How to implement .map() using for()? - javascript

I have these array and variable:
var arr = [['one','blue'], ['two','red'], ['three','green']]
var variable = 'thre';
Also I have this code:
arr.map(function(x){
if(x[0].indexOf(variable) >= 0)
{
alert('Number is found');
}
});
As you know, map works as a loop, and in the array above, there is three items and then map executes its statement 3 times. So that alert will be run.
Now I'm trying to limit the mapping, I mean I want to execute a statement 2 times. So I user for() like this:
for ( var c = 0; c < 2; c++ ) {
if ( arr[c][0].indexOf(variable) >= 0 )
{
alert('number is found');
}
}
But ^ doesn't work, It gives me this error:
Uncaught TypeError: Cannot read property '0' of undefined {in line 2}
How can I fix it?
EDIT: Here is my code in reality:
ZippedArray.map(function(x){
if(x[0].indexOf(name) >= 0)
{
MatchesNames.push(x[0]);
MatchesIds.push(x[1]);
}
});
I want this output:
MatchesNames = MatchesNames.slice(0,2);
MatchesIds = MatchesIds.slice(0,2);
How to limit .map() ? I want something like break; after 2 times.


Based on your comments, it seems you want to loop until you've found two matches in the if condition.
In that case, you can use .some(), which will halt the loop as soon as you return true (or any truthy value).
ZippedArray.some(function(x){
if(x[0].indexOf(name) >= 0)
{
MatchesNames.push(x[0]);
MatchesIds.push(x[1]);
}
return MatchesNames.length == 2; // Breaks when this returns `true`
});
This example assumes that MatchesNames was empty before you called .some().
If there could be other items in the array, and you just want to push two more in at the most, then you could keep a count.
var found = 0;
ZippedArray.some(function(x){
if(x[0].indexOf(name) >= 0)
{
MatchesNames.push(x[0]);
MatchesIds.push(x[1]);
found++;
}
return found == 2;
});
If you want to use a traditional for loop, then do this:
var found = 0;
for (var i = 0; i < ZippedArray.length; i++) {
var x = ZippedArray[i];
if(x[0].indexOf(name) >= 0)
{
MatchesNames.push(x[0]);
MatchesIds.push(x[1]);
found++;
}
if (found == 2) {
break;
}
}


The code you posted does not throw an error.
But when you're limiting the loop count of iterating an array, you should add a range check for the index:
for (var c = 0; c < 2 && c < arr.length; c++) {
// or alternatively
for (var c = 0, l = Math.min(2, arr.length); c < l; c++) {

Related

How to limit a number between several numbers (get the most nearest small number)? [duplicate]

Example: I have an array like this: [0,22,56,74,89] and I want to find the closest number downward to a different number. Let's say that the number is 72, and in this case, the closest number down in the array is 56, so we return that. If the number is 100, then it's bigger than the biggest number in the array, so we return the biggest number. If the number is 22, then it's an exact match, just return that. The given number can never go under 0, and the array is always sorted.
I did see this question but it returns the closest number to whichever is closer either upward or downward. I must have the closest one downward returned, no matter what.
How do I start? What logic should I use?
Preferably without too much looping, since my code is run every second, and it's CPU intensive enough already.

You can use a binary search for that value. Adapted from this answer:
function index(arr, compare) { // binary search, with custom compare function
var l = 0,
r = arr.length - 1;
while (l <= r) {
var m = l + ((r - l) >> 1);
var comp = compare(arr[m]);
if (comp < 0) // arr[m] comes before the element
l = m + 1;
else if (comp > 0) // arr[m] comes after the element
r = m - 1;
else // arr[m] equals the element
return m;
}
return l-1; // return the index of the next left item
// usually you would just return -1 in case nothing is found
}
var arr = [0,22,56,74,89];
var i=index(arr, function(x){return x-72;}); // compare against 72
console.log(arr[i]);
Btw: Here is a quick performance test (adapting the one from #Simon) which clearly shows the advantages of binary search.

var theArray = [0,22,56,74,89];
var goal = 56;
var closest = null;
$.each(theArray, function(){
if (this <= goal && (closest == null || (goal - this) < (goal - closest))) {
closest = this;
}
});
alert(closest);
jsFiddle http://jsfiddle.net/UCUJY/1/

Array.prototype.getClosestDown = function(find) {
function getMedian(low, high) {
return (low + ((high - low) >> 1));
}
var low = 0, high = this.length - 1, i;
while (low <= high) {
i = getMedian(low,high);
if (this[i] == find) {
return this[i];
}
if (this[i] > find) {
high = i - 1;
}
else {
low = i + 1;
}
}
return this[Math.max(0, low-1)];
}
alert([0,22,56,74,89].getClosestDown(75));

Here's a solution without jQuery for more effiency. Works if the array is always sorted, which can easily be covered anyway:
var test = 72,
arr = [0,56,22,89,74].sort(); // just sort it generally if not sure about input, not really time consuming
function getClosestDown(test, arr) {
var num = result = 0;
for(var i = 0; i < arr.length; i++) {
num = arr[i];
if(num <= test) { result = num; }
}
return result;
}
Logic: Start from the smallest number and just set result as long as the current number is smaller than or equal the testing unit.
Note: Just made a little performance test out of curiosity :). Trimmed my code down to the essential part without declaring a function.

Here's an ES6 version using reduce, which OP references. Inspired by this answer get closest number out of array
lookup array is always sorted so this works.
const nearestBelow = (input, lookup) => lookup.reduce((prev, curr) => input >= curr ? curr : prev);
const counts = [0,22,56,74,89];
const goal = 72;
nearestBelow(goal, counts); // result is 56.
Not as fast as binary search (by a long way) but better than both loop and jQuery grep https://jsperf.com/test-a-closest-number-function/7

As we know the array is sorted, I'd push everything that asserts as less than our given value into a temporary array then return a pop of that.
var getClosest = function (num, array) {
var temp = [],
count = 0,
length = a.length;
for (count; count < length; count += 1) {
if (a[count] <= num) {
temp.push(a[count]);
} else {
break;
}
}
return temp.pop();
}
getClosest(23, [0,22,56,74,89]);

Here is edited from #Simon.
it compare closest number before and after it.
var test = 24,
arr = [76,56,22,89,74].sort(); // just sort it generally if not sure about input, not really time consuming
function getClosest(test, arr) {
var num = result = 0;
var flag = 0;
for(var i = 0; i < arr.length; i++) {
num = arr[i];
if(num < test) {
result = num;
flag = 1;
}else if (num == test) {
result = num;
break;
}else if (flag == 1) {
if ((num - test) < (Math.abs(arr[i-1] - test))){
result = num;
}
break;
}else{
break;
}
}
return result;
}

Happy numbers - recursion

I have an issue with a recursive algorithm, that solves the problem of finding the happy numbers.
Here is the code:
function TestingFunction(number){
sumNumberContainer = new Array(0);
CheckIfNumberIsHappy(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
console.log(sumOfTheNumbers);
if(sumOfTheNumbers == 1){
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
//return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
//return false;
}
}
}
}
CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
Algorithm is working ALMOST fine. I've tested it out by calling function with different numbers, and console was displaying correct results. The problem is that I almost can't get any value from the function. There are only few cases in which I can get any value: If the number is build out of ,,0", and ,,1", for example 1000.
Because of that, I figured out, that I have problem with returning any value when the function is calling itself again.
Now I ended up with 2 results:
Returning the
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
which is giving an infinity looped number. For example when the number is happy, the function is printing in the console number one again and again...
Returning the
//return true
or
//return false
which gives me an undefined value
I'm a little bit in check by this problem, and I'm begging you guys for help.

I would take a step back and reexamine your problem with recursion in mind. The first thing you should think about with recursion is your edge cases — when can you just return a value without recursing. For happy numbers, that's the easy case where the sum of squares === 1 and the harder case where there's a cycle. So test for those and return appropriately. Only after that do you need to recurse. It can then be pretty simple:
function sumSq(num) {
/* simple helper for sums of squares */
return num.toString().split('').reduce((a, c) => c * c + a, 0)
}
function isHappy(n, seen = []) {
/* seen array keeps track of previous values so we can detect cycle */
let ss = sumSq(n)
// two edge cases -- just return
if (ss === 1) return true
if (seen.includes(ss)) return false
// not an edge case, save the value to seen, and recurse.
seen.push(ss)
return isHappy(ss, seen)
}
console.log(isHappy(23))
console.log(isHappy(22))
console.log(isHappy(7839))

Here's a simplified approach to the problem
const digits = x =>
x < 10
? [ x ]
: [ ...digits (x / 10 >> 0), x % 10 ]
const sumSquares = xs =>
xs.reduce ((acc, x) => acc + x * x, 0)
const isHappy = (x, seen = new Set) =>
x === 1
? true
: seen.has (x)
? false
: isHappy ( sumSquares (digits (x))
, seen.add (x)
)
for (let n = 1; n < 100; n = n + 1)
if (isHappy (n))
console.log ("happy", n)
// happy 1
// happy 7
// happy 10
// ...
// happy 97
The program above could be improved by using a technique called memoization

Your code is almost correct. You just forgot to return the result of the recursive call:
function TestingFunction(number){
sumNumberContainer = new Array(0);
if (CheckIfNumberIsHappy(number))
console.log(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
if(sumOfTheNumbers == 1){
return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return false;
}
}
}
}
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
for (let i=0; i<100; ++i)
TestingFunction(i.toString()); // 1 7 10 13 ... 91 94 97

I've got the solution, which was given to me in the comments, by the user: Mark_M.
I just had to use my previous
return true / return false
also I had to return the recursive statement in the function, and return the value of the CheckIfTheNumberIsHappy function, which was called in TestingFunction.
The working code:
function TestingFunction(number){
sumNumberContainer = new Array(0);
return CheckIfNumberIsHappy(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
console.log(sumOfTheNumbers);
if(sumOfTheNumbers == 1){
return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return false;
}
}
}
}
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
Thanks for the great support :)

checking if a substring of length 5 is in another string (refactored)

I'm trying to check if string b contains any 5-character substring a.
This works, but is a little messy:
var a = "1eabcde";
var b = "12abcde12fg";
for(var i=0; i<a.length; i++){
for(var j=i;j<a.length-i;j++){
if(a.charAt(j) == b.charAt(i) && a.charAt(j+1) == b.charAt(i+1) && a.charAt(j+2) == b.charAt(i+2) && a.charAt(j+3) == b.charAt(i+3) && a.charAt(j+4) == b.charAt(i+4)){
alert("ya");
}
}
}
Are there any other cleaner options?

You can use substring and indexOf:
var a = "1eabcde";
var b = "12abcde12fg";
for (var i = 0; i <= a.length - 5; i++) {
if (b.indexOf(a.substring(i, i + 5)) >= 0) {
alert("ya");
}
}
(You could use a.substr(i, 5) instead of a.substring(i, i + 5). Those two calls behave identically.)
Note that if you loop from 0 to a.length (as in your original code), then all suffixes of a of length 5 or less will be searched for in b.
In one respect, this code does not behave the same as your original: it will alert only once for each matching substring of a, regardless of how many times that particular substring may occur in b. Thus, if a = 'abcde' and b = '01abcde23abcde45, your original code would pop up two alerts (one for each occurrence of 'abcde'), whereas the above will only alert once. If you want the original behavior, change the if to a while like this:
for (var i = 0; i <= a.length - 5; i++) {
var j = -1;
while ((j = b.substring(j+1).indexOf(a.substr(i, 5))) >= 0) {
alert("ya");
}
}

This is the cleanest approach :
var a = "1eabcde";
var b = "12abcde12fg";
for (var i = 0; i <= a.length - 5; i++) {
if(b.indexOf(a.substr(i, 5)) > -1) {
alert("ya");
}
}

try to not add the elements already in the array

function neighbor(color_indices) {
var neighbor_face = [];
var neighbor_index_temp = [];
//initialize to the given length
var color_indices_length = color_indices.length;
for (var i = 0; i < color_indices_length; i++) {
if (color_indices[i] % 2 == 0 ) {
if (color_indices[i] % 10 == 8) {
neighbor_index_temp[0] = (color_indices[i]) + 1;
neighbor_index_temp[1] = (color_indices[i]) - 17;
neighbor_index_temp[2] = (color_indices[i]) - 19;
//check if it is in the array
for (var k = 0; k < 3; k++){
if ($.inArray(neighbor_index_temp[k],color_indices) != -1){
color_indices.push(neighbor_index_temp[k]);
}
}
The input : color_indices would be an array of the global variable. I am trying to push neighbor_index_temp of only new to the color_indices. I tried to implement $.inArray but it doesn't seem to work. Any suggestions?
Thanks!

You can use indexOf to achieve this. Ie.
if(color_indices.indexOf(neighbor_index_temp[k]) == -1) {
// value neighbor_index_temp[k] is not in color_indices array
}
Edit : $.inArray might do the same as indexOf, in this case you might want to change '!= -1' to '== -1'

Javascript if statement less than 'x' object in array don't show data

Here I am showing related documents. I only want this to show if there is more than one related document in the array.
displayTable = function(data) {
var data = [];
for (var i = 0; i < data.results.data.length; i++) {
if (data.results.data[i].length < 0) {
console.log('dont show');
} else if (data.results.data[i]) {
data.push(data.results.data[i]);
}
}

pubData.results.data[i].length < 0 means less than zero, not less than one.
Use < 1 for less than one:
if (pubData.results.data[i].length < 1)
Or optionally === 0, if it's a true array (the length of a true JavaScript array is never negative):
if (pubData.results.data[i].length === 0)
Or (again for a true array) you can use !:
if (!pubData.results.data[i].length)

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