Get Unique Key-Value count as an object - javascript

I've got the following response from the server:
I want to get the unique key with the occurrence count.
In the following format:
0:{"name":"physics 1","count":2}
1:{"name":"chem 1","count":6}
I've already checked How to count the number of occurrences of each item in an array? but that is not I want.

Here is an es6 solution.
const data = [{
id: 0,
name: 'physics 1',
questionId: 1,
questionNr: 1
}, {
name: 'physics 1',
}, {
name: 'chem 1',
}, {
name: 'chem 1',
}, {
name: 'chem 2',
}];
const grouped = data.reduce((groups, cur) => {
const key = cur.name;
groups[key] = (groups[key] || 0) + 1;
return groups;
}, {});
const result = Object.keys(grouped).map(key => ({name: key, count: grouped[key]}));
console.log(result);

You could do it this way:
var source = [
{'section_name': 'test1'},
{'section_name': 'test2'},
{'section_name': 'test1'},
];
var temp = {};
for (var i = source.length - 1; i >= 0; i--) {
var key = source[i].section_name;
if (!temp[key]) {
temp[key] = 0;
}
temp[key] += 1;
}
var keys = Object.keys(temp);
var result = [];
for (var i = keys.length - 1; i >= 0; i--) {
var key = keys[i];
result.push({"name":key,"count":temp[key]});
}
console.log(result);

use this function this uses map and filter
t.reduce((f,l)=>{
var k=f.filter(elem=>elem.section_name==l.section_name);
if(k.length==1) k[0].count++;
else f.push({section_name:l.section_name,count:1})
return f;
},[] )
you can check this against this to verify
var t=[{section_name:"Physics"},{section_name:"Physics"},{section_name:"Chemistry"},{section_name:"Chemistry"},{section_name:"Physics"}]

Related

Divide object array elements into groups of n each javascript

I have an Object as below:
const boxOfFruits = {
apples: [
{
name: "Kashmiri",
},
{
name: "Washington",
},
{
name: "Himalayan",
},
{
name: "Fuji",
}
],
oranges: [
{
name: "Nagpur",
},
{
name: "Clementine",
},
],
mangoes: [
{
name: "Totapuri",
},
{
name: "Alphonso",
},
{
name: "Langda",
},
],
}
I want to divide these fruits into boxes; maximum of n each, let's say where n is 3 and apples, oranges and mangoes are equally distributed.
So the output in this case would be:
box_1 = [{name: "Kashmiri"}, {name: "Nagpur"},{name: "Totapuri"}];
box_2 = [{name: "Washington"}, {name: "Clementine"},{name: "Alphonso"}];
box_3 = [{name: "Himalayan"},{name: "Langda"}, {name: "Fuji"}];
The type of fruits(apple,oranges,etc)/keys in object can increase/decrease and n is also variable. In case total fruits are less than n, then it would be just 1 box of fruits.
What I have tried so far:
Using Lodash, I am calculating the minimum and the maximum fruits in a single type:
const minFruitType = _.min(Object.values(basket).map((eachBasket: any) => eachBasket.length));
Total teams will the sum of the fruits / n
Will distribute the minimum fruits (l) in the first l boxes and fill the rest with the remaining fruits at every iteration while at the start of every iteration will calculate the minimum type of fruits again.
You can use Object.values(), array#reduce and array#forEach to transform your object.
const boxOfFruits = { apples: [ { name: "Kashmiri", }, { name: "Washington", }, { name: "Himalayan", }, ], oranges: [ { name: "Nagpur", }, { name: "Clementine", }, ], mangoes: [ { name: "Totapuri", }, { name: "Alphonso", }, { name: "Langda", }, ], },
result = Object.values(boxOfFruits).reduce((r, arr) => {
arr.forEach((o,i) => {
const key = `box_${i+1}`;
r[key] ??= r[key] || [];
r[key].push(o)
});
return r;
},{});
console.log(result);
The easiest way would be to use lodash.js's zip() function:
const boxes = _.zip( Object.values(boxOfFruits) );
Note that _.zip() will give you undefined values when the source arrays are different lengths, so you'll need/want to filter those out:
const boxes == _.zip( Object.values(boxOfFruits) )
.map(
box => box.filter(
x => x !== undefined
)
);
But that will not distribute the fruits evenly. For that, it shouldn't get much for difficult than this:
function distribute(boxOfFruits, n) {
const boxes = [];
const fruits = Object.keys(boxOfFruits);
for ( const fruit of fruits ) {
let i = 0;
const items = boxOfFruits[fruit];
for (const item of items) {
boxes[i] = !boxes[i] ?? [];
boxes[i] = boxes[i].push(item);
++i;
i = i < n ? i : 0 ;
}
}
return boxes;
}
A modified version of #Nicholas Carey's answer worked for me:
function distribute(boxOfFruits, n) {
let boxes = [];
let totalFruits = Object.values(boxOfFruits)
.reduce((content, current) => content + current.length, 0);
let maxBoxes = Math.ceil(totalFruits / 4);
Object.values(boxOfFruits).forEach((fruits) => {
let i = 0;
fruits.forEach((fruit) => {
boxes[i] ??= boxes[i] || [];
boxes[i].push(fruit);
++i;
i = i < (n+1) ? i : 0;
});
});
// Extra boxes created, redistribute them to
// starting boxes
let newBoxes = teams.slice(0, maxBoxes);
let pendingBoxes = teams.slice(maxBoxes);
let pendingFruits = pendingBoxes.flat();
let distributedBoxes = newBoxes.map((eachBox) => {
let required = n - eachBox.length;
if (required > 0) {
eachBox.push(...pendingFruits.splice(0, required));
}
return eachBox;
});
return distributedBoxes;
}
Code is pretty much the same as Nicholas's accept the below changes:
Directly fetched the values and iterated over those
empty array creation was failing, this way works
and checking on the max box size with n+1 instead of n

React.js: How to find duplicates for properties in an array of object and put a progressive number on that field

I have an array of object and each object is for example :
const myArr=[{name:"john",id:1}{name:"john",id:2}{name:"mary",id:3}]
for the first 2 element for the property "name" I have the name "john" that is duplicate.
How can I modify the rendered names like that:
const myArr=[{name:"john (1 of 2)",id:1}{name:"john (2 of 2)",id:2}{name:"mary",id:3}]
Thanks in advance!
Reduce the input array into a map by name (i.e. group by name property), and map the array of values to the result array. If the group array has more than 1 element in it then sub-map the group to include the numbering. Flatten the overall result.
const myArr = [
{ name: "john", id: 1 },
{ name: "john", id: 2 },
{ name: "mary", id: 3 }
];
const res = Object.values(
myArr.reduce((groups, current) => {
if (!groups[current.name]) {
groups[current.name] = [];
}
groups[current.name].push(current);
return groups;
}, {})
).flatMap((value) => {
if (value.length > 1) {
return value.map((current, i, arr) => ({
...current,
name: `${current.name} (${i + 1} of ${arr.length})`
}));
}
return value;
});
console.log(res);
You can do use reduce(), filter(), and flat() and do this:
const myArr = [
{name:"john", id:1},
{name:"john", id:2},
{name:"mary", id:3}
]
const res = Object.values(myArr.reduce((acc, curr) => {
const total = myArr.filter(({ name }) => name === curr.name).length;
if(!acc[curr.name]) {
acc[curr.name] = [
{...curr}
]
} else {
const currentSize = acc[curr.name].length;
if(currentSize === 1) {
acc[curr.name][0].name = `${acc[curr.name][0].name} (1 of ${total})`
}
acc[curr.name].push({
...curr,
name: `${curr.name} (${currentSize + 1} of ${total})`
})
}
return acc;
}, {})).flat();
console.log(res);
const myArr = [{name:"john",id:1}, {name:"john",id:2}, {name:"mary",id:3}];
const namesArray = myArr.map(elem => elem.name);
const namesTraversed = [];
let currentCountOfName = 1;
let len = 0;
myArr.forEach(elem => {
len = namesArray.filter(name => name === elem.name).length;
if (len > 1) {
if (namesTraversed.includes(elem.name)) {
namesTraversed.push(elem.name);
currentCountOfName = namesTraversed.filter(name => name === elem.name).length;
elem.name = `${elem.name} (${currentCountOfName} of ${len})`;
} else {
namesTraversed.push(elem.name);
currentCountOfName = 1;
elem.name = `${elem.name} (${currentCountOfName} of ${len})`;
}
}
});
console.log(myArr);
Check if this helps you
const myArr = [{
name: "john",
id: 1
}, {
name: "john",
id: 2
}, {
name: "mary",
id: 3
}]
// to keep a track of current copy index
let nameHash = {}
const newMyArr = myArr.map(ele => {
const noOccurence = myArr.filter(obj => obj.name ===ele.name).length;
if(noOccurence > 1){
// if there are multiple occurences get the current index. If undefined take 1 as first copy index.
let currentIndex = nameHash[ele.name] || 1;
const newObj = {
name: `${ele.name} (${currentIndex} of ${noOccurence})`,
id: ele.id
}
nameHash[ele.name] = currentIndex+ 1;
return newObj;
}
return ele;
})
console.log(newMyArr);

Finding objects in a nested array along with their position

I've taken the following sample from a different question. And I am able to identify the object. But I also need to find our the position of that object. For example:
var arr = [{
Id: 1,
Categories: [{
Id: 1
},
{
Id: 2
},
]
},
{
Id: 2,
Categories: [{
Id: 100
},
{
Id: 200
},
]
}
]
If I want to find the object by the Id of the Categories, I can use the following:
var matches = [];
var needle = 100; // what to look for
arr.forEach(function(e) {
matches = matches.concat(e.Categories.filter(function(c) {
return (c.Id === needle);
}));
});
However, I also need to know the position of the object in the array. For example, if we are looking for object with Id = 100, then the above code will find the object, but how do I find that it's the second object in the main array, and the first object in the Categories array?
Thanks!
Well, if every object is unique (only in one of the categories), you can simply iterate over everything.
var arr = [{
Id: 1,
Categories: [{Id: 1},{Id: 2}]
},
{
Id: 2,
Categories: [{Id: 100},{Id: 200}]
}
];
var needle = 100;
var i = 0;
var j = 0;
arr.forEach(function(c) {
c.Categories.forEach(function(e) {
if(e.Id === needle) {
console.log("Entry is in position " + i + " of the categories and in position " + j + " in its category.");
}
j++;
});
j = 0;
i++;
});
function findInArray(needle /*object*/, haystack /*array of object*/){
let out = [];
for(let i = 0; i < haystack.lenght; i++) {
if(haystack[i].property == needle.property) {
out = {pos: i, obj: haystack[i]};
}
}
return out;
}
if you need the position and have to filter over an property of the object you can use a simple for loop. in this sample your result is an array of new object because there can be more mathches than 1 on the value of the property.
i hope it helps
Iterate over the array and set index in object where match found
var categoryGroups = [{
Id : 1,
Categories : [{
Id : 1
}, {
Id : 2
},
]
}, {
Id : 2,
Categories : [{
Id : 100
}, {
Id : 200
},
]
}
]
var filterVal = [];
var needle = 100;
for (var i = 0; i < categoryGroups.length; i++) {
var subCategory = categoryGroups[i]['Categories'];
for (var j = 0; j < subCategory.length; j++) {
if (subCategory[j]['Id'] == findId) {
filterVal.push({
catIndex : i,
subCatIndex : j,
id : needle
});
}
}
}
console.log(filterVal);
Here is solution using reduce:
var arr = [{ Id: 1, Categories: [{ Id: 1 }, { Id: 2 }, ] }, { Id: 2, Categories: [{ Id: 100 }, { Id: 200 }, ] } ]
const findPositions = (id) => arr.reduce((r,c,i) => {
let indx = c.Categories.findIndex(({Id}) => Id == id)
return indx >=0 ? {mainIndex: i, categoryIndex: indx} : r
}, {})
console.log(findPositions(100)) // {mainIndex: 1, categoryIndex: 0}
console.log(findPositions(1)) // {mainIndex: 0, categoryIndex: 0}
console.log(findPositions(200)) // {mainIndex: 1, categoryIndex: 1}
console.log(findPositions(0)) // {}
Beside the given answers with fixt depth searh, you could take an recursive approach by checking the Categories property for nested structures.
function getPath(array, target) {
var path;
array.some(({ Id, Categories = [] }) => {
var temp;
if (Id === target) {
path = [Id];
return true;
}
temp = getPath(Categories, target);
if (temp) {
path = [Id, ...temp];
return true;
}
});
return path;
}
var array = [{ Id: 1, Categories: [{ Id: 1 }, { Id: 2 },] }, { Id: 2, Categories: [{ Id: 100 }, { Id: 200 }] }];
console.log(getPath(array, 100));
.as-console-wrapper { max-height: 100% !important; top: 0; }

Get all possible options for a matrix in javascript

I have an 'item' object in JavaScript, and the item can have settings like
color, size, etc.
I need to get all possible combinations in an array.
So lets say we have an item that looks like this:
var newItem = {
name: 'new item',
Settings: [
{name: 'color', values: ['green', 'blue', 'red']},
{name: 'size', values: ['15', '18', '22']},
{name: 'gender',values: ['male', 'female']}
]
};
I need to somehow get this:
[
[{SettingName:'color',value:'green'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'green'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'green'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'male'}],
[{SettingName:'color',value:'green'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'15'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'green'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'18'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'green'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'blue'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'female'}],
[{SettingName:'color',value:'red'},{SettingName:'size',value:'22'},{SettingName:'gender',value:'female'}]
]
This can be a good interview question.
See JS Bin for running example.
getAllPermutations(newItem);
function getAllPermutations(item) {
var permutations = [];
getAllPermutations0(item, permutations, []);
console.log(permutations);
}
function getAllPermutations0(item, permutations, array) {
if (array && array.length === item.Settings.length) {
permutations.push(array.slice()); // The slice clone the array
return;
}
var index = array.length;
var setting = item.Settings[index];
for (var i = 0; i < setting.values.length; i++) {
if (index === 0)
array = [];
var currValue = setting.values[i];
array.push({
SettingName: setting.name,
value: currValue
});
getAllPermutations0(item, permutations, array);
array.pop(); // pop the old one first
}
}
Here is a none recursive solution. It takes an empty or existing settings "matrix" and a values array, and return a new matrix as a combination of existing matrix content cloned for each new value, appended with pairs of new value setting items.
[A] -> [1,2] gives [A][1][A][2]
[A][1][A][2] -> [X,Y] gives [A][1][X][A][2][Y][A][2][X][A][1][Y]
and so on
function processSettings(settings, name, values) {
if (settings.length == 0) {
values.forEach(function(value) {
settings.push( [{ SettingName: name, value: value }] )
})
} else {
var oldSettings = JSON.parse(JSON.stringify(settings)), settings = [], temp, i = 0
for (i; i<values.length; i++) {
temp = JSON.parse(JSON.stringify(oldSettings))
temp.forEach(function(setting) {
setting.push( { SettingName: name, value: values[i] } )
settings.push(setting)
})
}
}
return settings
}
You can now create the desired settings literal this way :
var settings = []
for (var i=0; i<newItem.Settings.length; i++) {
var item = newItem.Settings[i]
settings = processSettings(settings, item.name, item.values)
}
demo -> http://jsfiddle.net/b4ck98mf/
The above produces this :
[
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"male"}],
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"15"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"18"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"green"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"blue"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"female"}],
[{"SettingName":"color","value":"red"},{"SettingName":"size","value":"22"},{"SettingName":"gender","value":"female"}]
]
You can use Array.prototype.map(), for loop, while loop, Array.prototype.concat(). Iterate gender values; select each of color, size value in succession beginning at index 0 of either; iterating the furthest adjacent array from current gender, increment the index of the closest adjacent array; merge the resulting two gender arrays to form a single array containing all combinations of gender, color, size
var colors = newItem.Settings[0].values;
var sizes = newItem.Settings[1].values;
var gen = newItem.Settings[2].values;
var i = sizes.length;
var res = [].concat.apply([], gen.map(function(value, key) {
var next = -1;
var arr = [];
for (var curr = 0; curr < i; curr++) {
while (next < i - 1) {
arr.push([{
SettingName: "gender",
value: value
}, {
SettingName: "size",
value: sizes[curr]
}, {
SettingName: "color",
value: colors[++next]
}])
}
next = -1;
}
return arr
}))
var newItem = {
"name": "new item",
"Settings": [{
"name": "color",
"values": [
"green",
"blue",
"red"
]
}, {
"name": "size",
"values": [
"15",
"18",
"22"
]
}, {
"name": "gender",
"values": [
"male",
"female"
]
}]
}
var colors = newItem.Settings[0].values;
var sizes = newItem.Settings[1].values;
var gen = newItem.Settings[2].values;
var i = sizes.length;
var res = [].concat.apply([], gen.map(function(value, key) {
var next = -1;
var arr = [];
for (var curr = 0; curr < i; curr++) {
while (next < i - 1) {
arr.push([{
SettingName: "gender",
value: value
}, {
SettingName: "size",
value: sizes[curr]
}, {
SettingName: "color",
value: colors[++next]
}])
}
next = -1;
}
return arr
}))
document.querySelector("pre").textContent = JSON.stringify(res, null, 2)
<pre></pre>
plnkr http://plnkr.co/edit/C2fOJpfwOrlBwHLQ2izh?p=preview
An approach using Array.prototype.reduce(), Array.prototype.sort(), Object.keys(), for loop, while loop
var newItem = {
name: 'new item',
Settings: [
{
name: 'color',
values: ['green', 'blue', 'red']
},
{
name: 'size',
values: ['15', '18', '22']
},
{
name: 'gender',
values: ['male', 'female']
}
]
};
var props = ["SettingName", "value"];
var settings = newItem.Settings;
function p(settings, props) {
var data = settings.reduce(function(res, setting, index) {
var name = setting.name;
var obj = {};
obj[name] = setting.values;
res.push(obj);
return res.length < index ? res : res.sort(function(a, b) {
return a[Object.keys(a)[0]].length - b[Object.keys(b)[0]].length
})
}, []);
var key = data.splice(0, 1)[0];
return [].concat.apply([], key[Object.keys(key)].map(function(value, index) {
return data.reduce(function(v, k) {
var keys = [v, k].map(function(obj) {
return Object.keys(obj)[0]
});
var i = Math.max.apply(Math, [v[keys[0]].length, k[keys[1]].length]);
var next = -1;
var arr = [];
for (var curr = 0; curr < i; curr++) {
while (next < i - 1) {
var a = {};
a[props[0]] = keys[0];
a[props[1]] = v[keys[0]][++next];
var b = {};
b[props[0]] = keys[1];
b[props[1]] = k[keys[1]][next];
var c = {};
c[props[0]] = Object.keys(key)[0];
c[props[1]] = value;
arr.push([a, b, c]);
};
next = -1;
}
return arr
});
}));
}
document.querySelector("pre").textContent = JSON.stringify(
p(settings, props), null, 2
);
<pre></pre>

Split array of objects into new array or objects based on age value

Split array of objects into new array or objects based on age value in Javascript
var items = [
{name:"Foo", age:16, color:"w"},
{name:"Bar", age:18, color:"b"},
{name:"foo", age:16, color:"w"},
{name:"bar", age:18, color:"w"},
{name:"foobar", age:18, color:"b"},
{name:"barfoo", age:20, color:"w"}
];
How can I return a list like:
var items = [
{age:16,name:"Foo"|"foo",gender:"w"|"w"},
{age:18,name:"Bar"|"bar"|"foobar",gender:"b"|"w"|"b"},
{age:20,name:"barfoo",gender:"w"}
];
I have worked but i got output with 'undefined' in name. Below is my code.
var data = [{age: 21,name: "Walter",color: "black"},{age: 25,name: "sentinel",color: "black"
},{age: 21,name: "Micah",color: "purple"},{age: 25,name: "mike",color: "black"},{age: 21,name: "Danny",color: "white"},{age: 25,name: "mike",color: "black"}];
var obj=data;
var arrayobj = obj.length;
var i, row, arr = obj, ss = {};
for (i = 0; i < arr.length; i++) {
row = arr[i];
ss[row.age] = ss[row.age] || {count: 0};
if (ss[row.age][row.age] === undefined) {
ss[row.age][row.name] = row.name;
ss[row.age]['name']+=row.name+'|';
ss[row.age]['color']+=row.color+'|';
ss[row.age]['count'] += 1;
}
}
console.table(ss);
I'm assuming you want to group the items by their age. Here is one way:
(fiddle)
items.reduce(function(buckets,item){
if(!buckets[item.age]) buckets[item.age] = [];
buckets[item.age].push(item);
return buckets;
},{});
Let's explain:
For each item, if we don't already have a 'bucket' for it, create a new empty one
Add it to the bucket
return the new updated bucket list.
The method returns an object with 3 properties: 16,18 and 20, each containing the objects with that age.
This will work. The output is in different format than one provided by exebook .
Please check and confirm. Here's a fiddle....
** UX Manager
var buckets = [];
for (var item in items) {
var currentAge = items[item].age;
if(!buckets[currentAge]) {
buckets[currentAge] = [];
for (var i in items) {
if (currentAge === items[i].age) {
buckets[currentAge].push(items[i]);
}
}
}
}
var items = [
{name:"Foo", age:16, color:"w"},
{name:"Bar", age:18, color:"b"},
{name:"foo", age:16, color:"w"},
{name:"bar", age:18, color:"w"},
{name:"foobar", age:18, color:"b"},
{name:"barfoo", age:20, color:"w"}
];
var result = [] // THIS IS THE RESULTING ARRAY THAT YOU WANT
function find(age) {
for (var i = 0; i < result.length; i++)
if (result[i].age == age) return i
return -1
}
function append(i, obj) {
result[i].name.push(obj.name)
result[i].color.push(obj.color)
}
for (var i = 0; i < items.length; i++) {
var x = find(items[i].age)
if (x < 0) result.push({ age: items[i].age, name: [items[i].name], color : [items[i].color]})
else append(x, items[i])
}
console.log(result) // PRINT THE RESULT, alternatively you can use alert(result)
The output
[ { age: 16, name: [ 'Foo', 'foo' ], color: [ 'w', 'w' ] },
{ age: 18, name: [ 'Bar', 'bar', 'foobar' ], color: [ 'b', 'w', 'b' ] },
{ age: 20, name: [ 'barfoo' ], color: [ 'w' ] } ]

Categories

Resources