I have an array
[
{ price: 10 },
{ price: 10 },
{ price: 10 },
{ price: 10 },
{ price: 20 },
{ price: 20 },
]
and I want it transformed into
[
{ numElements: 4, price: 10 },
{ numElements: 2, price: 20 },
]
I have tried using arr.reduce((prev, curr) => ..., []) to accomplish this, but I can't figure out how to do it.
A traditional method might use a for/loop to wrangle the data, but these days JavaScript has a number of functional methods that can help. This code uses reduce and map. To get your data in the format you want is a two stage process.
First, use reduce to create a hash table using the price as a key (because you know the each price is going to be unique:
const obj = arr.reduce((p, c) => {
// If price exists as a key its value by 1
// otherwise set it to 1.
p[c.price] = ++p[c.price] || 1;
return p;
}, {});
OUTPUT
{
"10": 4,
"20": 2
}
As it stands you've got a perfectly good object that you can access by the key/price and I would probably just stop there:
obj['10'] // 4
But if you want to get that data into the format in your question, map over the object keys to return an array of new objects.
const out = Object.keys(obj).map(key => {
return { price: +key, numElements: obj[key] };
});
DEMO
var hash = {}, result = [];
arr.forEach(function(el){
if(hash[el.price]){
hash[el.price].numElements++;
}else{
result.push(hash[el.price]={price:el.price,numElements:1});
}
});
Run
May use a hash table for price lookup. Or with reduce and find:
arr.reduce((res,{price})=>
(( res.find(el=>el.price===price) || res[res.push({price,numElements:0})-1] )
.numElements++,res)
);
Run
You can use try this:
let arr = [
{ price: 10 },
{ price: 10 },
{ price: 10 },
{ price: 10 },
{ price: 20 },
{ price: 20 },
]
let result = []
let counter = {}
arr.forEach( el => {
if (!counter[el.price]) counter[el.price] = 1
else counter[el.price]++
console.log(counter[el.price])
})
for (let id in counter) {
result.push({numElements: counter[id], price: id})
}
Assuming that the data comes sorted on price property, with a single .reduce() you may do as follows;
var data = [{ price: 10 }, { price: 10 }, { price: 10 }, { price: 10 }, { price: 20 }, { price: 20 }],
result = data.reduce((r,d,i) => i ? r[r.length-1].price === d.price ? (r[r.length-1].numElemenets++, r)
: (r.push(Object.assign({}, d, {numElemenets: 1})),r)
: [Object.assign({}, d, {numElemenets: 1})], {});
console.log(result);
You could look up the price in the result array and if not found insert a new object.
var data = [{ price: 10 }, { price: 10 }, { price: 10 }, { price: 10 }, { price: 20 }, { price: 20 }],
grouped = data.reduce((r, { price }) => {
var t = r.find(p => price === p.price);
t || r.push(t = { numElements: 0, price });
t.numElements++;
return r;
}, []);
console.log(grouped);
Related
Assuming I have 2 arrays:
const products = [
{
name: 'prod1',
category: 'Meat'
},
{
name: 'prod2',
category: 'Meat'
},
{
name: 'prod3',
category: 'Dairy'
}];
const rate = [
{
name: 'prod1',
rate: 23,
},
{
name: 'prod2',
rate: 36
},
{
name: 'prod3',
rate: 50,
}];
How would you get the category that has the highest sum rate? For example, prod1 and prod2 share the same category 'Meat" and hence the rate for meat is 36 + 23 = 59.
The way I thought about it is to create an adjusted Array of products where each entry will contain the rate from the second array and then I will create a result array and push an object of category and sumRate after iterating the adjustedArray.
So if the result Array has an object with category, I would adjust the sum and add the new rate, if not I'll create a new entry with category: rate.
Can we do this in a very optimal way?
As the OP probably knows, canonical grouping goes like this...
const prodsByCategory = products.reduce((acc, p) => {
let cat = p.category;
if (!acc[cat]) acc[cat] = [];
acc[cat].push(p);
return acc;
}, {});
Modify this a little to add the data which will be needed to optimize.
const prodsByCategory = products.reduce((acc, p) => {
let cat = p.category;
// instead of just an array, keep an array and a total
if (!acc[cat]) acc[cat] = { products: [], totalRate: 0 };
// instead of just pushing, push and increment total with a lookup
acc[cat].products.push(p);
acc[cat].totalRate += rateForProduct(p) || 0;
return acc;
}, {});
We need a lookup for rateForProduct, like this:
const rateForProduct = product => {
return rate.find(r => r.name === product.name)?.rate || 0;
}
That should produce an object keyed by category with values that have a prop called totalRate. Sort those entries so that the first one is maximized. Here's a demo...
const products = [{
name: 'prod1',
category: 'Meat'
},
{
name: 'prod2',
category: 'Meat'
},
{
name: 'prod3',
category: 'Dairy'
}
];
const rate = [{
name: 'prod1',
rate: 23,
},
{
name: 'prod2',
rate: 36
},
{
name: 'prod3',
rate: 50,
}
];
const rateForProduct = product => {
return rate.find(r => r.name === product.name)?.rate || 0;
}
const prodsByCategory = products.reduce((acc, p) => {
let cat = p.category;
if (!acc[cat]) acc[cat] = {
products: [],
totalRate: 0
};
acc[cat].products.push(p);
acc[cat].totalRate += rateForProduct(p);
return acc;
}, {});
const sortedEntries = Object.entries(prodsByCategory).sort((a, b) => b[1].totalRate - a[1].totalRate);
const bestEntry = {
category: sortedEntries[0][0],
rate: sortedEntries[0][1].totalRate
}
console.log(bestEntry);
This question already has answers here:
Better way to sum a property value in an array
(20 answers)
Closed 2 years ago.
I've got an array of objects:
let recipe = [
{
Type:"AAAA",
Total: 20
},
{
Type:"BBBB",
Total: 20,
},
{
Type:"BBBB",
Total: 20,
},
{
Type:"AAAA",
Total: 20,
},
{
Type:"AAAA",
Total: 20,
}
]
And I want to get, using javascript (no libraries), the final form like this:
let finalRecipe = [["AAAA",60],["BBBB",40]]
I tried to used this but not working
let recipeModified = recipe.reduce((r, timeoff) => {
const { Type, Total } = timeoff;
const totalSum = Object.values({
Total,
}).reduce((acc, val) => ([...acc, ...val], []));
r[Type] = [...(r[Type] || []), totalSum];
return r;
}, {})
https://codesandbox.io/s/cranky-ishizaka-i7m6v?file=/src/index.js
You can use the function Array.prototype.reduce for grouping the objects by Type and the function Object.values to extract the grouped objects.
let recipe = [ { Type:"AAAA", Total: 20 }, { Type:"BBBB", Total: 20, }, { Type:"BBBB", Total: 20, }, { Type:"AAAA", Total: 20, }, { Type:"AAAA", Total: 20, }],
result = Object.values(recipe.reduce((r, {Type, Total}) => {
(r[Type] || (r[Type] = [Type, 0]))[1] += Total;
return r;
}, {}));
console.log(result);
I currently have an array that has the following structure:
data = [
{
time: 100,
info: [{
name: "thing1",
count: 3
}, {
name: "thing2",
count: 2
}, {
}]
},
{
time: 1000,
info: [{
name: "thing1",
count: 7
}, {
name: "thing2",
count: 0
}, {
}]
}
];
But I would like to restructure the array to get something like this:
data = [
{
name: "thing1",
info: [{
time: 100,
count: 3
}, {
time: 1000,
count: 7
}, {
}]
},
{
name: "thing2",
info: [{
time: 100,
count: 2
}, {
time: 1000,
count: 0
}, {
}]
}
];
So basically the key would have to be switched from time to name, but the question is how. From other posts I have gathered that using the map function might work, but since other posts had examples to and from different structures I am still not sure how to use this.
There are a number of ways to achieve this however, the key idea will be to perform a nested looping of both data items and their (nested) info items. Doing that allows your algorithm to "visit" and "map" each piece of input data, to a corresponding value in the resulting array.
One way to express that would be to use nested calls to Array#reduce() to first obtaining a mapping of:
name -> {time,count}
That resulting mapping would then be passed to a call to Object.values() to transform the values of that mapping to the required array.
The inner workings of this mapping process are summarized in the documentation below:
const data=[{time:100,info:[{name:"thing1",count:3},{name:"thing2",count:2},{}]},{time:1e3,info:[{name:"thing1",count:7},{name:"thing2",count:0},{}]}];
const result =
/* Obtain array of values from outerMap reduce result */
Object.values(
/* Iterate array of data items by reduce to obtain mapping of
info.name to { time, count} value type */
data.reduce((outerMap, item) =>
/* Iterate inner info array of current item to compound
mapping of info.name to { time, count} value types */
item.info.reduce((innerMap, infoItem) => {
if(!infoItem.name) {
return innerMap
}
/* Fetch or insert new { name, info } value for result
array */
const nameInfo = innerMap[ infoItem.name ] || {
name : infoItem.name, info : []
};
/* Add { time, count } value to info array of current
{ name, info } item */
nameInfo.info.push({ count : infoItem.count, time : item.time })
/* Compound updated nameInfo into outer mapping */
return { ...innerMap, [ infoItem.name] : nameInfo }
}, outerMap),
{})
)
console.log(result)
Hope that helps!
The approach I would take would be to use an intermediate mapping object and then create the new array from that.
const data = [{time: 100, info: [{name: "thing1", count: 3}, {name: "thing2", count: 2}, {}]}, {time: 1e3, info: [{name: "thing1", count: 7}, {name: "thing2", count: 0}, {}]} ];
const infoByName = {};
// first loop through and add entries based on the name
// in the info list of each data entry. If any info entry
// is empty ignore it
data.forEach(entry => {
if (entry.info) {
entry.info.forEach(info => {
if (info.name !== undefined) {
if (!infoByName[info.name]) {
infoByName[info.name] = [];
}
infoByName[info.name].push({
time: entry.time,
count: info.count
});
}
});
}
});
// Now build the resulting list, where name is entry
// identifier
const keys = Object.keys(infoByName);
const newData = keys.map(key => {
return {
name: key,
info: infoByName[key]
};
})
// newData is the resulting list
console.log(newData);
Well, the other guy posted a much more elegant solution, but I ground this one out, so I figured may as well post it. :)
var data = [
{
time: 100,
info: [{
name: "thing1",
count: 3
}, {
name: "thing2",
count: 2
}, {
}]
},
{
time: 1000,
info: [{
name: "thing1",
count: 7
}, {
name: "thing2",
count: 0
}, {
}]
}
];
var newArr = [];
const objInArray = (o, a) => {
for (var i=0; i < a.length; i += 1) {
if (a[i].name === o)
return true;
}
return false;
}
const getIndex = (o, a) => {
for (var i=0; i < a.length; i += 1) {
if (a[i].name === o) {
return i;
}
}
return false;
}
const getInfoObj = (t, c) => {
let tmpObj = {};
tmpObj.count = c;
tmpObj.time = t;
return tmpObj;
}
for (var i=0; i < data.length; i += 1) {
let t = data[i].time;
for (var p in data[i].info) {
if ("name" in data[i].info[p]) {
if (objInArray(data[i].info[p].name, newArr)) {
let idx = getIndex(data[i].info[p].name, newArr);
let newInfoObj = getInfoObj(t, data[i].info[p].count);
newArr[idx].info.push(newInfoObj);
} else {
let newObj = {};
newObj.name = data[i].info[p].name;
let newInfo = [];
let newInfoObj = getInfoObj(t, data[i].info[p].count);
newInfo.push(newInfoObj);
newObj.info = newInfo;
newArr.push(newObj);
}}
}
}
console.log(newArr);
try to use Object.keys() to get the key
I am trying to add the unique values for the fiddle below in the output I want
{ category: 'fos', value: 70 },
{ category: 'nyedva', value: 30 }
I am able to get the unique values in the array not sure where to add the values
http://jsfiddle.net/mj3q0sk3/
var catalog={
products : [
{ category: 'fos', value: 10 },
{ category: 'fos', value: 20 },
{ category: 'nyedva', value: 30 },
{ category: 'fos', value: 40 },
]
};
var categories = [];
var sum=[];
$.each(catalog.products, function(index, value) {
if ($.inArray(value.category, categories)==-1) {
categories.push(value.category);
}
else {
console.log("CAt Val:" +value.category);
var total=value.value;
sum.push(total);
}
});
console.log(categories);
console.log(sum);
You can use forEach() loop to return desired result.
var catalog = {"products":[{"category":"fos","value":10},{"category":"fos","value":20},{"category":"nyedva","value":30},{"category":"fos","value":40}]}
var result = [];
catalog.products.forEach(function(e) {
var c = e.category;
!this[c] ? (this[c] = e, result.push(this[c])) : this[c].value += e.value
}, {})
console.log(result)
You can do this without the need for jQuery:
var res = catalog.products.reduce(function(res, product) {
if (!res.hasOwnProperty(product.category)) {
res[product.category] = 0;
}
res[product.category] += product.value;
return res;
}, {});
console.log(res);
This yields:
{ fos: 70, nyedva: 30 }
If you want it as an array of categories:
console.log(Object.keys(res).map(function(key) { return { category: key, value: res[key] }; }));
This will give you:
[ { category: 'fos', value: 70 },
{ category: 'nyedva', value: 30 } ]
Here is a dummy example. I have an array of objects:
var cars = [
{
name: "Hyundai",
plans: [
{
name: "Something",
add-ons: [
{
cost: 100
},
{
cost: 75
}
]
}, { ... }
]
},
{
name: "Jeep",
plans: [
{
name: "Something",
add-ons: [
{
cost: 50
},
{
cost: 75
}
]
}, { ... }
]
},
{
name: "Buick",
plans: [
{
name: "Something",
add-ons: [
{
cost: 35
},
{
cost: 50
}
]
}, {...}
]
}
]
What I'm trying to do is find the top 2 cars that have the cheapest add-on and reference them via another variable.
Like this:
var top2 = findTopTwo(cars);
findTopTwo(arr) {
return arr.sort(function(a, b) {
// My trouble spot
}).slice(0, 2);
}
With my simple example, the result for top2 would be:
Buick ( cheapest add-on was $35, the value used to compare against )
Jeep ( cheapest add-on was $50, value used to compare against )
So what I would do is feed all of them into an array and then sort it on the cost. That would be my naive approach. The more optimal solution would be to only store 2 objects at a given time instead of a list of all items.
The naive approach would be as simple as:
var items = [];
for ( var i in cars ){
var car = cars[i];
for (var i in car["plans"]){
for (var j = 0; j < car["plans"][i]["add-ons"]){
items.push({"name": car.name, "cost": car["plans"][i]["add-ons"][j]["cost"]});
}
}
}
return items.sort(function(a,b){ return a.cost < b.cost }).slice(0,2);
That will return a list of 2 objects, the object contains the name of the car and the cost. The more effecient thing would be to do something like this:
var biggest = function(arr){
if (arr.length < 2 ) return -1;
return arr[0].cost > arr[1].cost ? 0 : 1;
}
var items = [];
for ( var i in cars ){
var car = cars[i];
for (var i in car["plans"]){
for (var j = 0; j < car["plans"][i]["add-ons"]){
var obj = {"name": car.name, "cost": car["plans"][i]["add-ons"][j]["cost"]};
}
var index = biggest(items)
if (index < 0){
items.push(obj);
}else{
if (items[index].cost > obj.cost)
items[index] = obj;
}
}
}
return items;
this more interesting design will push the first 2 into the list, but then it will find the biggest of the 2 costs and then checks to see if the new one is smaller than it. If the new one is smaller than item[index] it will be replaced.
This will never have the array larger than 2 so it takes up less memory
Another approach. By this approach your original data will not be sorted or modified.
var cars=[{name:"Hyundai",plans:[{name:"Something","add-ons":[{cost:100},{cost:75}]}]},
{name:"Jeep",plans:[{name:"Something","add-ons":[{cost:50},{cost:75}]}]},
{name:"Buick",plans:[{name:"Something","add-ons":[{cost:35},{cost:50}]}]}];
function findTopTwo(cars) {
return cars.map(
car =>
car.plans.reduce(
(prevPlan, plan) =>
plan['add-ons'].reduce((prevAddOn, addOn) => {
if (prevAddOn.cost > addOn.cost) {
prevAddOn.cost = addOn.cost;
}
return prevAddOn;
}, prevPlan), {
cost: Number.MAX_VALUE,
name: car.name
})
)
.sort((a, b) => a.cost - b.cost)
.slice(0, 2)
.map(item => item.name);
}
console.log(findTopTwo(cars));
I had to play around with the object, but here is the gist of it -
var cars = [{
name: "Hyundai",
plans: {
addons: [{
cost: 100
}, {
cost: 75
}]
}
}, {
name: "Jeep",
plans: {
addons: [{
cost: 50
}, {
cost: 75
}]
}
}, {
name: "Buick",
plans: {
addons: [{
cost: 35
}, {
cost: 50
}]
}
}];
var top2 = findTopTwo(cars);
console.log(top2);
function findTopTwo(arr) {
return arr.sort(function (a, b) {
// this map outputs array of costs: [35, 40]
// and Math.min takes the lowest value of each
var a_max_cost = Math.min.apply(null, a.plans.addons.map(function(i){i.cost})),
b_max_cost = Math.min.apply(null, b.plans.addons.map(function(i){i.cost}));
return a_max_cost - b_max_cost;
})
.slice(0, 2);
}
Basically, you need to return a-b in the sort function, where a and b are the lowest addon values. So I calculated the max of both cars on comparison, and used those values to decide which goes where.
Edit: I see you've updated the JS object, the answer should be similar to min, you will only need to figure out which plan to use for a and b. You can do so similar to my use of the Math.max function
One simple way of doing it is to first sort the addons by price (if you don't mind the side effect that addons then remain sorted by price).
function findTopTwo(arr) {
arr.forEach(function (elem) {
elem.plans.addons = elem.plans.addons.sort(function (a, b) {
return a.cost > b.cost;
});
});
return arr.sort(function(a, b) {
return a.plans.addons[0].cost > b.plans.addons[0].cost;
}).slice(0, 2);
}
jsbin example
Using #casraf's data:
const sortedCars = cars.map(car => {
car.plans.addons.sort((a, b) => a.cost - b.cost);
return car;
}).sort((a, b) => {
return a.plans.addons[0].cost - b.plans.addons[0].cost;
});
Line 2 sorts each cars' addons array from low to high. Line 5 sorts the cars from low to high based on the first index of their respective addons property.
If the ES6 syntax is confusing, here's a translation to ES5
I suggest to use sorting with map, then take the top 2 entries and get the data from cars.
var cars = [{ name: "Hyundai", plans: [{ 'add-ons': [{ cost: 100 }, { cost: 75 }] }] }, { name: "Jeep", plans: [{ 'add-ons': [{ cost: 50 }, { cost: 75 }] }] }, { name: "Buick", plans: [{ 'add-ons': [{ cost: 35 }, { cost: 50 }] }] }],
cost = cars.
map(function (a, i) {
return {
index: i,
cost: a.plans.reduce(function (r, b) {
return Math.min(r, b['add-ons'].reduce(function (s, c) {
return Math.min(s, c.cost);
}, Infinity));
}, Infinity)
};
}).
sort(function (a, b) { return a.cost - b.cost; }),
top2 = cost.slice(0, 2).map(function (a) {
return cars[a.index];
});
console.log(top2);
.as-console-wrapper { max-height: 100% !important; top: 0; }