We Keep Coding

Is there any way to sum all sub array item and then multiply all sub arrays using a 'for' or 'for-of' loop?

i want to create a function using for() or for of() loop, which take an nested array as argument then add and multiply its item.
Suppose, myArray = [[5,6],[9,2],[4,8]]
Now i want to process it like: [[5+6] * [9+2] * [4+8]]
I solve it using .map() and .reduce(), but is there any way to do same using classic for() or for of() loop. this is my trial.
let myArray = [[1,2],[3,4],[5,6]]
function multyPlus(array) {
resul = 0
for (const subArray of array) {
for (const num of subArray) {
resul += num
}
resul *= subArray
}
return resul
}
console.log(multyPlus(myArray));
//Nan

I would try a two step system that first adds the numbers, then multiplies it to the previous numbers:
function sum(array) {
var total = 0;
for (var item of array)
total += item;
return total;
}
var myArray = [[5,6],[9,2],[4,8]];
var output = 1;
for (var item of myArray)
output *= sum(item);

Maybe Like This:
let myArray = [[1,2],[3,4],[5,6]]
function multyPlus(_array){
var out = 1;
for(var key1 in _array){
var out2 = 0;
for(var key2 in _array[key1]){
out2 += _array[key1][key2];
}
out = out * out2;
}
return out;
}
console.log(multyPlus(myArray));

You can define separate adder and multiplier functions -
const adder = (nums = []) =>
{ let r = 0
for (const n of nums)
r += n
return r
}
const multiplier = (nums = []) =>
{ let r = 1
for (const n of nums)
r *= n
return r
}
const myCalc = (input = []) =>
{ const r = []
for (const x of input)
r.push(adder(x))
return multiplier(r)
}
const result =
myCalc([[1,2],[3,4],[5,6]])
console.log(result) // 231
That said, I think the functional approach is superior when you use named functions. Each function is highly reusable and there's virtually no room for bugs to hide -
const add = (x = 0, y = 0) =>
x + y
const mult = (x = 0, y = 0) =>
x * y
const sum = (nums = []) =>
nums.reduce(add, 0)
const product = (nums = []) =>
nums.reduce(mult, 1)
const myCalc = (input = []) =>
product(input.map(sum)) // <-- easy peasy!
const result =
myCalc([[1,2],[3,4],[5,6]])
console.log(result) // 231
If you have something against map and reduce, you can write myCalc and sum by hand using simple recursion -
const sum = ([ x, ...more ]) =>
x === undefined
? 0
: x + sum(more)
const myCalc = ([ x, ...more ]) =>
x === undefined
? 1
: sum(x) * myCalc(more)
const result =
myCalc([[1,2],[3,4],[5,6]])
console.log(result) // 231

How to populate an array with integers

Please, how do you populate an array say ‘num’ with numbers not in a second array say ‘fig’? I’m trying to use a loop to have the values of the already populated array ‘fig’ compared to ‘num’ which is to be populated with integers not found in ‘fig’. I’m a bit confused.

If you need to do an array with n numbers you can use this two ways.
const arrayLength = 100;
const numberArray = [...new Array(arrayLength).keys()]
const anotherWay = new Array(arrayLength).fill().map((_, idx) => idx + 1);
console.log(numberArray, anotherWay)

so to do this we have to do a few things:
1) define an existing array with numbers to avoid
2) define length on new array
3) generate a random number and make it an integer
4) check to see if we need to avoid
5) if it's a new value add it to the second array
var first=[55,45,35,1,2,3,4,5];
var second = [];
var i = 7;
var x;
while (i != 0){
x = ~~(Math.random()*100);
var check = false;
for(let j=0; j<first.length;j++){
if(x == first[j]){
check = true;
}
}
if(!check){
second.push(x);
i--;
}
}
console.log(second);

const fig = [-21, 0, 3, 6, 7, 42]
const min = Math.min(...fig) // or fig[0] if the the array is already sorted
const max = Math.max(...fig) // or fig[fig.length - 1]
const num = Array.from({ length: max - min }, (_, i) => i + min)
.filter(el => !fig.includes(el))
or, saving one loop
const num = Array.from({ length: max - min }).reduce((acc, _, i) => {
const curr = i + min
if (!fig.includes(curr)) {
return acc.concat(curr)
}
return acc
}, [])
This is assuming your range is from the smallest number in fig to the largest in fig.

JavaScript how to filter two arrays at the same time

Say I have two arrays:
var one = [1,2,3,4,5];
var two = ["A","B","C","D","E"];
If i were to filter the first array so that it returns this result:
var resultOne = one.filter(v => v == 2 || v == 3 );
The result would return [2,3] but how would I filter the other array to return [B,C]... based on the result of this first one ?

var one = [1,2,3,4,5];
var two = ["A","B","C","D","E"];
var resultOne = two.filter((v,i) => one[i] == 2 || one[i] == 3 );
console.log(resultOne)

You can implement filter on the array by passing the index, return if index + 1 includes in the resulted array.
Please Note: Since the solution is based on the index, it will not work for the random numbers of array (not sequential).
var one = [1,2,3,4,5];
var two = ["A","B","C","D","E"];
var resultOne = one.filter(v => v == 2 || v == 3 );
var resultTwo = two.filter((v,i) => resultOne.includes(i+1));
console.log(resultOne);
console.log(resultTwo);
OR: In Single line:
var one = [1,2,3,4,5];
var two = ["A","B","C","D","E"];
var result = two.filter((v,i) => one.filter(v => v == 2 || v == 3).includes(i+1));
console.log(result);

You can use Array.reduce():
var one = [1,2,3,4,5];
var two = ["A","B","C","D","E"];
const [resultOne, resultTwo] = one.reduce((acc, v, i) => (v == 2 || v == 3) ? (acc[0].push(v), acc[1].push(two[i]), acc) : acc, [[], []]);
console.log(resultOne);
console.log(resultTwo);
Or use this version if the two arrays haven't the same length:
var one = [1,2,3,4,5,6];
var two = ["A","B","C","D","E"];
const [resultOne, resultTwo] = one.reduce((acc, v, i) => (v == 2 || v == 6) ? (acc[0].push(v), (i in two) ? acc[1].push(two[i]) : '', acc) : acc, [[], []]);
console.log(resultOne);
console.log(resultTwo);

A good old for loop or use the second parameter of the filter callback which gives you the current array index. Simply push the corresponding element of two onto resultTwo by index.
var resultTwo = [];
var resultOne = one.filter((v, i) => {
var isMatch = v == 2 || v == 3;
if (isMatch) resultTwo.push(two[i]);
return isMatch;
});

EDIT: Just realized somone posted the same answer... The only thing I can add is that you need to subtract 1 from your original array to arrive at the correct index.
If you use map you can map the results from your filtering. This is assuming the result will always be an index for your second collection.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map
var one = [1,2,3,4,5];
var two = ["A","B","C","D","E"];
var resultOne = one.filter(v => v == 2 || v == 3 );
var resultTwo = resultOne.map((indexFromResultOne) => two[indexFromResultOne - 1]);
// Because you aren't using a zero based index.
console.log(resultTwo);

Not the most efficient, but this is the most general and readable way I can think of to filter multiple corresponding arrays:
var a = [1, 2, 3, 4, 5];
var b = ["A", "B", "C", "D", "E"];
var indices = [];
for (var i = 0; i < a.length; i++) {
var v = a[i];
if (v == 2 || v == 3) {
indices.push(i);
}
}
console.log(indices.map(i => a[i]));
console.log(indices.map(i => b[i]));

Combine values from two arrays in JavaScript

I have an array which looks like:
var data = [{"year":[1981],"weight":[3]},
{"year":[1982],"weight":[4]},
{"year":[1985],"weight":[7]}]
My data series starts with year 1980 and ends with year 1986. My task is to input all missing values into the array; in my case the final array should be:
var data = [{"year":[1980],"weight":[0]},
{"year":[1981],"weight":[3]},
{"year":[1982],"weight":[4]},
{"year":[1983],"weight":[0]},
{"year":[1984],"weight":[0]},
{"year":[1985],"weight":[7]},
{"year":[1986],"weight":[0]}]
I implemented this task in two steps. First I created an empty array with length of seven elements (for years 1980 - 1986) and initialize each element with value {"year": $CURRENT_YEAR, "weight": 0}. Then I loop through data array, find index of current year in the empty array and replace year and weight fields with current values. My code is pasted below.
I wonder if the code could be rewritten in a more elegant way.
// Create empty array
var my_array = []
var length = 7
// 1st step
year = 1980
for (var i = 0; i < length; i++) {
my_array.push({"year": year, "weight": 0});
year++
}
// 2nd step
for (var j = 0; j < data.length; j++) {
curr_year = data[j]["year"][0];
curr_weight = data[j]["weight"][0]
var index = my_array.findIndex(function(item, i) {return item.year === curr_year})
my_array[index] = {"year": curr_year, "weight": curr_weight}
}

It's best to do this job by .map() Besides if you have a large input array it might be wise to set up a hash (lut) in the first place such as;
var data = [{"year":[1981],"weight":[3]},
{"year":[1982],"weight":[4]},
{"year":[1985],"weight":[7]}],
lut = data.reduce((p,c) => p[c.year[0]] ? p : (p[c.year[0]] = c, p), {});
range = [1980,1986],
result = Array(range[1]-range[0] + 1).fill()
.map((_,i) => lut[i+range[0]] ? lut[i+range[0]] : {year: [i+range[0]], weight: [0]});
console.log(result);

You can combine the 2 loops and do both steps in one loop
// Create empty array
var my_array = []
var length = 7
year = 1980
for (var i = 0; i < length; i++) {
// check if there is data for the year
var index = data.findIndex(function(item, i) {return item.year === year});
if(index > -1){ //if there is data, use it
my_array.push({"year": data[index]["year"][0], "weight": data[index]["weight"][0]});
}else{ //put in default data
my_array.push({"year": year, "weight": 0});
}
year++;
}

Find index of element in array each time is bad performance for large data. I can suggest the following algorithm:
// Create empty object and fill it with values where keys are years
var years = {};
data.forEach(item => {
years[item.year[0]] = item.weight[0];
});
// Result array with all years
var result = [];
var startYear = 1980;
var endYear = 1986;
// Generate our result array
for (var i = startYear; i <= endYear; i++) {
// If property for given year (i) exists in "years" object then add it to "result" array
// in other case add default object with weight 0
var o = years[i] ? { year: [i], weight: [years[i]] } : { year: [i], weight: [0] };
result.push(o);
}

You could do this with just find() and while loop.
var data = [{"year":[1981],"weight":[3]},{"year":[1982],"weight":[4]},{"year":[1985],"weight":[7]}];
var i = 1980;
var result = [];
while(i <= 1986) {
var find = data.find(e => e.year[0] == i);
(find) ? result.push(find) : result.push({year: [i], weight: [0]});
i++;
}
console.log(result)
You could also first use map() to get array of years and then use while loop with indexOf().
var data = [{"year":[1981],"weight":[3]},{"year":[1982],"weight":[4]},{"year":[1985],"weight":[7]}];
var i = 1980;
var result = [];
var years = data.map(e => e.year[0]);
while(i <= 1986) {
var ind = years.indexOf(i);
(ind != -1) ? result.push(data[ind]) : result.push({year: [i], weight: [0]});
i++;
}
console.log(result)

Hashing array of strings in javascript

Just wondering if there is some other way than this.
var hashStringArray = function(array) {
array.sort();
return array.join('|');
};
I don't like sorting much and using that delimiter is not safe either if it's contained in one of the strings. In overall I need to produce same hash no matter the order of strings. It will be rather short arrays (up to 10 items), but it will be required very often so it shouldn't be too slow.
I intend to use it with ES6 Map object and I need to easily find same array collection.
Updated example of use
var theMap = new Map();
var lookup = function(arr) {
var item = null;
var hashed = hashStringArray(arr);
if (item = theMap.get( hashed )) {
return item;
}
theMap.set( hashed, itemBasedOnInput );
return itemBasedOnInput;
}
var arr1 = ['alpha','beta','gama'];
var arr2 = ['beta','alpha','gama'];
lookup(arr1) === lookup(arr2)
Performance tests
http://jsperf.com/hashing-array-of-strings/5

Two things occurred to me as the basis of a solution:
summing doesn't depend on order, which is actually a flaw in simple checksums (they don't catch changes in block order within a word), and
we can convert strings to summable numbers using their charcodes
Here's a function to do (2) :
charsum = function(s) {
var i, sum = 0;
for (i = 0; i < s.length; i++) {
sum += (s.charCodeAt(i) * (i+1));
}
return sum
}
Here's a version of (1) that computes an array hash by summing the charsum values:
array_hash = function(a) {
var i, sum = 0
for (i = 0; i < a.length; i++) {
var cs = charsum(a[i])
sum = sum + (65027 / cs)
}
return ("" + sum).slice(0,16)
}
Fiddle here: http://jsfiddle.net/WS9dC/11/
If we did a straight sum of the charsum values, then the array ["a", "d"] would have the same hash as the array ["b", "c"] - leading to undesired collisions. So based on using non-UTF strings, where charcodes go up to 255, and allowing for 255 characters in each string, then the max return value of charsum is 255 * 255 = 65025. So I picked the next prime number up, 65027, and used (65027 / cs) to compute the hash. I am not 100% convinced this removes collisions... perhaps more thought needed... but it certainly fixes the [a, d] versus [b, c] case.
Testing:
var arr1 = ['alpha','beta','gama'];
var arr2 = ['beta','alpha','gama'];
console.log(array_hash(arr1))
console.log(array_hash(arr2))
console.log(array_hash(arr1) == array_hash(arr2))
Outputs:
443.5322979371356
443.5322979371356
true
And testing a case that shows different hashes:
var arr3 = ['a', 'd'];
var arr4 = ['b', 'c'];
console.log(array_hash(arr3))
console.log(array_hash(arr4))
console.log(array_hash(arr3) == array_hash(arr4))
outputs:
1320.651443298969
1320.3792001649144
false
Edit:
Here's a revised version, which ignore duplicates from the arrays as it goes, and return the hash based on unique items only:
http://jsfiddle.net/WS9dC/7/
array_hash = function(a) {
var i, sum = 0, product = 1
for (i = 0; i < a.length; i++) {
var cs = charsum(a[i])
if (product % cs > 0) {
product = product * cs
sum = sum + (65027 / cs)
}
}
return ("" + sum).slice(0, 16)
}
testing:
var arr1 = ['alpha', 'beta', 'gama', 'delta', 'theta', 'alpha', 'gama'];
var arr2 = ["beta", "gama", "alpha", "theta", "delta", "beta"];
console.log(array_hash(arr1))
console.log(array_hash(arr2))
console.log(array_hash(arr1) === array_hash(arr2))
returns:
689.878503111701
689.878503111701
true
Edit
I've revised the answer above to account for arrays of words that have the same letters. We need these to return different hashes, which they now do:
var arr1 = ['alpha', 'beta']
var arr2 = ['alhpa', 'ateb']
The fix was to add a multiplier to the charsum func based on the char index:
sum += (s.charCodeAt(i) * (i+1));

If you calculate a numeric hash code for each string, then you can combine them with an operator where the order doesn't matter, like the ^ XOR operator, then you don't need to sort the array:
function hashStringArray(array) {
var code = 0;
for (var i = 0; i < array.length; i++) {
var n = 0;
for (var j = 0; j < array[i].length; j++) {
n = n * 251 ^ array[i].charCodeAt(j);
}
code ^= n;
}
return code
};

You can do this:
var hashStringArray = function(array) {
return array.sort().join('\u200b');
};
The \u200b character is an unicode character that also means null, but is not the same as the \0 character, which is most widely used.
'\u200b' == '\0'
> false

An idea to have very fast hash if your set of possible string is less than 32 items long : hash the string with a built-in hash function that will return power-of two as hash :
function getStringHash(aString) {
var currentPO2 = 0;
var hashSet = [];
getStringHash = function ( aString) {
var aHash = hashSet[aString];
if (aHash) return aHash;
aHash = 1 << currentPO2++;
hashSet[aString] = aHash;
return aHash;
}
return getStringHash(aString);
}
Then use this hash on your string array, ORing the hashes ( | ) :
function getStringArrayHash( aStringArray) {
var aHash = 0;
for (var i=0; i<aStringArray.length; i++) {
aHash |= getStringHash(aStringArray[i]);
}
return aHash;
}
So to test a bit :
console.log(getStringHash('alpha')); // 1
console.log(getStringHash('beta')); // 2
console.log(getStringHash('gamma')); // 4
console.log(getStringHash('alpha')); // 1 again
var arr1 = ['alpha','beta','gama'];
var arr2 = ['beta','alpha','gama'];
var arr3 = ['alpha', 'teta'];
console.log(getStringArrayHash(arr1)); // 11
console.log(getStringArrayHash(arr2)); // 11 also, like for arr1
var arr3 = ['alpha', 'teta'];
console.log(getStringArrayHash(arr3)); // 17 : a different array has != hashset
jsbin is here : http://jsbin.com/rozanufa/1/edit?js,console
RQ !!! with this method, arrays are considered as set, meaning that a repeated item won't change the hash of an array !!!
This HAS to be faster since it uses only 1) function call 2) lookup 3) integer arithmetic.
So no sort, no (long) string, no concat.
jsperf confirms that :
http://jsperf.com/hashing-array-of-strings/4
EDIT :
version with prime numbers, here : http://jsbin.com/rozanufa/3/edit?js,console
// return the unique prime associated with the string.
function getPrimeStringHash(aString) {
var hashSet = [];
var currentPrimeIndex = 0;
var primes = [ 2, 3, 5, 7, 11, 13, 17 ];
getPrimeStringHash = function ( aString) {
var aPrime = hashSet[aString];
if (aPrime) return aPrime;
if (currentPrimeIndex == primes.length) aPrime = getNextPrime();
else aPrime = primes[currentPrimeIndex];
currentPrimeIndex++
hashSet[aString] = aPrime;
return aPrime;
};
return getPrimeStringHash(aString);
// compute next prime number, store it and returns it.
function getNextPrime() {
var pr = primes[primes.length-1];
do {
pr+=2;
var divides = false;
// discard the number if it divides by one earlier prime.
for (var i=0; i<primes.length; i++) {
if ( ( pr % primes[i] ) == 0 ) {
divides = true;
break;
}
}
} while (divides == true)
primes.push(pr);
return pr;
}
}
function getStringPrimeArrayHash( aStringArray) {
var primeMul = 1;
for (var i=0; i<aStringArray.length; i++) {
primeMul *= getPrimeStringHash(aStringArray[i]);
}
return primeMul;
}
function compareByPrimeHash( aStringArray, anotherStringArray) {
var mul1 = getStringPrimeArrayHash ( aStringArray ) ;
var mul2 = getStringPrimeArrayHash ( anotherStringArray ) ;
return ( mul1 > mul2 ) ?
! ( mul1 % mul2 )
: ! ( mul2 % mul1 );
// Rq : just test for mul1 == mul2 if you are sure there's no duplicates
}
Tests :
console.log(getPrimeStringHash('alpha')); // 2
console.log(getPrimeStringHash('beta')); // 3
console.log(getPrimeStringHash('gamma')); // 5
console.log(getPrimeStringHash('alpha')); // 2 again
console.log(getPrimeStringHash('a1')); // 7
console.log(getPrimeStringHash('a2')); // 11
var arr1 = ['alpha','beta','gamma'];
var arr2 = ['beta','alpha','gamma'];
var arr3 = ['alpha', 'teta'];
var arr4 = ['alpha','beta','gamma', 'alpha']; // == arr1 + duplicate 'alpha'
console.log(getStringPrimeArrayHash(arr1)); // 30
console.log(getStringPrimeArrayHash(arr2)); // 30 also, like for arr1
var arr3 = ['alpha', 'teta'];
console.log(getStringPrimeArrayHash(arr3)); // 26 : a different array has != hashset
console.log(compareByPrimeHash(arr1, arr2) ); // true
console.log(compareByPrimeHash(arr1, arr3) ); // false
console.log(compareByPrimeHash(arr1, arr4) ); // true despite duplicate