I have an Array full of transactions and I want to divide it by day. It will be an array of date that is and array of transations. It may be a little messy but I want to return this structure.
What I tried to do returns me the structure I want, but I don't know how to merge duplicated key values.
This is the array
const transactions = [
{
name: "Salário",
receiveDate: "2020-05-12T00:00:00.000Z",
value: "1000",
},
{
name: "Pagamento ",
receiveDate: "2020-05-12T00:00:00.000Z",
value: "2350",
},
{
name: "Passagem no VEM",
paidDate: "2020-05-02T00:00:00.000Z",
value: "130",
},
{
name: "Almoço",
paidDate: "2020-05-08T00:00:00.000Z",
value: "50",
},
];
This is what I already tried by now
const days = [];
const finalArray = [];
for (let i = 0; i < transactions.length; i++) {
transactions[i].day = transactions[i].receiveDate || transactions[i].paidDate;
days.push(transactions[i].day);
}
const datesToMatch = [...new Set(days)].map((date) => {
return { [date]: null };
});
transactions.map((transaction) => {
datesToMatch.map((dayObject) => {
const day = Object.keys(dayObject).toString();
if (day === transaction.day) {
finalArray.push({ [day]: [transaction] });
}
});
});
The output
[ { '2020-05-12T00:00:00.000Z': [ [Object] ] },
{ '2020-05-12T00:00:00.000Z': [ [Object] ] },
{ '2020-05-02T00:00:00.000Z': [ [Object] ] },
{ '2020-05-08T00:00:00.000Z': [ [Object] ] } ]
Expected output
[ { '2020-05-12T00:00:00.000Z': [ [Object, Object] ] },
{ '2020-05-02T00:00:00.000Z': [ [Object] ] },
{ '2020-05-08T00:00:00.000Z': [ [Object] ] } ]
Thanks!
Explanation:
dates : extract dates from both fields
uniqueDates : build a Set and convert it into an array so it only has uniqueDates
dateToTransactions : map every unique date to an object with one key (itself) and filter every transaction that is equal to it.
const transactions = [{
name: "Salário",
receiveDate: "2020-05-12T00:00:00.000Z",
value: "1000",
},
{
name: "Pagamento ",
receiveDate: "2020-05-12T00:00:00.000Z",
value: "2350",
},
{
name: "Passagem no VEM",
paidDate: "2020-05-02T00:00:00.000Z",
value: "130",
},
{
name: "Almoço",
paidDate: "2020-05-08T00:00:00.000Z",
value: "50",
},
];
const dates = transactions.map(x => {
const received = x.receiveDate || [];
const paid = x.paidDate || [];
return received + paid;
});
const uniqueDates = [...new Set(dates)];
const dateToTransactions =
uniqueDates.map(
date => {
sameDate = transactions.filter(x => x.receiveDate === date || x.paidDate == date);
return {[date]: sameDate};
});
console.log(dateToTransactions);
I would do something like this:
const days = [];
for (let i = 0; i < transactions.length; i++) {
transactions[i].day = transactions[i].receiveDate || transactions[i].paidDate;
days.push(transactions[i].day);
}
const result = new Map();
days.forEach((day) => {
result.set(day, [])
});
transactions.forEach((transaction) => {
let r = result.get(transaction.day);
r.push(transaction);
result.set(transaction.day, r);
});
Then, in the result map you have a list of the transactions that were made for each day.
This will give the result you expect
const days = {};
const finalArray = transactions.forEach((transaction) => {
let date = (transaction.receiveDate || transaction.paidDate)
if (!days[date]) { days[date] = [transaction]}
else {days[date].push(transaction)}
});
console.log(days);
Related
I'm working on school-app. person enter students marks from frontend and I've to store it in my backend. I know my data-structure is quite bad. but this is only way I can comfortly use and fit it in my front end application/website.
codeSandbox link
Full Code:
//This data is already set need to push information in this array.
let student = [{
"detail": {
"name": "Mark",
"surname":"widen"
},
}];
//formatting the query in json.
const keys = Object.keys(query)[0].split(",")
const values = Object.values(query)[0].split(",")
const newObj = {}
for (let i = 0; i < keys.length; i++) {
newObj[keys[i]] = values[i]
}
// I've to push it along with "academic-year". so,
for (let a = 0; a < newObj.length; a++) {
const year = a + "st-Year"
console.log(year) // Expected Output: 1st-Year and 2nd-Year
}
// How to run this both for-loop synchronously way ?? AND
//pushing with "ObtainedMarks" and "year" (Error over here)
student.push({
ObtainedMarks: {
year : [
{ physics: newObj }
],
year : [
{ physics: newObj }
]
}
})
console.log(student) //Here's I want expected Output
Expected Output:
let student = [{
"detail": {
"name": "Mark",
"surname":"widen"
},
ObtainedMarks: {
"1st-Year": [
{ physics: { "marks": "500" } } //Physics subject is default.
],
"2nd-Year": [
{ physics: { "mark": "200" } } //Physics subject is default.
]
}
}];
I want to push returned data in student array. with 1st-Year
and 2nd-Year's for-loop.
You can do the conversion in your for-loop
let student = [{
"detail": {
"name": "Mark",
"surname": "widen"
},
}];
let query = {
"marks,mark": "500,200"
}
const keys = Object.keys(query)[0].split(",");
const values = Object.values(query)[0].split(",");
const marks = {}
for (let i = 0; i < keys.length; i++) {
marks[i === 0 ? `${i+1}st-year` : `${i+1}nd-year`] = [{
physics: {
[keys[i]]: values[i]
}
}];
}
student.push({
obtainedMarks: marks
});
console.log(student);
Alternative way: map through the keys and create an object from entries after manipulating the data.
let student = [{
"detail": {
"name": "Mark",
"surname": "widen"
},
}];
let query = {
"marks,mark": "500,200"
}
const keys = Object.keys(query)[0].split(",");
const values = Object.values(query)[0].split(",");
const marks = Object.fromEntries(keys.map((k, i) => {
return [
i === 0 ? `${i+1}st-year`: `${i+1}nd-year`,
[{ physics: { [k]: values[i] }}]
];
}));
student.push({
obtainedMarks: marks
});
console.log(student);
I have an array like this.
let arr = [
{
"ABBRIVATION":"ISB",
"name":"ISLAMABAD",
},
{
"ABBRIVATION":"RAW",
"name":"PINDI",
},
{
"ABBRIVATION":"SWB",
"name":"SWABI",
},
{
"ABBRIVATION":"AQ",
"name":"AQEEL",
},
]
I want to change it to like this
let me explain it a little. I want to assign the abbreviation directly to the name and the iterate through that array
let outout = [
{
"ISB":"ISLAMABAD"
},
{
"RAW":"ISLAMABAD"
},
{
"SWB":"SWABI"
},
{
"AQ":"AQEEL"
},
]
that is what I tried
let k = arr.map((item) => {
return item.ABB = item.name
})
console.log(k)
and here is the output
[ 'ISLAMABAD', 'PINDI', 'SWABI', 'AQEEL' ]
Here you go, use array map, simples
let arr = [
{
"ABBRIVATION":"ISB",
"name":"ISLAMABAD",
},
{
"ABBRIVATION":"RAW",
"name":"PINDI",
},
{
"ABBRIVATION":"SWB",
"name":"SWABI",
},
{
"ABBRIVATION":"AQ",
"name":"AQEEL",
},
]
let outout = arr.map(({ABBRIVATION, name}) => ({[ABBRIVATION]: name}));
console.log(outout);
Nothing more than a simple Array.prototype.map() needed.
let arr = [
{
ABBRIVATION: "ISB",
name: "ISLAMABAD",
},
{
ABBRIVATION: "RAW",
name: "PINDI",
},
{
ABBRIVATION: "SWB",
name: "SWABI",
},
{
ABBRIVATION: "AQ",
name: "AQEEL",
},
];
const result = arr.map(e => ({ [e.ABBRIVATION]: e.name }));
console.log(result);
map over the array of objects (map returns a new array) and assign the name to a new key defined by the abbreviation.
You code works the way it does because item.ABB is undefined, but you're also assigning item.name to it which does get returned, so you just get an array of names returned.
const arr=[{ABBRIVATION:"ISB",name:"ISLAMABAD"},{ABBRIVATION:"RAW",name:"PINDI"},{ABBRIVATION:"SWB",name:"SWABI"},{ABBRIVATION:"AQ",name:"AQEEL"}];
const out = arr.map(obj => {
return { [obj.ABBRIVATION]: obj.name };
});
console.log(out);
Hi I have seen people answer, but most of them use the map function, I provide some other solutions, hoping to expand the thinking
Use forEach function
const datas = [
{
"ABBRIVATION":"ISB",
"name":"ISLAMABAD",
},
{
"ABBRIVATION":"RAW",
"name":"PINDI",
},
{
"ABBRIVATION":"SWB",
"name":"SWABI",
},
{
"ABBRIVATION":"AQ",
"name":"AQEEL",
}
];
datas.forEach((obj, i, arr) => {
const{'ABBRIVATION':k, 'name':v} = obj;
arr[i] = {[k]:v};
});
console.log(datas);
Use flatMap function
const datas = [
{
"ABBRIVATION":"ISB",
"name":"ISLAMABAD",
},
{
"ABBRIVATION":"RAW",
"name":"PINDI",
},
{
"ABBRIVATION":"SWB",
"name":"SWABI",
},
{
"ABBRIVATION":"AQ",
"name":"AQEEL",
}
];
const result = datas.flatMap(obj => {
const {'ABBRIVATION':k, 'name':v} = obj;
return {[k]:v};
});
console.log(result);
this is how you suppose to do it.
arr.reduce((d, c)=>([...d, {[c.ABBRIVATION]: c.name}]),[])
let arr = [
{
"ABBRIVATION":"ISB",
"name":"ISLAMABAD",
},
{
"ABBRIVATION":"RAW",
"name":"PINDI",
},
{
"ABBRIVATION":"SWB",
"name":"SWABI",
},
{
"ABBRIVATION":"AQ",
"name":"AQEEL",
},
]
console.log(arr.reduce((data, current)=>([...data, {[current.ABBRIVATION]: current.name}]),[]))
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);
I have an array:
let ar = [
{
uid:1,
flat_no: 1
},
{
uid:2,
flat_no: 2
},
{
uid:1,
flat_no:3
}
];
If uid are same then I want to remove duplicate uid and concatenate its flat_no. The output array should be like this:
[
{
uid:1,
flat_no: [1,3]
},
{
uid:2,
flat_no: 2
}
];
You can use a combination of Array.reduce and Array.find.
If you find an existing item in your accumulator array, just update it's flat_no property, otherwise push it to the accumulator array.
let arr = [
{
uid: 1,
flat_no: 1
},
{
uid: 2,
flat_no: 2
},
{
uid: 1,
flat_no: 3
}
]
arr = arr.reduce((arr, item) => {
const existing = arr.find(innerItem => innerItem.uid === item.uid)
if (existing) {
existing.flat_no = Array.isArray(existing.flat_no)
? existing.flat_no
: [existing.flat_no]
existing.flat_no.push(item.flat_no)
} else {
arr.push(item)
}
return arr
}, [])
console.log(arr)
You can iterate over your array and fill an object (used as a hashmap here).
Once done, you extract the values to get your result.
let hashResult = {}
ar.forEach(element => {
if (hashResult[element.uid] == undefined) {
hashResult[element.uid] = { uid: element.uid, flat_no: [] }
}
hashResult[element.uid].flat_no.push(element.flat_no)
})
let result = Object.values(hashResult)
console.log(new Date(), result)
You can do this in a concise way with a single Array.reduce and Object.values to match your desired output:
let data = [ { uid:1, flat_no: 1 }, { uid:2, flat_no: 2 }, { uid:1, flat_no:3 } ];
const result = data.reduce((r, {uid, flat_no}) => {
r[uid] ? r[uid].flat_no = [r[uid].flat_no, flat_no] : r[uid] = {uid, flat_no}
return r
}, {})
console.log(Object.values(result))
1)Reduce the initial array to an object which has uid as the key and the flat_no as the value.
2)Then run a map on the keys to convert it into an array of objects with uid and flat_no.
1) First Step Code
let ar = [{uid:1, flat_no: 1},{uid:2, flat_no: 2},{uid:1, flat_no:3}];
let outputObj = ar.reduce((outputObj,currObj,currIndex) => {
let {uid,flat_no} = currObj
if (outputObj[uid]) {
outputObj[uid].push(flat_no)
}
else {
outputObj[uid] = [flat_no]
}
return outputObj
},{})
2)
let finalOutput = Object.keys(outputObj).map(key =>
({uid:key,flat_no:outputObj[key]}))
console.log(finalOutput)
I have an array like bellow each index contains different set of objects,I want to create an uniformal data where object missing in each index will with Value:0 ,
var d = [
[
{axis:"Email",value:59,id:1},
{axis:"Social Networks",value:56,id:2},
],
[
{axis:"Sending Money",value:18,id:6},
{axis:"Other",value:15,id:7},
]
];
how can I get an array like bellow using above above array
var d = [
[
{axis:"Email",value:59,id:1},
{axis:"Social Networks",value:56,id:2},
{axis:"Sending Money",value:0,id:6},
{axis:"Other",value:0,id:7},
],
[
{axis:"Email",value:0,id:1},
{axis:"Social Networks",value:0,id:2},
{axis:"Sending Money",value:18,id:6},
{axis:"Other",value:15,id:7},
]
];
There are two functions:
getAllEntries that find all objects and stores them into a variable accEntries. Then accEntries is used to search for all occurrences in a sub-array of d. This whole process is done in checkArray.
checkArray is used to fetch all found and not-found entries in d. Both Arrays (found and not-found) are then used to build a new sub-array that contains either found entries with certain values and/or not-found entries with values of 0.
Hope this helps:
var d = [
[
{
axis: 'Email',
value: 59,
id: 1
},
{
axis: 'Social Networks',
value: 56,
id: 2
},
],
[
{
axis: 'Sending Money',
value: 18,
id: 6
},
{
axis: 'Other',
value: 15,
id: 7
},
]
];
function getAllEntries(array) {
var uniqueEntries = [];
array.forEach(function (subarray) {
subarray.forEach(function (obj) {
if (uniqueEntries.indexOf(obj) === - 1) uniqueEntries.push(obj);
});
});
return uniqueEntries;
}
function checkArray(array, acceptedEntries) {
var result = [];
array.forEach(function (subArray) {
var subResult = [];
var foundEntries = [];
subArray.forEach(function (obj) {
if (foundEntries.indexOf(obj.axis) === - 1) foundEntries.push(obj.axis);
});
var notFound = acceptedEntries.filter(function (accepted) {
return foundEntries.indexOf(accepted.axis) === - 1;
});
foundEntries.forEach(function (found) {
subArray.forEach(function (obj) {
if (obj.axis === found) subResult.push(obj);
});
});
notFound.forEach(function (notfound, index) {
subResult.push({
axis: notfound.axis,
value: 0,
id: notfound.id
});
});
result.push(subResult);
});
return result;
}
var accEntries = getAllEntries(d);
var result = checkArray(d, accEntries);
console.log(result);
You can loop over the array to find all the unique objects and then again loop over to push the values that are not present comparing with the array of objects of unique keys.
You can use ES6 syntax to find if an object with an attribute is present like uniKeys.findIndex(obj => obj.axis === val.axis); and the to push with a zero value use the spread syntax like d[index].push({...val, value: 0});
Below is the snippet for the implementation
var d = [
[
{axis:"Email",value:59,id:1},
{axis:"Social Networks",value:56,id:2},
],
[
{axis:"Sending Money",value:18,id:6},
{axis:"Other",value:15,id:7},
{axis:"Social Networks",value:89,id:2},
]
];
var uniKeys = [];
$.each(d, function(index, item) {
$.each(item, function(idx, val){
const pos = uniKeys.findIndex(obj => obj.axis === val.axis);
if(pos == - 1) {
uniKeys.push(val);
}
})
})
$.each(d, function(index, item) {
var temp = [];
$.each(uniKeys, function(idx, val){
const pos = item.findIndex(obj => obj.axis === val.axis);
if(pos == - 1) {
d[index].push({...val, value: 0});
}
})
})
console.log(d);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
How about a short shallowCopy function (Object.assign is not available in IE) and otherwise less than 10 new lines of code?
var d = [
[
{axis:"Email",value:59,id:1},
{axis:"Social Networks",value:56,id:2}
],
[
{axis:"Sending Money",value:18,id:6},
{axis:"Other",value:15,id:7}
]
];
var newD_0 = [shallowCopy(d[0][0]), shallowCopy(d[0][1]), shallowCopy(d[1][0]), shallowCopy(d[1][1])];
var newD_1 = [shallowCopy(d[0][0]), shallowCopy(d[0][1]), shallowCopy(d[1][0]), shallowCopy(d[1][1])];
newD_0[2].id = 0;
newD_0[3].id = 0;
newD_1[0].id = 0;
newD_1[1].id = 0;
d = [newD_0, newD_1];
function shallowCopy(obj) {
var copy = {};
for (var key in obj) {
if (obj.hasOwnProperty(key)) {
copy[key] = obj[key];
}
}
return copy;
}
console.log(JSON.stringify(d));
RESULT:
[
[
{
"axis":"Email",
"value":59,
"id":1
},
{
"axis":"Social Networks",
"value":56,
"id":2
},
{
"axis":"Sending Money",
"value":18,
"id":0
},
{
"axis":"Other",
"value":15,
"id":0
}
],
[
{
"axis":"Email",
"value":59,
"id":0
},
{
"axis":"Social Networks",
"value":56,
"id":0
},
{
"axis":"Sending Money",
"value":18,
"id":6
},
{
"axis":"Other",
"value":15,
"id":7
}
]
]