Adding up the elements of an array that match with elements in another array

Question

Adding up the elements of an array that match with elements in another array

In my JavaScript code, I am facing a challenge that I have been struggling to solve. I have attempted various methods, but none have proven successful for me so far. The situation is as follows: I have two arrays of equal length. For example:

var years = [2010, 2011, 2009, 2008, 2010, 2011, 2007, 2008, 2008];
var money = [2, 3, 6, 5, 13, 8, 3, 9, 7];

Each element in the money array corresponds to an element at the same index in the years array, representing funds spent in that year. My goal is to create a new array of distinct years (removing duplicate values from the original years array) and a new money array with elements that are sums of the initial money values associated with each year.

years_new = [2010, 2011, 2009, 2008, 2007];
money_new = [2 + 13, 3 + 8, 6, 5 + 9 + 7, 3];

I'm looking for guidance on how to achieve this task efficiently. Any help would be greatly appreciated. Thank you!

javascript arrays

Answer 1

Answer №1

To create an object, you can employ the Array.prototype.reduce method. This will result in keys representing years and their corresponding values being the sum of the elements in the money array.

var newObj = years.reduce(function(obj, year, index) {
    obj[year] = (obj[year] || 0) + money[index];
    return obj;
}, {});

To obtain all the years, the Object.keys method can be utilized:

var newYears = Object.keys(newObj);
var newMoney = newYears.map(function(year) { return newObj[year] });

Answer 2

To create an object, you can employ the Array.prototype.reduce method. This will result in keys representing years and their corresponding values being the sum of the elements in the money array.

var newObj = years.reduce(function(obj, year, index) {
    obj[year] = (obj[year] || 0) + money[index];
    return obj;
}, {});

To obtain all the years, the Object.keys method can be utilized:

var newYears = Object.keys(newObj);
var newMoney = newYears.map(function(year) { return newObj[year] });

Answer 3

Answer №2

Here is an alternative method that utilizes the following helper functions (please note that any other similar uniq function can be used, as efficiency is not the primary focus here).

Helpers.

Array.prototype.uniq = function() {
  return this.reduce(function(result, current) {
    return result.indexOf(current) < 0 ? result.concat([current]) : result;
  }, []);
};

Array.prototype.indicesOf = function(element) {
  return this.reduce(function(indicesArray, currentItem, index) {
    return (currentItem === element) ? indicesArray.concat(index) : indicesArray;
  }, []);
};

Usage.

Now to select the unique years, all you need to do is call uniq:

var new_years = years.uniq();

To map the old money to the corresponding new ones, we iterate through each new year and sum up the money values based on their indices in the original array:

var new_money = new_years.map(function(new_year) {
  return years.indicesOf(new_year).reduce(function(total, ind) {
    return total + money[ind];
  }, 0);
});

Answer 4

Here is an alternative method that utilizes the following helper functions (please note that any other similar uniq function can be used, as efficiency is not the primary focus here).

Helpers.

Array.prototype.uniq = function() {
  return this.reduce(function(result, current) {
    return result.indexOf(current) < 0 ? result.concat([current]) : result;
  }, []);
};

Array.prototype.indicesOf = function(element) {
  return this.reduce(function(indicesArray, currentItem, index) {
    return (currentItem === element) ? indicesArray.concat(index) : indicesArray;
  }, []);
};

Usage.

Now to select the unique years, all you need to do is call uniq:

var new_years = years.uniq();

To map the old money to the corresponding new ones, we iterate through each new year and sum up the money values based on their indices in the original array:

var new_money = new_years.map(function(new_year) {
  return years.indicesOf(new_year).reduce(function(total, ind) {
    return total + money[ind];
  }, 0);
});

Answer 5

Answer №3

Here is a potential solution:

let years = [2010, 2011, 2009, 2008, 2010, 2011, 2007, 2008, 2008];
let money = [2, 3, 6, 5, 13, 8, 3, 9, 7];

const fixed_years = {};

for(let i = 0; i < years.length; i++) {
    if(fixed_years[years[i]]) {
        fixed_years[years[i]] += money[i]; 
    } else {
        fixed_years[years[i]] = money[i];
    }
}

console.log(fixed_years);

const new_years = Object.keys(fixed_years);
const new_money = Object.values(fixed_years);

console.log(new_money);

Answer 6

Here is a potential solution:

let years = [2010, 2011, 2009, 2008, 2010, 2011, 2007, 2008, 2008];
let money = [2, 3, 6, 5, 13, 8, 3, 9, 7];

const fixed_years = {};

for(let i = 0; i < years.length; i++) {
    if(fixed_years[years[i]]) {
        fixed_years[years[i]] += money[i]; 
    } else {
        fixed_years[years[i]] = money[i];
    }
}

console.log(fixed_years);

const new_years = Object.keys(fixed_years);
const new_money = Object.values(fixed_years);

console.log(new_money);

Answer 7

Answer №4

To keep track of counted values, consider using a temporary object.

var years = [2010, 2011, 2009, 2008, 2010, 2011, 2007, 2008, 2008],
    money = [2, 3, 6, 5, 13, 8, 3, 9, 7],
    count = {}, i = 0,
    years_new = [],
    money_new = [];

while (i < years.length && i < money.length) {
    count[years[i]] = (count[years[i]] || 0) + money[i];
    i++;
}
years_new = Object.keys(count);
money_new = years_new.map(function (a) { return count[a]; });

document.write('<pre>' + JSON.stringify(years_new, 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(money_new, 0, 4) + '</pre>');

If you prefer to count without any temporary storage, another method maintains the order of the original years array.

var years = [2010, 2011, 2009, 2008, 2010, 2011, 2007, 2008, 2008],
    money = [2, 3, 6, 5, 13, 8, 3, 9, 7],
    i = 0, p,
    years_new = [],
    money_new = [];

while (i < years.length && i < money.length) {
    p = years_new.indexOf(years[i]);
    if (~p) {
        money_new[p] += money[i];
    } else {
        years_new.push(years[i]);
        money_new.push(money[i]);
    }
    i++;
}
document.write('<pre>' + JSON.stringify(years_new, 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(money_new, 0, 4) + '</pre>');

A different approach involves reducing both arrays by comparing indexes of years and utilizing Array.prototype.indexOf. This method splices both arrays and incorporates the spliced value accordingly.

var years = [2010, 2011, 2009, 2008, 2010, 2011, 2007, 2008, 2008],
    money = [2, 3, 6, 5, 13, 8, 3, 9, 7],
    i = years.length, p,
    years_new = years,
    money_new = money;

while (i--) {
    p = years_new.indexOf(years_new[i]);
    if (p !== i) {
        years_new.splice(i, 1);
        money_new[p] += +money_new.splice(i, 1);
    }
}
document.write('<pre>' + JSON.stringify(years_new, 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(money_new, 0, 4) + '</pre>');

Answer 8

To keep track of counted values, consider using a temporary object.

var years = [2010, 2011, 2009, 2008, 2010, 2011, 2007, 2008, 2008],
    money = [2, 3, 6, 5, 13, 8, 3, 9, 7],
    count = {}, i = 0,
    years_new = [],
    money_new = [];

while (i < years.length && i < money.length) {
    count[years[i]] = (count[years[i]] || 0) + money[i];
    i++;
}
years_new = Object.keys(count);
money_new = years_new.map(function (a) { return count[a]; });

document.write('<pre>' + JSON.stringify(years_new, 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(money_new, 0, 4) + '</pre>');

If you prefer to count without any temporary storage, another method maintains the order of the original years array.

var years = [2010, 2011, 2009, 2008, 2010, 2011, 2007, 2008, 2008],
    money = [2, 3, 6, 5, 13, 8, 3, 9, 7],
    i = 0, p,
    years_new = [],
    money_new = [];

while (i < years.length && i < money.length) {
    p = years_new.indexOf(years[i]);
    if (~p) {
        money_new[p] += money[i];
    } else {
        years_new.push(years[i]);
        money_new.push(money[i]);
    }
    i++;
}
document.write('<pre>' + JSON.stringify(years_new, 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(money_new, 0, 4) + '</pre>');

A different approach involves reducing both arrays by comparing indexes of years and utilizing Array.prototype.indexOf. This method splices both arrays and incorporates the spliced value accordingly.

var years = [2010, 2011, 2009, 2008, 2010, 2011, 2007, 2008, 2008],
    money = [2, 3, 6, 5, 13, 8, 3, 9, 7],
    i = years.length, p,
    years_new = years,
    money_new = money;

while (i--) {
    p = years_new.indexOf(years_new[i]);
    if (p !== i) {
        years_new.splice(i, 1);
        money_new[p] += +money_new.splice(i, 1);
    }
}
document.write('<pre>' + JSON.stringify(years_new, 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(money_new, 0, 4) + '</pre>');

Adding up the elements of an array that match with elements in another array

Answer №1

Answer №2

Answer №3

Answer №4

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