodrigo Over a year ago

@Tex no problem! I see your edit just when I finish commenting! Looks fine now!!

tex Over a year ago

@CathalO'Donnell Take a look at my corrected example and Jose's correct one using map(). That's definitely the method I prefer over the for loop.

brk Over a year ago

how do you know p20 will always have lowest value and p90 will have highest value?

tex Over a year ago

@user2181397 It seems very likely from the fact that he was pulling min from final[0] and was trying to pull max from the last key in each object in his original code, from his explanation of how to calculate Q1, Q2 and median, and from his data, which appears to be some sort of statistical dataset.

|

Davide Lorenzo MARINO · Accepted Answer · 2016-06-15 15:22:51Z

1

You have objects with different namings.

What you call

final[i][0]

Should be something like

final[i].p20

or similar. Same for all other references, infact in final[i] doesn't exists a property 0. So final[i][0] is undefined.

answered Jun 15, 2016 at 15:22

Davide Lorenzo MARINO

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Comments

Jose Hermosilla Rodrigo · Accepted Answer · 2016-06-15 15:25:53Z

1

You can use map function for this. Map return a new Array :

const array = [
{name : "0006", p20 : 13367.8, p30: 15654.2, p40 : 17199.8, p50 : 18715, p60 : 20285.4, p70 : 22654.6, p80 : 24910.2, p90 : 30394.2}
];

var formated = array.map(el=>({
  name: el.name,
  min: el.p20,
  max: el.p90,
  Q1: ((el.p30 + el.p40) / 2),
  Q2: ((el.p70 + el.p80) / 2),
  median: ((el.p50 + el.p60) / 2)
}));

console.log(formated);

answered Jun 15, 2016 at 15:25

Jose Hermosilla Rodrigo

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