how to calculate a sum of an array in an object - javascript

I had a variable like that
const data = {
code: 1,
items: [
{ nickname: 1, name: [
{id : "A"},
{id : "B"}
]
},
{
nickname: 2, name: [
{id: "A"},
{id: "C"}
]
}
]
}
after that, I want to show how many characters: A:2, B:1, C:1

You can do that is following steps:
Use flatMap() on the array data.items
Inside flatMap() use map() to convert all the object to their id and return it from flatMap(). This way you will array ["A","B","A","C"]
Then use reduce() and get an object with count of all the letters.
const data = { code: 1, items: [ { nickname: 1, name: [ {id : "A"}, {id : "B"} ] }, { nickname: 2, name: [ {id: "A"}, {id: "C"} ] } ] }
const res = data.items.flatMap(x =>
x.name.map(a => a.id)
).reduce((ac,a) => (ac[a] = ac[a] + 1 || 1,ac),{});
console.log(res)

const data = {
code: 1,
items: [
{
nickname: 1,
name: [
{ id: "A" },
{ id: "B" }
]
},
{
nickname: 2,
name: [
{ id: "A" },
{ id: "C" }
]
}
]
};
const res = data.items.reduce((acc, next) => {
next.name.forEach(({ id }) => {
acc[id] = acc[id] + 1 || 1;
});
return acc;
}, {});
console.log(res);
You can do that in a single shot using reduce.
Reducing data.items will allow you to add to the accumulator (initially an empty object), the value of the currently looped name property item.
The result will be an object owning all the occurences of each encountered letter in the name property of each array.
Relevant lines explained:
data.items.reduce((acc, next) will call the reduce method on data.items. acc is the reduce accumulator (initially an empty object), next is the currently looped item of data.items.
next.name.forEach(({id}) in this line, we loop the name property of the currently looped item (data.items[n]). ({id}) is a short syntax to acquire the id property of the looped item in the foreach. It's equivalent to (item => item.id).
acc[id] = acc[id] + 1 || 1; tries to increase the property [id] of the accumulator (example: "A" of {}) by 1. If it does not exist, it sets the value to 1.
return acc; returns the accumulator.

You could iterate name and take id in a loop for assigning the count.
const
data = { code: 1, items: [{ nickname: 1, name: [{ id : "A" }, { id : "B" }] }, { nickname: 2, name: [{ id: "A" }, { id: "C" }] }] },
result = data.items.reduce(
(r, { name }) => (name.forEach(({ id }) => r[id] = (r[id] || 0 ) + 1), r),
{}
);
console.log(result);

Related

Array of arrays of objects - find object value occurance and return true/false (js)

I have array of array of object as follows:
[
[
{
id: 1,
itemName: 'xxx',
...
},
{
id: 1,
itemName: 'yyy',
...
},
...
],
[
{
id: 2,
itemName: 'aaa',
...
},
{
id: 2,
itemName: 'kkk',
...
},
...
],
[
{
id: 3,
itemName: 'kkk',
...
},
{
id: 3,
itemName: 'yyy',
...
},
...
]
]
I am trying to check if any itemName from objects inside arrays equals given string, but I stuck at the solution that keeps these arrays with such object in one array. Here is my solution:
function isNameAcrossData(givenString){
return arr.map(arrItem =>
arrItem.find(item => item.itemId === givenString)
);
}
My solution doesn't return boolean but just one array with objects, that contain givenString and undefined as last array element. How to modify it to return just true/false value?
Use a .some inside a .some, to see if some of the arrays have at least one element inside matching the condition:
const isNameAcrossData = givenString => arr.some(
subarr => subarr.some(
({ itemName }) => itemName === givenString
)
);
const arr=[[{id:1,itemName:"xxx"},{id:1,itemName:"yyy"}],[{id:2,itemName:"aaa"},{id:2,itemName:"kkk"}],[{id:3,itemName:"kkk"},{id:3,itemName:"yyy"}]];
console.log(isNameAcrossData('xxx'));
console.log(isNameAcrossData('doesntexist'));
You could also flatten the outer array first:
const isNameAcrossData = givenString => arr.flat().some(
({ itemName }) => itemName === givenString
);
const arr=[[{id:1,itemName:"xxx"},{id:1,itemName:"yyy"}],[{id:2,itemName:"aaa"},{id:2,itemName:"kkk"}],[{id:3,itemName:"kkk"},{id:3,itemName:"yyy"}]];
console.log(isNameAcrossData('xxx'));
console.log(isNameAcrossData('doesntexist'));
You could check with some and return an array of boolean with using the wanted property.
function mapHasValue(key, value) {
return data.map(array => array.some(item => item[key] === value));
}
var data = [[{ id: 1, itemName: 'xxx' }, { id: 1, itemName: 'yyy' }], [{ id: 2, itemName: 'aaa' }, { id: 2, itemName: 'kkk' }], [{ id: 3, itemName: 'kkk' }, { id: 3, itemName: 'yyy' }]];
console.log(mapHasValue('id', 3));
Your code returns
[undefined, undefined, undefined]
because map returns an array so this approach won't work
You have first to loop through all the data and check inside then outside the loop assign to some variable true if there is a match.
Basically you have to return after you loop the data.
Working example for both cases:
const arr=[[{id:1,itemName:"xxx"},{id:1,itemName:"yyy"}],[{id:2,itemName:"aaa"},{id:2,itemName:"kkk"}],[{id:3,itemName:"kkk"},{id:3,itemName:"yyy"}]];
function isNameAcrossData(givenString){
let isMatch = false;
arr.map(childArr => {
childArr.map(obj => obj.itemName === givenString ? isMatch = true : null);
});
return isMatch;
}
console.log(isNameAcrossData('kkk'));
console.log(isNameAcrossData('bbb'));

filter array of object using an array [duplicate]

This question already has an answer here:
Filter array of object from another array
(1 answer)
Closed 3 years ago.
I want to filter an array of objects using an array but I want the results on the basis of array index and the result should be repeated when the array index value is repeated.
const data = [{
id='1',
name:'x'
},
{
id='4',
name:'a'
},
{
id='2',
name:'y'
},
{
id='3',
name:'z'
}
]
cons idArray = [1,4,3,2,4,3,2]
I have tried following code and get the result only once
const filteredData = data.filter(arrayofObj => idArray.includes(arrayofObj.id))
console.log(filteredData)
expected output is
expected output is =
[{id = '1,name:'x'},{id='4',name:'a'},{
id='3',
name:'z'
},
{
id='2',
name:'y'
},{
id='4',
name:'a'
},
{
id='3',
name:'z'
},{
id='2',
name:'y'
}]
First convert data array into Object with id's as keys.
Second, use map method over idArray and gather objects from above object.
const data = [
{
id: "1",
name: "x"
},
{
id: "4",
name: "a"
},
{
id: "2",
name: "y"
},
{
id: "3",
name: "z"
}
];
const dataObj = data.reduce((acc, curr) => {
acc[curr.id] = { ...curr };
return acc;
}, {});
const idArray = [1, 4, 3, 2, 4, 3, 2];
const results = idArray.map(id => ({ ...dataObj[id] }));
console.log(results);
You could map with a Map.
const
data = [{ id: '1', name: 'x' }, { id: '4', name: 'a' }, { id: '2', name: 'y' }, { id: '3', name: 'z' }],
idArray = [1, 4, 3, 2, 4, 3, 2],
result = idArray.map(Map.prototype.get, new Map(data.map(o => [+o.id, o])));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

Get count from Array of arrays

I have an array of arrays below. With ES6, how can I get a count of each value Good, Excellent & Wow into a new array e.g [{name: Good, count: 4} {name: Excellent, count: 5}, {name:Wow, count:2}] in dynamic style. I am attempting to use Object.assign but I am failing to "unique" out the count of the key plus instead, I need to use an array as I am trying to render this out on the front end. Do I need to use reduce? how?
let k = 0
const stats = {}
const remarks = [
[{name: "Good"}],
[{name: "Good"}, {name: "Excellent"}],
[{name: "Good"}, {name: "Excellent"}, {name: "Wow"}],
[{name: "Good"}, {name: "Excellent"}, {name: "Wow"}],
[{name: "Excellent"}],
[{name: "Excellent"}]
]
remarks.forEach((arr) => {
arr.map((e) => {
Object.assign(stats, { [e.name]: k = k + 1 })
})
})
console.log(stats);
Output:
stats: {Good: 8, Excellent: 11, Wow: 9}
Which is Incorrect plus I need to use an array.
Expected output:
[{name: Good, count: 4} {name: Excellent, count: 5}, {name:Wow, count:2}]
Flatten the array of arrays and reduce it starting with an object like : { Good: 0, Excellent: 0, Wow: 0}
then .map the Object.entries of the result to transform it to an array :
const remarks = [
[{ name: "Good" }],
[{ name: "Good" }, { name: "Excellent" }],
[{ name: "Good" }, { name: "Excellent" }, { name: "Wow" }],
[{ name: "Good" }, { name: "Excellent" }, { name: "Wow" }],
[{ name: "Excellent" }],
[{ name: "Excellent" }]
];
const result = Object.entries(
remarks.flat().reduce(
(all, { name }) => {
all[name] += 1;
return all;
},
{ Good: 0, Excellent: 0, Wow: 0 }
)
).map(([name, count]) => ({ name, count }));
console.log(result);
You can try below logic:
var data = [[{name: "Good"}],[{name: "Good"}, {name:"Excellent"}],[{name: "Good"}, {name:"Excellent"}, {name:"Wow"}],[{name: "Good"}, {name:"Excellent"}, {name:"Wow"}],[{name:"Excellent"}],[{name:"Excellent"}]]
var nData = [];
(data || []).forEach( e => {
(e || []).forEach(ei => {
var i = (index = nData.findIndex(d => d.name === ei.name)) >=0 ? index : nData.length;
nData[i] = {
name: ei.name,
count : (nData[i] && nData[i].count ? nData[i].count : 0)+1
}
});
});
console.log(nData);
Hope this helps!
You can use reduce, then convert the result into an array of objects:
const counts = remarks.reduce((result, list) => {
list.forEach(remark => {
result[remark.name] = (result[remark.name] || 0) + 1;
});
}, {});
const finalResult = [];
for (let name in counts) {
finalResult.push({name, count: counts[name]});
}
You could achieve this pretty easily by:
1) Flattening the nested array into 1 single level array.
2) Iterating over the flat array and create a "count map" by using Array.prototype.reduce
For example:
const remarks = [
[{
name: 'Good'
}],
[{
name: 'Good'
}, {
name: 'Excellent'
}],
[{
name: 'Good'
}, {
name: 'Excellent'
}, {
name: 'Wow'
}],
[{
name: 'Good'
}, {
name: 'Excellent'
}, {
name: 'Wow'
}],
[{
name: 'Excellent'
}],
[{
name: 'Excellent'
}]
]
const flatten = arr => arr.reduce((accum, el) => accum.concat(el), [])
const map = flatten(remarks).reduce((accum, el) => {
if (accum[el.name]) {
accum[el.name] += 1;
} else {
accum[el.name] = 1;
}
return accum;
}, {});
console.log(map)
First find the counts using reduce than pass that to another function to get the desired view structure:
const Good = 1,
Excellent = 2,
Wow = 3;
const remarks = [
[{name: Good}],
[{name: Good}, {name:Excellent}],
[{name: Good}, {name:Excellent}, {name:Wow}],
[{name: Good}, {name:Excellent}, {name:Wow}],
[{name:Excellent}],
[{name:Excellent}]
];
/*
[{name: Good, count: 4} {name: Excellent, count: 5}, {name:Wow, count:2}]
*/
function counts(remarks) {
return remarks.flat().reduce((acc, v) => {
const name = v.name;
let count = acc[name] || 0;
return {
...acc,
[name]: count + 1
}
}, {});
}
function view(counts) {
return Object.keys(counts).map(key => {
let count = counts[key];
return { name: key, count };
})
}
console.log(view(counts(remarks)));
Any time you are making a smaller set of data, or transforming data, in JavaScript reduce should be the first method you attempt to use. In this case, you may want to pair it with an indexer (hence preloading with an array of index and an array of result).
This works in one pass without needing to know the name values up front.
const remarks = [
[{name: "Good"}],
[{name: "Good"}, {name: "Excellent"}],
[{name: "Good"}, {name: "Excellent"}, {name: "Wow"}],
[{name: "Good"}, {name: "Excellent"}, {name: "Wow"}],
[{name: "Excellent"}],
[{name: "Excellent"}]
];
const stats = remarks.reduce((p,c) => (
c.forEach( ({name}) => {
if(!p[0].hasOwnProperty(name)){
p[1].push({name:name,count:0});
p[0][name] = p[1].length - 1;
}
p[1][p[0][name]].count++;
}),p),[{},[]])[1];
console.log(stats);
A slightly more concise and definitely less readable approach (but it's worth to mention) could be:
const remarks = [
[{ name: "Good" }],
[{ name: "Good" }, { name: "Excellent" }],
[{ name: "Good" }, { name: "Excellent" }, { name: "Wow" }],
[{ name: "Good" }, { name: "Excellent" }, { name: "Wow" }],
[{ name: "Excellent" }],
[{ name: "Excellent" }]
];
const stats = Object.entries(
remarks
.flat()
.reduce((acc, {name}) => (acc[name] = -~acc[name], acc), {})))
).map(([name, count]) => ({ name, count }));
console.log(stats);
It uses the comma operator in the reducer to returns the accumulator; and the bitwise operator NOT to create a counter without the needs to initialize the object upfront with all the names.
const flattenedRemarks = _.flatten(remarks);
const groupedRemarks = _.groupBy(flattenedRemarks, (remark) => remark.name);
const remarkCounts = _.mapValues(groupedRemarks, (group) => group.length);
const data = {
"mchale": {
"classes":["ESJ030", "SCI339"], // get the length
"faculty":["Hardy", "Vikrum"] // get the length
},
"lawerence":{
"classes":["ENG001"], // get the length
"faculty":["Speedman", "Lee", "Lazenhower"] // get the length
}
};
const count = Object.keys(data).map(campusName => {
const campus = data[campusName];
return Object.keys(campus).map(key => campus[key].length).reduce((p, c) => p + c, 0);
}).reduce((p, c) => p + c, 0);
console.log(count);

What happen when i run array.some in array filter

I just learned a new trick to find the same object in 2 array object, it works very well. it uses array.filter and array.some, as code bellow, but I don't understand how filter() can run when some() will return true or false.
const similarity = (arr, values) => arr.filter(item => values.some(m => (m.id === item.id) && (m.name === item.name)));
my input:
let arr1 = [
{
id: 1,
name: "kiet"
},
{
id: 2,
name: 'phan'
},
{
id: 3,
name: 'tuan'
}]
let arr2 = [
{
id: 1,
name: "kiet"
},
{
id: 2,
name: 'haha'
},
{
id: 5,
name: 'tuan'
}
]
my result :
[ { id: 1, name: 'kiet' } ]
You are Filtering the array, then passing this condition inside the some
(m.id === item.id) && (m.name === item.name)
If the id of the second array (here, m is the second array's objects) is equal to the first array's id (here, item is the array's objects) and the name is equal to the first array's name property. If yes then it will return true. So the filter gets a true and if is does, it will return that particular object for which it gets a true
let arr1 = [{ id: 1, name: "kiet" }, { id: 2, name: 'phan' }, { id: 3, name: 'tuan'}]
let arr2 = [ { id: 1, name: "kiet" }, { id: 2, name: 'haha' }, { id: 5, name: 'tuan' } ]
const similarity = (arr, values) => arr.filter(item => values.some(m => (m.id === item.id) && (m.name === item.name)));
console.log(similarity(arr1,arr2))
You get the elements from the array which have the same properties id/name as the values array.
Array#filter needs a (kind of) boolean value and if true or truthy, then the item is taken to the new array.
Array#some checks an item and if the callback returns a truthy value, then it returns true, if not then false.
const
similarity = (array, values) => array.filter(item => values.some(m => m.id === item.id && m.name === item.name)),
arr1 = [ { id: 1, name: "kiet" }, { id: 2, name: 'phan' }, { id: 3, name: 'tuan' }],
arr2 = [ { id: 1, name: "kiet" }, { id: 2, name: 'haha' }, { id: 5, name: 'tuan' }],
result = similarity(arr1, arr2);
console.log(result);
For larger data sets, you could take a Set, this is iterated once with a combined value and checked against for filtering.
const
similarity = (array, values) => {
const
getKey = ({ id, name }) => [id, name].join('|'),
keys = new Set(values.map(getKey));
return array.filter(o => keys.has(getKey(o)));
},
arr1 = [ { id: 1, name: "kiet" }, { id: 2, name: 'phan' }, { id: 3, name: 'tuan' }],
arr2 = [ { id: 1, name: "kiet" }, { id: 2, name: 'haha' }, { id: 5, name: 'tuan' }],
result = similarity(arr1, arr2);
console.log(result);
This is a good technique if you want to check for array intersection.
const vowels = [...'AEIOU'];
const isVowel = x => vowels.some(v => v === x);
const hasVowel = string => [...string].filter(isVowel).length > 0;

JavaScript - Filter a array by iterating on another array

The first object{array}, the one i want to filter :
const object1 = {
"count" : 2,
"result" : [
{ "id": 1 },
{ "id": 2 }
]
}
The second array :
const array2 = [{
"id": 1,
"id": 44
}]
I want to filter the first array object1.result (or create a new array apart from it) if object.result[i].id is equal to array2[i].id , an reduce the number of array count object1.count based on the number of filter element.
in the example above I should have a new object :
theNewObject = {
"count" : 1,
"result" : [
{ "id": 2 }
]
}
You could use a Set an collect all id for filtering.
var object = { count : 2, result : [{ id: 1 }, { id: 2 }] },
array = [{ id: 1 }, { id: 44 }],
ids = new Set(array.map(({ id }) => id));
object.result = object.result.filter(({ id }) => ids.has(id));
object.count = object.result.length;
console.log(object);
An approach which counts down the count.
var object = { count : 2, result : [{ id: 1 }, { id: 2 }] },
array = [{ id: 1 }, { id: 44 }],
ids = new Set(array.map(({ id }) => id));
object.result = object.result.filter(({ id }) => ids.has(id) || !object.count--);
console.log(object);
You can combine the filter() method with the find() method:
const object1 = {
count: 2,
result: [
{ id: 1 },
{ id: 2 },
],
};
const array2 = [
{ id: 1 },
{ id: 4 },
];
const filteredResult = object1.result.filter(({ id }) => !array2.find(x => x.id === id));
const object3 = {
count: filteredResult.length,
result: filteredResult,
};
console.log(object3);
The { id } syntax is destructuring assignment.
You can also use reduce():
const object1 = {
count: 2,
result: [
{ id: 1 },
{ id: 2 },
],
};
const array2 = [
{ id: 1 },
{ id: 4 },
];
const object3 = object1.result.reduce(
({ count, result }, { id }) => array2.find(x => x.id === id)
? ({ count, result })
: ({ count: count + 1, result: result.concat([{ id }]) }),
{ count: 0, result: [] },
);
console.log(object3);

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