I have 3 two dimensional arrays as given below which are series data to plot lines on a graph with the key being the timestamp.
const arr1 = [[1641013200000,1881],[1643691600000,38993],[1646110800000,41337],[1648785600000,78856],[1651377600000,117738],[1654056000000,119869],[1656648000000,157799],[1659326400000,196752],[1662004800000,199061],[1664596800000,237034],[1667275200000,239153],[1669870800000,269967]]
const arr2 = [[1641013200000,1302],[1643691600000,3347],[1646110800000,4754],[1648785600000,6948],[1651377600000,9725],[1654056000000,11314],[1656648000000,13787],[1659326400000,16666],[1662004800000,18370],[1664596800000,20876],[1667275200000,22384],[1669870800000,23560]]
const arr3 = [[1643691600000,67350],[1648785600000,134700],[1651377600000,202148],[1654056000000,202270],[1656648000000,269843],[1659326400000,337346],[1662004800000,337470],[1664596800000,404861],[1667275200000,404889],[1669870800000,472239]]
I want to plot another series line which gives the cumulative total of all three arrays values
(Note: if a timestamp is not present in either of the arrays, add the previous index value)
const totalArray = [
[1641013200000,3183],[1643691600000, 109690],[1646110800000, 113441],[1648785600000, 220504],
[1651377600000, 329611],[1654056000000, 333453],[1656648000000, 441429],[1659326400000, 550764],
[1662004800000, 554901],[1664596800000, 662771],[1667275200000, 666426],[1669870800000, 765766]
]
I have tried this, but some values are incorrect due to the timestamp not being present in either one
Approach:
const arr1 = [
[1641013200000, 1881],
[1643691600000, 38993],
[1646110800000, 41337],
[1648785600000, 78856],
[1651377600000, 117738],
[1654056000000, 119869],
[1656648000000, 157799],
[1659326400000, 196752],
[1662004800000, 199061],
[1664596800000, 237034],
[1667275200000, 239153],
[1669870800000, 269967]
];
const arr2 = [
[1641013200000, 1302],
[1643691600000, 3347],
[1646110800000, 4754],
[1648785600000, 6948],
[1651377600000, 9725],
[1654056000000, 11314],
[1656648000000, 13787],
[1659326400000, 16666],
[1662004800000, 18370],
[1664596800000, 20876],
[1667275200000, 22384],
[1669870800000, 23560]
];
const arr3 = [
[1643691600000, 67350],
[1648785600000, 134700],
[1651377600000, 202148],
[1654056000000, 202270],
[1656648000000, 269843],
[1659326400000, 337346],
[1662004800000, 337470],
[1664596800000, 404861],
[1667275200000, 404889],
[1669870800000, 472239]
];
const calculateTotal = () => {
var ret;
for (let a3 of arr3) {
var index = arr1.map(function(el) {
return el[0];
}).indexOf(a3[0]);
console.log(index);
if (index === -1) {
ret = arr1[index][0];
console.log(ret);
}
}
let unsortedArr = arr1.concat(arr2, arr3);
var sortedArray = unsortedArr.sort((a, b) => a[0] - b[0]);
var added = addArray(sortedArray);
console.log("Curent Output: " + JSON.stringify(added));
}
const addArray = (tuples) => {
var hash = {},
keys = [];
tuples.forEach(function(tuple) {
var key = tuple[0],
value = tuple[1];
if (hash[key] === undefined) {
keys.push(key);
hash[key] = value;
} else {
hash[key] += value;
}
});
return keys.map(function(key) {
return ([key, hash[key]]);
});
}
calculateTotal();
Is it possible to achieve this?
In your code there is this:
if (index === -1) {
ret = arr1[index][0];
But that assignment will fail as arr1[-1] is not defined.
Then when you do:
let unsortedArr = arr1.concat(arr2, arr3);
...you end up with an array that does not have the knowledge to use default values (from a previous index) when any of the three arrays has a "missing" time stamp.
I would suggest this approach:
Collect all the unique timestamps (from all arrays) into a Map, and associate arrays to each of those keys: these will be empty initially.
Populate those arrays with the timestamps from the original arrays
Get the sorted list of entries from that map
Fill the "gaps" by carrying forward values to a next array when the corresponding slot is undefined. At the same time sum up these values for the final output.
Here is an implementation:
function mergeArrays(...arrays) {
const map = new Map(arrays.flatMap(arr => arr.map(([stamp]) => [stamp, []])));
arrays.forEach((arr, i) => {
for (const [timeStamp, value] of arr) {
map.get(timeStamp)[i] = value;
}
});
const state = Array(arrays.length).fill(0);
return Array.from(map).sort(([a], [b]) => a - b).map(([timeStamp, arr], i) =>
[timeStamp, state.reduce((sum, prev, j) => sum + (state[j] = arr[j] ?? prev), 0)]
);
}
// Example run
const arr1 = [[1641013200000,1881],[1643691600000,38993],[1646110800000,41337],[1648785600000,78856],[1651377600000,117738],[1654056000000,119869],[1656648000000,157799],[1659326400000,196752],[1662004800000,199061],[1664596800000,237034],[1667275200000,239153],[1669870800000,269967]];
const arr2 = [[1641013200000,1302],[1643691600000,3347],[1646110800000,4754],[1648785600000,6948],[1651377600000,9725],[1654056000000,11314],[1656648000000,13787],[1659326400000,16666],[1662004800000,18370],[1664596800000,20876],[1667275200000,22384],[1669870800000,23560]];
const arr3 = [[1643691600000,67350],[1648785600000,134700],[1651377600000,202148],[1654056000000,202270],[1656648000000,269843],[1659326400000,337346],[1662004800000,337470],[1664596800000,404861],[1667275200000,404889],[1669870800000,472239]];
const result = mergeArrays(arr1, arr2, arr3);
console.log(result);
For input:
["abc","def","okg","fed","bca"]
expected output should be:
["abc","bca"],["def","fed"],["okg"]
here "abc", "bca" and "def", "fed" contains same character and "okg" there is no element which contains these character
const arr = ["abc", "def", "okg", "fed", "bca"];
let find = (arr) => {
let res = [];
for (let i = 0; i < arr.length; i++) {
for (let j = 1; j < arr.length; j++) {
if (arr[i].search(arr[j])) {
res.push(arr[j]);
}
}
}
return res;
}
console.log(find(arr))
A reduce will do the trick - it seems the shortest code here (apart from the one using lodash)
const arr = ["abc", "def", "okg", "fed", "bca"],
res = Object.values(arr.reduce((acc, ele) => {
const key = ele.split("").sort();
(acc[key] = acc[key] || []).push(ele)
return acc
}, {}))
console.log(res)
.search returns a number indicating the index of where the match was found. Check that the result isn't -1 instead of checking that the result is truthy. But...
.search isn't the right tool here anyway, because it won't find different combinations of the same character. You need a different approach. One way would be to create an object whose keys are the characters found, and the values are the number of occurrences, then use a sorted representation of that object as a key. For example, have both abc and bca turn into something like:
a,1-b,1-c,1
Iterate through the input array, generating a key for each string, and putting the string on an object with that key. At the end, take the object's values.
const strToKey = (str) => {
const grouped = {};
for (const char of str) {
grouped[char] = (grouped[char] || 0) + 1;
}
return Object.entries(grouped)
.sort((a, b) => a[0].localeCompare(b[0]))
.join('-');
};
let find = (arr) => {
const grouped = {};
for (const str of arr) {
const key = strToKey(str);
grouped[key] ??= [];
grouped[key].push(str);
}
return Object.values(grouped);
}
console.log(find(["abc", "def", "okg", "fed", "bca"]));
Another option, when creating the keys, instead of sorting the object afterwards, you could sort the string first:
const strToKey = (str) => {
const grouped = {};
for (const char of [...str].sort()) {
grouped[char] = (grouped[char] || 0) + 1;
}
return Object.entries(grouped).join('-');
};
let find = (arr) => {
const grouped = {};
for (const str of arr) {
const key = strToKey(str);
grouped[key] ??= [];
grouped[key].push(str);
}
return Object.values(grouped);
}
console.log(find(["abc", "def", "okg", "fed", "bca"]));
const input = ["abc","def","okg","fed","bca"]
function getSortedString (str) {
return [...str].sort().join('');
};
function groupBy(input) {
const grouped = [];
while(input.length) {
const nextInput = [];
const first = input[0];
const matched = [first];
for (let i = 1; i < input.length; i++) {
if(getSortedString(first) === getSortedString(input[i])) {
matched.push(input[i])
} else {
nextInput.push(input[i])
}
}
input = nextInput;
grouped.push(matched);
}
console.log(grouped);
}
groupBy(input);
Using Object.values and groupBy (from lodash), you can get a straightforward solution:
You group your array elements by their "sorted" form and then use Object.values to get the output array.
const arr = ["abc", "def", "okg", "fed", "bca"];
const sortString = (str) => str.split("").sort().join("")
const result = Object.values(_.groupBy(arr, sortString));
console.log(result);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.21/lodash.min.js"></script>
I want to sort an array values in an ascending or descending order without using sort().
I have created a function, however I am not satisfied with it.
I believe the code below could be much shorter and more concise.
Please let me know where to modify or you may entirely change the code too. Thank you in advance.
const func = arg => {
let flip = false;
let copy = [];
for(let val of arg) copy[copy.length] = val;
for(let i=0; i<arg.length; i++) {
const previous = arg[i-1];
const current = arg[i];
if(previous > current) {
flip = true;
copy[i] = previous;
copy[i-1] = current;
}
}
if(flip) return func(copy);
return copy;
};
l(func([5,2,8,1,9,4,7,3,6]));
If your input is composed of whole numbers, as in the example, pne option is to reduce the array into an object, whose keys are the numbers, and whose values are the number of times those values have occured so far. Then, iterate over the object (whose Object.entries will iterate in ascending numeric key order, for whole number keys), and create the array to return:
const func = arr => {
const valuesObj = {};
arr.forEach((num) => {
valuesObj[num] = (valuesObj[num] || 0) + 1;
});
return Object.entries(valuesObj)
.flatMap(
([num, count]) => Array(count).fill(num)
);
};
console.log(
func([5,2,8,1,9,10,10,11,4,7,3,6])
);
This runs in O(N) time.
To account for negative integers as well while keeping O(N) runtime, create another object for negatives:
const func = arr => {
const valuesObj = {};
const negativeValuesObj = {};
arr.forEach((num) => {
if (num >= 0) valuesObj[num] = (valuesObj[num] || 0) + 1;
else negativeValuesObj[-num] = (negativeValuesObj[-num] || 0) + 1;
});
return [
...Object.entries(negativeValuesObj).reverse()
.flatMap(
([num, count]) => Array(count).fill(-num)
),
...Object.entries(valuesObj)
.flatMap(
([num, count]) => Array(count).fill(num)
)
];
};
console.log(
func([5,2,8,1,-5, -1, 9,10,10,11,4,7,3,6, -10])
);
For non-integer items, you'll have to use a different algorithm with higher computational complexity.
How to write a function to remove certain elements into a new array and leave the original array with only the remaining elements?
the first part is easy using a for loop pushing the even numbers into a new array but mutating the original array to leave only the odd numbers is hard
function remove(arr, cb){
var removed = [];
var newArr = [];
for(var i = 0; i < arr.length; i++) {
if(cb(arr[i], i, arr)) {
removed.push(arr[i]);
}
}
return removed;
}
Use an else statement to fill newArr with values that should stay in the original arr, then empty it using splice() before copying the items from newArr back into it.
function remove (arr, cb) {
var removed = [];
var newArr = [];
for (var i = 0; i < arr.length; i++) {
if (cb(arr[i], i, arr)) {
removed.push(arr[i]);
} else {
newArr.push(arr[i]);
}
}
arr.splice(0);
for (var i = 0; i < newArr.length; i++) {
arr.push(newArr[i]);
}
return removed;
}
Welcome to Stackoverflow!
Personally, I'd avoid anything that mutates an input parameter, as this increases code complexity and makes it hard to reason about what's happening from the calling side.
Instead, I'd write a method that returns an array of two arrays. This can be easily split into two variables at the calling end using by using array destructuring.
See the example below:
const splitArr = (arr, pred) =>
arr.reduce(
(prev, curr, idx) => {
prev[+pred(curr, idx, arr)].push(curr);
return prev;
}, [[], []]
);
// usage //
const myArr = [1, 2, 3, 4];
const [arr1, arr2] = splitArr(myArr, x => x > 2);
console.log(arr1);
console.log(arr2);
Because pred is a function that returns a boolean value, we can co-erce this value to 0 or 1 using +someBoolean. We can then use this value as an index to decide into which of the two output arrays the value should be pushed.
You were definitely on the right track with your solution, a couple tweaks and we can make it very readable and also very easy to work with. I tried to keep the format of what it looked like you were doing.
I do take advantage of destructuring here, this could be returned as just an object, and then reference the properties.
const myArr = [0,1,2,3,4,5,6,7,8,9,10];
const splitItems = (arr, logicFunc) => {
let secondSet = []
const firstSet = arr.filter(v => {
if (logicFunc(v)) return true
else secondSet.push(v)
})
return { firstSet, secondSet }
}
const myLogicFunc = v => (v < 3 || v == 9)
const { firstSet, secondSet } = splitItems(myArr, myLogicFunc)
console.log(`My first set: ${firstSet}`) // My first set: 0,1,2,9
console.log(`My second set: ${secondSet}`) // My second set: 3,4,5,6,7,8,10
/* OR without destructuring:
const myArrays = splitItems(myArr, myLogicFunc)
console.log(`My first set: ${myArrays.firstSet}`)
console.log(`My second set: ${myArrays.secondSet}`)
*/
Please let me know if you have any questions
In modern JavaScript apps we do not mutate arrays we create new array, this avoids side effects, so what we do is create two new arrays
const split = (source, conditionFunc) = [ source.filter(i => conditionFunc(i)), source.filter(i => !conditionFunc(i))];
Then you have an array of two arrays of the values that meed condition and those that don't and you have not caused any side effects.
const odssAndEvens = split(source, i => i % 2 === 1);
Or with reduce so you don't iterate the array twice
const split = (source, conditionFunc) = source.reduce((results, item) => {
if (conditionFunc(item)) {
results[0].push(item);
} else {
results[1].push(item);
}
return results;
}, [[],[]]);