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Issue with Implementing a hit counter that records hits for last 5 min

I've encountered this problem and still trying to solve this problem:
You want to log the number of hits to a site.
Implement two functions,
log_hit() which gets called when a hit is registered, and
get_hits_in_last_five_minutes() which returns the total number of hits
in the last five minutes.
Assume that all timestamps come in increasing order.
My idea (attempt to solve the problem):
Datastructure: Array.
My Logic:
When log_hit() is called, we basically store the time in ms (Date().getTime() in ms) in the array and store the hits on the particular second in hashmap.
function Counter() {
this.map = {};
this.store = [];
}
Counter.prototype.log_hit = function() {
const d = new Date().getTime()/1000;
this.store.push(d);
this.map[d] = this.map[d] ? this.map[d]++ : 1;
}
Counter.prototype.get_hits_in_last_five_minutes = function() {
const fiveMin = 60 * 60; //seconds.
const initalPointer = new Date().getTime() / 1000 - fiveMin;
// Somehow retrieve the value and return it back
}
However, I don't think this is the most optimal way of solving it , if I wanted to extend the solution for hour or some other granularity.
How would I solve this kind of problem?

Using queue would be the right way to deal with this problem.
var HitCounter = function() {
this.queue = [];
};
/**
* Record a hit.
#param timestamp - The current timestamp (in seconds granularity).
* #param {number} timestamp
* #return {void}
*/
HitCounter.prototype.hit = function(timestamp) {
this.queue.push(timestamp);
};
/**
* Return the number of hits in the past 5 minutes.
#param timestamp - The current timestamp (in seconds granularity).
* #param {number} timestamp
* #return {number}
*/
HitCounter.prototype.getHits = function(timestamp) {
while(this.queue.length && this.queue[0] <= timestamp-300) {
this.queue.shift();
}
return this.queue.length;
};
const counter = new HitCounter();
counter.hit(1);
counter.hit(2);
counter.hit(3);
counter.getHits(4);
counter.hit(300);
counter.getHits(300);
console.log(counter.getHits(301)); // should output 3.

We need to exploit this fact-
Assume that all timestamps come in increasing order.
Algorithm:
We record every hit in an array and increase it's size gradually by 1.
To get the no. of hits in the last 5 minutes(excluding current hit), we do a binary search to get our answer since all timestamps come in increasing order, which would be sorted by default.
First, we do a binary search on the array to get the last valid upper bound index with respect to time t provided to get_hits_in_last_five_minutes().
Upper bound used is the limit till which we could use the hits to judge the no. of calls in the last 5 minutes. This is necessary because the problem statement says get_hits_in_last_five_minutes() which returns the total number of hits in the last five minutes. So, technically it means that it will be used as an API to check how many calls were made till 5 minutes prior to the parameter passed time t. It doesn't guarantee that the time t passed to this method will always be the last inserted timestamp in the counter. Due to this, we need to search for the upper bound in the array, i.e, till which index the hits stored in the array could be counted as valid for our answer.
Second, we do a binary search from 0 till upper_bound to get all the valid hits that were under last 5 minutes prior to t.
Space complexity: O(n) where n is the no. of hits registered.
Time Complexity:
O(log(n)) to search for the upper bound.
O(log(n)) to get the actual hits registered in last 5 minutes.
Total Complexity = O(log(n)) + O(log(n)) = 2 * O(log(n)) = O(log(n))
Note: I converted time t in to seconds while storing and searching.
Code:
function Counter() {
this.hits = [];
this.hits_size = 0;
}
Counter.prototype.log_hit = function(t) {
this.hits[this.hits_size++] = t * 60;
}
Counter.prototype.get_hits_in_last_five_minutes = function(t) {
if (this.hits_size < 2) return 0;
t *= 60;
var upper_bound = this.getUpperBound(t);
this.last_call_type = 2;
var low = 0,
high = upper_bound;
while (low <= high) {
var mid = low + parseInt((high - low) / 2);
if (this.hits[mid] > t - 300) high = mid - 1;
else if (this.hits[mid] < t - 300) low = mid + 1;
else return upper_bound - mid + 1;
}
return upper_bound - low + 1;
}
Counter.prototype.getUpperBound = function(t) {
var low = 0,
high = t > this.hits[this.hits_size - 1] ? this.hits_size - 1 : this.hits_size - 2;
var ans = 0;
while (low <= high) {
var mid = low + parseInt((high - low) / 2);
if (this.hits[mid] >= t) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
if (high < 0) return -1;
return ans;
}
console.log("*****Counter 1******");
var c1 = new Counter();
c1.log_hit(1);
console.log("Registered, 1 = " + c1.get_hits_in_last_five_minutes(1));
c1.log_hit(2);
console.log("Registered, 2 = " + c1.get_hits_in_last_five_minutes(2));
c1.log_hit(3);
console.log("Registered, 3 = " + c1.get_hits_in_last_five_minutes(3));
c1.log_hit(4);
console.log("Registered, 4 = " + c1.get_hits_in_last_five_minutes(4));
c1.log_hit(5);
console.log("Registered, 5 = " + c1.get_hits_in_last_five_minutes(5));
c1.log_hit(6);
console.log("Registered, 6 = " + c1.get_hits_in_last_five_minutes(6));
c1.log_hit(7);
console.log("Registered, 7 = " + c1.get_hits_in_last_five_minutes(7));
c1.log_hit(8);
console.log("Registered, 8 = " + c1.get_hits_in_last_five_minutes(8));
c1.log_hit(9);
console.log("Registered, 9 = " + c1.get_hits_in_last_five_minutes(9));
c1.log_hit(10);
console.log("Registered, 10 = " + c1.get_hits_in_last_five_minutes(10));
console.log("*****Counter 2******");
var c2 = new Counter();
c2.log_hit(2);
console.log("Registered, 2 = " + c2.get_hits_in_last_five_minutes(2));
c2.log_hit(7);
console.log("Registered, 7 = " + c2.get_hits_in_last_five_minutes(7));
c2.log_hit(8);
console.log("Registered, 8 = " + c2.get_hits_in_last_five_minutes(8));
c2.log_hit(9);
console.log("Registered, 9 = " + c2.get_hits_in_last_five_minutes(9));
c2.log_hit(10);
console.log("Registered, 10 = " + c2.get_hits_in_last_five_minutes(10));
c2.log_hit(11);
console.log("Registered, 11 = " + c2.get_hits_in_last_five_minutes(11));
c2.log_hit(12);
console.log("Registered, 12 = " + c2.get_hits_in_last_five_minutes(12));
c2.log_hit(17);
console.log("Registered, 17 = " + c2.get_hits_in_last_five_minutes(17));
console.log("Unregistered, 18 = " + c2.get_hits_in_last_five_minutes(18));
c2.log_hit(19);
console.log("Registered, 19 = " + c2.get_hits_in_last_five_minutes(19));
console.log("Unregistered, 20 = " + c2.get_hits_in_last_five_minutes(20));
c2.log_hit(21);
console.log("Registered, 21 = " + c2.get_hits_in_last_five_minutes(21));
console.log("Unregistered, 6 = " + c2.get_hits_in_last_five_minutes(6));
console.log("Unregistered, 500 = " + c2.get_hits_in_last_five_minutes(500));
console.log("Unregistered, 15 = " + c2.get_hits_in_last_five_minutes(15));
console.log("Registered, 17 = " + c2.get_hits_in_last_five_minutes(17));
Why binary search?
You might wonder as to why not loop backwards and get all those values' count that are under t - 300. This way, it could be O(1) per call.
Note that our search space is just under t - 300. If this increases, for example, t - 9000 then our backward iteration also increases till 9000, and if the number of calls for get_hits_in_last_five_minutes() happens to be 10000 or more, then the complexity of just looping multiplies to the overall complexity. So, it could be
10000 calls to get_hits_in_last_five_minutes() * 9000
If we use the algorithm described above, it will be
10000 calls to get_hits_in_last_five_minutes() * log(n)
What if the calls are never ending(Infinite)?
This depends upon how we choose to use the get_hits_in_last_five_minutes() method.
If time t passed to the calls made to get_hits_in_last_five_minutes() will always be in non-decreasing manner, then we could truncate/remove hits from our storage.
To do that, we can again do a binary search to get the index of the maximum value from our storage that doesn't come under t - 300. After that, we could just do an array slice and reset this.hits_size to the new length.
We need to do this in log_hit() method and not get_hits_in_last_five_minutes() since time t passed to it need not necessarily be a part of our registered hits.
Using array slice could add up to the complexity a bit since it returns a shallow copy of the original array and it's complexity is O(N) where N is end - start. See array slice complexity. To avoid this, we can make a linked list and store data in it and use a map to store the nodes. This way, we could reset the list's head and make the trimming/truncating O(1). We will need to maintain a count of truncated nodes too as we can't reset the map with new values.
If the time t passed to the calls made to get_hits_in_last_five_minutes() for your website are in any random order, then we can't truncate anything since we need all data to give an answer. In that case, probably store data in the database. The good part is, you could just query your answer from DB and instead of doing computation in javascript.

Random integer in a certain range excluding one number

I would like get a random number in a range excluding one number (e.g. from 1 to 1000 exclude 577). I searched for a solution, but never solved my issue.
I want something like:
Math.floor((Math.random() * 1000) + 1).exclude(577);
I would like to avoid for loops creating an array as much as possible, because the length is always different (sometimes 1 to 10000, sometimes 685 to 888555444, etc), and the process of generating it could take too much time.
I already tried:
Javascript - Generating Random numbers in a Range, excluding certain numbers
How can I generate a random number within a range but exclude some?
How could I achieve this?

The fastest way to obtain a random integer number in a certain range [a, b], excluding one value c, is to generate it between a and b-1, and then increment it by one if it's higher than or equal to c.
Here's a working function:
function randomExcluded(min, max, excluded) {
var n = Math.floor(Math.random() * (max-min) + min);
if (n >= excluded) n++;
return n;
}
This solution only has a complexity of O(1).

One possibility is not to add 1, and if that number comes out, you assign the last possible value.
For example:
var result = Math.floor((Math.random() * 100000));
if(result==577) result = 100000;
In this way, you will not need to re-launch the random method, but is repeated. And meets the objective of being a random.

As #ebyrob suggested, you can create a function that makes a mapping from a smaller set to the larger set with excluded values by adding 1 for each value that it is larger than or equal to:
// min - integer
// max - integer
// exclusions - array of integers
// - must contain unique integers between min & max
function RandomNumber(min, max, exclusions) {
// As #Fabian pointed out, sorting is necessary
// We use concat to avoid mutating the original array
// See: http://stackoverflow.com/questions/9592740/how-can-you-sort-an-array-without-mutating-the-original-array
var exclusionsSorted = exclusions.concat().sort(function(a, b) {
return a - b
});
var logicalMax = max - exclusionsSorted.length;
var randomNumber = Math.floor(Math.random() * (logicalMax - min + 1)) + min;
for(var i = 0; i < exclusionsSorted.length; i++) {
if (randomNumber >= exclusionsSorted[i]) {
randomNumber++;
}
}
return randomNumber;
}
Example Fiddle
Also, I think #JesusCuesta's answer provides a simpler mapping and is better.
Update: My original answer had many issues with it.

To expand on #Jesus Cuesta's answer:
function RandomNumber(min, max, exclusions) {
var hash = new Object();
for(var i = 0; i < exclusions.length; ++i ) { // TODO: run only once as setup
hash[exclusions[i]] = i + max - exclusions.length;
}
var randomNumber = Math.floor((Math.random() * (max - min - exclusions.length)) + min);
if (hash.hasOwnProperty(randomNumber)) {
randomNumber = hash[randomNumber];
}
return randomNumber;
}
Note: This only works if max - exclusions.length > maximum exclusion. So close.

You could just continue generating the number until you find it suits your needs:
function randomExcluded(start, end, excluded) {
var n = excluded
while (n == excluded)
n = Math.floor((Math.random() * (end-start+1) + start));
return n;
}
myRandom = randomExcluded(1, 10000, 577);
By the way this is not the best solution at all, look at my other answer for a better one!

Generate a random number and if it matches the excluded number then add another random number(-20 to 20)
var max = 99999, min = 1, exclude = 577;
var num = Math.floor(Math.random() * (max - min)) + min ;
while(num == exclude || num > max || num < min ) {
var rand = Math.random() > .5 ? -20 : 20 ;
num += Math.floor((Math.random() * (rand));
}

import random
def rng_generator():
a = random.randint(0, 100)
if a == 577:
rng_generator()
else:
print(a)
#main()
rng_generator()

Exclude the number from calculations:
function toggleRand() {
// demonstration code only.
// this algorithm does NOT produce random numbers.
// return `0` - `576` , `578` - `n`
return [Math.floor((Math.random() * 576) + 1)
,Math.floor(Math.random() * (100000 - 578) + 1)
]
// select "random" index
[Math.random() > .5 ? 0 : 1];
}
console.log(toggleRand());
Alternatively, use String.prototype.replace() with RegExp /^(577)$/ to match number that should be excluded from result; replace with another random number in range [0-99] utilizing new Date().getTime(), isNaN() and String.prototype.slice()
console.log(
+String(Math.floor(Math.random()*(578 - 575) + 575))
.replace(/^(577)$/,String(isNaN("$1")&&new Date().getTime()).slice(-2))
);
Could also use String.prototype.match() to filter results:
console.log(
+String(Math.floor(Math.random()*10))
.replace(/^(5)$/,String(isNaN("$1")&&new Date().getTime()).match(/[^5]/g).slice(-1)[0])
);

Rounding to Significant Figures - Missing Zeros

I'm currently producing a JavaScript driven mathematics package, which focuses on rounding to various significant figures (S.F.) but I've run into a problem that I'm struggling to solve.
More on this problem later, but first some background for you.
The program is designed to select a completely random number within a given range and then automatically work out that number's relevant significant figures; for example:
Random Number: 0.097027 S.Fs: 9, 7, 0, 2, 7
Here is a screenshot of what I have produced to give you a visual representation:
As you can see, once the user has selected their number, they are then given the opportunity to click on four separate 'SF' buttons to view their random number presented to 1, 2, 3 and 4 S.Fs respectively.
For each S.F (1-4) the random number is rounded down, rounded up and rounded off to X SF and a scale below gives the user a more visual presentation to show why the SF value has been chosen by the program.
I've already written the vast majority of the code for this and tested it and so far the numbers are coming out how I'm expecting them to. Well nearly...
In the example I've given (0.097027); as you can see on the image I've included, the data for 4 S.F is absolutely correct and outputted accurately.
When I click on to the 3 SF button, I'd expect to see the following:
Random Number: 0.097027 3 S.F Rounded Up/Down/Off: 0.0970
However, what I'm actually getting is:
Random Number: 0.097027 3 S.F Rounded Up/Down/Off: 0.097
The program hasn't displayed the additional zero. This is a perfect example of a number in my program ending in a zero and in this case the zero is really significant and must be displayed.
The data is usually correct but there appears to be an issue with outputting significant zeros at the right time. I've researched the toFixed(x) method and if I assign toFixed(4) I get the correct required output, but because my numbers are generated randomly each time, they can range from a length of 5 figures, e.g. 89.404 up to > 10, e.g. `0.000020615.
So it looks like the toFixed method needs to be flexible/dynamic, e.g. toFixed(n) with a function run beforehand to determine exactly how many trailing zeros are needed?
Here are some key excerpts from my current solution for your consideration:
function generateNum() {
do {
genNumber = Math.random() * Math.pow (10, randomRange(-5, 5));
//Round
genNumber = roundToNSF(genNumber, 5, 0);
// This number must contain >1 digit which is 1 to 9 inclusive otherwise we may have e.g. 100. Rounding 100
}
while (!countNonZero(genNumber) || genNumber < 1E-05 || genNumber == 0);
//Round
genNumber = roundToNSF(genNumber, 5, 0);
genNumber = String(genNumber);
genNumber = Number(genNumber);
}
//----------------------------------------------------------------------------
function randomRange(min, max) {
/**
* Returns a random integer between min (inclusive) and max (inclusive)
* Using Math.round() will give you a non-uniform distribution!
*/
return Math.floor(Math.random() * (max - min + 1)) + min;
}
//---------------------------------------------------------------------------
//Click SF3 Button to reveal the data
function showSF3() {
//Remove any CSS properties on the buttons from previous use
removeButtonCSS();
document.getElementById('SFRounded').style.display = "block";
document.getElementById('scale').style.display = "block";
document.getElementById("SF3").className = document.getElementById("SF3").className + "buttonClick"; // this removes the blue border class
//Clear text
deleteRounded();
deleteScale();
//Run calculation
calculateAnswer();
//alert(genNumber.toFixed(4));
for (i = 3; i < 4; i++)
{
//Add The new data
sfRoundedTextBlock = document.getElementById('SFRounded');
//alert(downArray[i].toFixed(4));
//Data output to HTML.
sfRoundedTextBlock.innerHTML = sfRoundedTextBlock.innerHTML + '<p><strong>Number: </strong></br>' + String(genNumber) +
'</br>' + '<strong>Rounded down to ' + i + ' SF:</br></strong>' + downArray[i] + '</br>' +
'<strong>Rounded up to ' + i + ' SF:</br></strong>' + upArray[i] + '</br><strong>Rounded off to ' + i + ' SF:</br></strong>'
+ roundedArray[i] + '</br>' + '(See the scale below for why we choose <strong>' + roundedArray[i] + '</strong> as the rounded off value.)</p>';
}
}
//----------------------------------------------------------------------
var roundedArray = [];
var upArray = [];
var downArray = [];
var temp;
function calculateAnswer() {
//Clear Arrays
roundedArray = [];
upArray = [];
downArray = [];
// Work out the answer:
for (i = 0; i < 4; i++) {
var nSF = i + 1;
// Round OFF ...
temp = roundToNSF(genNumber, nSF, 0);
// We actually have to do this twice ...
roundedArray[nSF] = roundToNSF(temp, nSF, 0);
// Round UP ...
upArray[nSF] = roundToNSF(genNumber, nSF, 1);
// Round DOWN ...
downArray[nSF] = roundToNSF(genNumber, nSF, -1);
// e.g. x = 0.0098 rounded to 1SF is 0.010 initially (take the log of 0.0098 and try it!).
};
};
//-------------------------------------------------------------------------
//Globals
var aNumber;
var digits;
var way;
function roundToNSF(aNumber, digits, way){
// Round a number to n significant figures (can use roundToNDP provided we know how many decimal places):
if (way == undefined) { way = 0; }; // default is round off
if (aNumber !=0) {
if (aNumber > 0)
{
z = log10(aNumber);
}
else
{
z = log10(-aNumber);
};
z = Math.floor(z);
var nDP = digits - z - 1; // Rounding to nDP decimal places is equivalent to rounding to digits significant figures ...
var roundedNumber = roundToNDP(aNumber, nDP, way);
}
else {
roundedNumber = aNumber; // Number is zero ...
};
return Number(roundedNumber);
};
//---------------------------------------------------------------------------------
Update:
I'm still continuing to try and find a solution for this problem and an approach I have recently taken is to convert my randomly generated number into a searchable string variable and then use the indexOf(".") command to find the position of the decimal point (dp).
Then I've searched through my number, starting from the position of the dp to find the first instance of a significant, non-zero number [1-9].
var genNumber = 0.097027;
var rString = String(genNumber);
var positionofDP = rString.indexOf(".");
var regexp = /[1-9]/;
var positionofNonZero = Number(rString.search(regexp, positionofDP)); // Output would be '5'
I have then been able to target my search further, to determine whether my first significant number has any 'problematic' zeros in the immediate digits after it.
If there are any, then I set a Boolean variable to 'true' and then in a separate function create further text strings of my rounded off/down/up numbers, so I can then physically choose to add a '0' on to the end of the existing numerical characters.
This approach does work for me in isolated cases, but with my random number length ranging from 5-12 digits long, it still isn't dealing with all scenarios.
Maybe I need to create a dynamic toFixed(i) function? Any ideas would be greatly welcomed.

Instead of playing with the fixed points on an Int, you could manage the string directly.
Here's a link to a little fiddle: http://jsfiddle.net/5rw5G/4/
This not intended to completely/accurately solve your problem, but might help you see another solution.
function getRoundedSFs(num, SFCount) {
// Match every "leading zeros" before and after the .
var matches = num.toString().match(/^-?(0+)\.(0*)/);
// starting with "0."
if (matches) {
var firstIndex = matches[0].length;
var prefix = matches[0];
sf = Number(num.toString().substring(firstIndex, firstIndex + SFCount + 1));
sf = Math.round(sf / 10);
sf = prefix + sf.toString();
return Number(sf).toFixed(matches[2].length+SFCount);
}
// starting with something else like -5.574487436097115
else {
matches = num.toString().match(/^(-?(\d+))\.(\d+)/);
var decimalShift = SFCount - matches[2].length;
var rounded = Math.round(num * Math.pow(10, decimalShift));
rounded /= Math.pow(10, decimalShift);
return rounded.toFixed(decimalShift);
}
}

I've gone away again and I think I have now finally managed solve my initial problem.
There was a degree of confusion on my part surrounding when to use toFixed and toPrecision. I had previously attempted to convert my rounded up, down and off numbers into strings and then subsequently search through each of these to find the decimal point (".") and then work out the amount of trailing numbers, in order to then generate the correct toFixed point.
However, this was very hit and miss, given that my random number could be up to 12 digits, so what I've now done is to properly utilise toPrecision instead. For each 'SF button' (1-4) I have used the corresponding toPrecision point, e.g for SF1:
sfRoundedTextBlock.innerHTML = sfRoundedTextBlock.innerHTML + '<p><strong>Number: </strong></br>' + String(genNumber) +
'</br>' + '<strong>Rounded down to ' + i + ' SF:</br></strong>' + downArray[i].toPrecision(1) + '</br>' +
'<strong>Rounded up to ' + i + ' SF:</br></strong>' + upArray[i].toPrecision(1) + '</br><strong>Rounded off to ' + i + ' SF:</br></strong>'
+ roundedArray[i].toPrecision(1) + '</br>' + '(See the scale below for why we choose <strong>' + roundedArray[i].toPrecision(1) + '</strong> as the rounded off value.)</p>';
//Add The new scale data (Rounded Down)
downTextBlock = document.getElementById('down');
document.getElementById("down").innerHTML = String(downArray[i].toPrecision(1));
//Add The new scale data (Rounded Up)
upTextBlock = document.getElementById('up');
document.getElementById("up").innerHTML = String(upArray[i].toPrecision(1));
This was now giving me accurate results on every occasion, but there was still one hurdle left to jump. Occasionally I would reach a random scenario where scientific notation would have to be included in my outputted answer, e.g. 21819 rounded down to 1 SF, would read out at 2e+4 instead of 20000.
To combat this I setup my up, down and rounded figures into searchable strings, and then looked through these to find any illegal/scientific characters [a-z]. If I found any, I executed a slightly different version of my output which made use of parseFloat, which stripped out the scientific notation and displayed the correct figures:
//Convert Up, Down and Rounded into Strings based on their precision
var upString = String(upArray[i].toPrecision(1));
var downString = String(downArray[i].toPrecision(1));
var roundedString = String(roundedArray[i].toPrecision(1));
//Set up a regexp to search for characters [a-z], i.e. non-numeric
var regexp = /[a-z]/g;
//Search the up, down and rounded strings for non-numeric characters
var upResult = upString.match(regexp);
var downResult = downString.match(regexp);
var roundedResult = roundedString.match(regexp);
//If any of these strings contain a letter (non-numeric) we need to add in parseFloat to strip away the scientific notation included.
var containsChar = false;
if (upResult != null || downResult != null || roundedResult != null)
{
containsChar = true;
//alert("There is SN included here");
}
//Add The new data
sfRoundedTextBlock = document.getElementById('SFRounded');
if (containsChar == true)
{
sfRoundedTextBlock.innerHTML = sfRoundedTextBlock.innerHTML + '<p><strong>Number: </strong></br>' + String(genNumber) +
'</br>' + '<strong>Rounded down to ' + i + ' SF:</br></strong>' + parseFloat(downArray[i].toPrecision(1)) + '</br>' +
'<strong>Rounded up to ' + i + ' SF:</br></strong>' + parseFloat(upArray[i].toPrecision(1)) + '</br><strong>Rounded off to ' + i + ' SF:</br></strong>'
+ parseFloat(roundedArray[i].toPrecision(1)) + '</br>' + '(See the scale below for why we choose <strong>' + parseFloat(roundedArray[i].toPrecision(1)) + '</strong> as the rounded off value.)</p>';
//Add The new scale data (Rounded Down)
downTextBlock = document.getElementById('down');
document.getElementById("down").innerHTML = String(parseFloat(downArray[i].toPrecision(1)));
//Add The new scale data (Rounded Up)
upTextBlock = document.getElementById('up');
document.getElementById("up").innerHTML = String(parseFloat(upArray[i].toPrecision(1)));
}
Having tested this extensively it seems to be working as hoped.

Javascript adding zeros to the beginning of a string (max length 4 chars)

var number = 1310;
should be left alone.
var number = 120;
should be changed to "0120";
var number = 10;
should be changed to "0010";
var number = 7;
should be changed to "0007";

In all modern browsers you can use
numberStr.padStart(4, "0");
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
function zeroPad(num) {
return num.toString().padStart(4, "0");
}
var numbers = [1310, 120, 10, 7];
numbers.forEach(
function(num) {
var paddedNum = zeroPad(num);
console.log(paddedNum);
}
);

function pad_with_zeroes(number, length) {
var my_string = '' + number;
while (my_string.length < length) {
my_string = '0' + my_string;
}
return my_string;
}

try these:
('0000' + number).slice(-4);
or
(number+'').padStart(4,'0');

Here's another way. Comes from something I did that needs to be done thousands of times on a page load. It's pretty CPU efficient to hard code a string of zeroes one time, and chop as many as you need for the pad as many times as needed. I do really like the power of 10 method -- that's pretty flexible.
Anyway, this is as efficient as I could come up with:
For the original question, CHOOSE ONE of the cases...
var number = 1310;
var number = 120;
var number = 10;
var number = 7;
then
// only needs to happen once
var zeroString = "00000";
// one assignment gets the padded number
var paddedNum = zeroString.substring((number + "").length, 4) + bareNum;
//output
alert("The padded number string is: " + paddedNum);
Of course you still need to validate the input. Because this ONLY works reliably under the following conditions:
Number of zeroes in the zeroString is desired_length + 1
Number of digits in your starting number is less than or equal to your desired length
Backstory:
I have a case that needs a fixed length (14 digit) zero-padded number. I wanted to see how basic I could make this. It's run tens of thousands of times on a page load, so efficiency matters. It's not quite re-usable as-is, and it's a bit inelegant. Except that it is very very simple.
For desired n digits padded string, this method requires a string of (at least) n+1 zeroes. Index 0 is the first character in the string, which won't ever be used, so really, it could be anything.
Note also that string.substring() is different from string.substr()!
var bareNum = 42 + '';
var zeroString = "000000000000000";
var paddedNum = zeroString.substring(bareNumber.length, 14) + bareNum
This pulls zeroes from zeroString starting at the position matching the length of the string, and continues to get zeroes to the necessary length of 14. As long as that "14" in the third line is a lower integer than the number of characters in zeroString, it will work.

function pad(n, len) {
return (new Array(len + 1).join('0') + n).slice(-len);
}
might not work in old IE versions.

//to: 0 - to left, 1 - to right
String.prototype.pad = function(_char, len, to) {
if (!this || !_char || this.length >= len) {
return this;
}
to = to || 0;
var ret = this;
var max = (len - this.length)/_char.length + 1;
while (--max) {
ret = (to) ? ret + _char : _char + ret;
}
return ret;
};
Usage:
someString.pad(neededChars, neededLength)
Example:
'332'.pad('0', 6); //'000332'
'332'.pad('0', 6, 1); //'332000'

An approach I like is to add 10^N to the number, where N is the number of zeros you want. Treat the resultant number as a string and slice off the zeroth digit. Of course, you'll want to be careful if your input number might be larger than your pad length, but it's still much faster than the loop method:
// You want to pad four places:
>>> var N = Math.pow(10, 4)
>>> var number = 1310
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"1310"
>>> var number = 120
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"0120"
>>> var number = 10
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"0010"
…
etc. You can make this into a function easily enough:
/**
* Pad a number with leading zeros to "pad" places:
*
* #param number: The number to pad
* #param pad: The maximum number of leading zeros
*/
function padNumber(number, pad) {
var N = Math.pow(10, pad);
return number < N ? ("" + (N + number)).slice(1) : "" + number
}

I wrote a general function for this. It takes an input control and pad length as input.
function padLeft(input, padLength) {
var num = $("#" + input).val();
$("#" + input).val(('0'.repeat(padLength) + num).slice(-padLength));
}

With RegExp/JavaScript:
var number = 7;
number = ('0000'+number).match(/\d{4}$/);
console.log(number);
With Function/RegExp/JavaScript:
var number = 7;
function padFix(n) {
return ('0000'+n).match(/\d{4}$/);
}
console.log(padFix(number));

No loop, no functions
let n = "" + 100;
let x = ("0000000000" + n).substring(n.length);//add your amount of zeros
alert(x + "-" + x.length);

Nate as the best way I found, it's just way too long to read. So I provide you with 3 simples solutions.
1. So here's my simplification of Nate's answer.
//number = 42
"0000".substring(number.toString().length, 4) + number;
2. Here's a solution that make it more reusable by using a function that takes the number and the desired length in parameters.
function pad_with_zeroes(number, len) {
var zeroes = "0".repeat(len);
return zeroes.substring(number.toString().length, len) + number;
}
// Usage: pad_with_zeroes(42,4);
// Returns "0042"
3. Here's a third solution, extending the Number prototype.
Number.prototype.toStringMinLen = function(len) {
var zeroes = "0".repeat(len);
return zeroes.substring(self.toString().length, len) + self;
}
//Usage: tmp=42; tmp.toStringMinLen(4)

Use String.JS librairy function padLeft:
S('123').padLeft(5, '0').s --> 00123

How to format a float in javascript?

In JavaScript, when converting from a float to a string, how can I get just 2 digits after the decimal point? For example, 0.34 instead of 0.3445434.

There are functions to round numbers. For example:
var x = 5.0364342423;
print(x.toFixed(2));
will print 5.04.
EDIT:
Fiddle

var result = Math.round(original*100)/100;
The specifics, in case the code isn't self-explanatory.
edit: ...or just use toFixed, as proposed by Tim Büthe. Forgot that one, thanks (and an upvote) for reminder :)

Be careful when using toFixed():
First, rounding the number is done using the binary representation of the number, which might lead to unexpected behaviour. For example
(0.595).toFixed(2) === '0.59'
instead of '0.6'.
Second, there's an IE bug with toFixed(). In IE (at least up to version 7, didn't check IE8), the following holds true:
(0.9).toFixed(0) === '0'
It might be a good idea to follow kkyy's suggestion or to use a custom toFixed() function, eg
function toFixed(value, precision) {
var power = Math.pow(10, precision || 0);
return String(Math.round(value * power) / power);
}

One more problem to be aware of, is that toFixed() can produce unnecessary zeros at the end of the number.
For example:
var x=(23-7.37)
x
15.629999999999999
x.toFixed(6)
"15.630000"
The idea is to clean up the output using a RegExp:
function humanize(x){
return x.toFixed(6).replace(/\.?0*$/,'');
}
The RegExp matches the trailing zeros (and optionally the decimal point) to make sure it looks good for integers as well.
humanize(23-7.37)
"15.63"
humanize(1200)
"1200"
humanize(1200.03)
"1200.03"
humanize(3/4)
"0.75"
humanize(4/3)
"1.333333"

var x = 0.3445434
x = Math.round (x*100) / 100 // this will make nice rounding

The key here I guess is to round up correctly first, then you can convert it to String.
function roundOf(n, p) {
const n1 = n * Math.pow(10, p + 1);
const n2 = Math.floor(n1 / 10);
if (n1 >= (n2 * 10 + 5)) {
return (n2 + 1) / Math.pow(10, p);
}
return n2 / Math.pow(10, p);
}
// All edge cases listed in this thread
roundOf(95.345, 2); // 95.35
roundOf(95.344, 2); // 95.34
roundOf(5.0364342423, 2); // 5.04
roundOf(0.595, 2); // 0.60
roundOf(0.335, 2); // 0.34
roundOf(0.345, 2); // 0.35
roundOf(551.175, 2); // 551.18
roundOf(0.3445434, 2); // 0.34
Now you can safely format this value with toFixed(p).
So with your specific case:
roundOf(0.3445434, 2).toFixed(2); // 0.34

There is a problem with all those solutions floating around using multipliers. Both kkyy and Christoph's solutions are wrong unfortunately.
Please test your code for number 551.175 with 2 decimal places - it will round to 551.17 while it should be 551.18 ! But if you test for ex. 451.175 it will be ok - 451.18. So it's difficult to spot this error at a first glance.
The problem is with multiplying: try 551.175 * 100 = 55117.49999999999 (ups!)
So my idea is to treat it with toFixed() before using Math.round();
function roundFix(number, precision)
{
var multi = Math.pow(10, precision);
return Math.round( (number * multi).toFixed(precision + 1) ) / multi;
}

If you want the string without round you can use this RegEx (maybe is not the most efficient way... but is really easy)
(2.34567778).toString().match(/\d+\.\d{2}/)[0]
// '2.34'

function trimNumber(num, len) {
const modulu_one = 1;
const start_numbers_float=2;
var int_part = Math.trunc(num);
var float_part = String(num % modulu_one);
float_part = float_part.slice(start_numbers_float, start_numbers_float+len);
return int_part+'.'+float_part;
}

There is no way to avoid inconsistent rounding for prices with x.xx5 as actual value using either multiplication or division. If you need to calculate correct prices client-side you should keep all amounts in cents. This is due to the nature of the internal representation of numeric values in JavaScript. Notice that Excel suffers from the same problems so most people wouldn't notice the small errors caused by this phenomen. However errors may accumulate whenever you add up a lot of calculated values, there is a whole theory around this involving the order of calculations and other methods to minimize the error in the final result. To emphasize on the problems with decimal values, please note that 0.1 + 0.2 is not exactly equal to 0.3 in JavaScript, while 1 + 2 is equal to 3.

Maybe you'll also want decimal separator? Here is a function I just made:
function formatFloat(num,casasDec,sepDecimal,sepMilhar) {
if (num < 0)
{
num = -num;
sinal = -1;
} else
sinal = 1;
var resposta = "";
var part = "";
if (num != Math.floor(num)) // decimal values present
{
part = Math.round((num-Math.floor(num))*Math.pow(10,casasDec)).toString(); // transforms decimal part into integer (rounded)
while (part.length < casasDec)
part = '0'+part;
if (casasDec > 0)
{
resposta = sepDecimal+part;
num = Math.floor(num);
} else
num = Math.round(num);
} // end of decimal part
while (num > 0) // integer part
{
part = (num - Math.floor(num/1000)*1000).toString(); // part = three less significant digits
num = Math.floor(num/1000);
if (num > 0)
while (part.length < 3) // 123.023.123 if sepMilhar = '.'
part = '0'+part; // 023
resposta = part+resposta;
if (num > 0)
resposta = sepMilhar+resposta;
}
if (sinal < 0)
resposta = '-'+resposta;
return resposta;
}

/** don't spend 5 minutes, use my code **/
function prettyFloat(x,nbDec) {
if (!nbDec) nbDec = 100;
var a = Math.abs(x);
var e = Math.floor(a);
var d = Math.round((a-e)*nbDec); if (d == nbDec) { d=0; e++; }
var signStr = (x<0) ? "-" : " ";
var decStr = d.toString(); var tmp = 10; while(tmp<nbDec && d*tmp < nbDec) {decStr = "0"+decStr; tmp*=10;}
var eStr = e.toString();
return signStr+eStr+"."+decStr;
}
prettyFloat(0); // "0.00"
prettyFloat(-1); // "-1.00"
prettyFloat(-0.999); // "-1.00"
prettyFloat(0.5); // "0.50"

I use this code to format floats. It is based on toPrecision() but it strips unnecessary zeros. I would welcome suggestions for how to simplify the regex.
function round(x, n) {
var exp = Math.pow(10, n);
return Math.floor(x*exp + 0.5)/exp;
}
Usage example:
function test(x, n, d) {
var rounded = rnd(x, d);
var result = rounded.toPrecision(n);
result = result.replace(/\.?0*$/, '');
result = result.replace(/\.?0*e/, 'e');
result = result.replace('e+', 'e');
return result;
}
document.write(test(1.2000e45, 3, 2) + '=' + '1.2e45' + '<br>');
document.write(test(1.2000e+45, 3, 2) + '=' + '1.2e45' + '<br>');
document.write(test(1.2340e45, 3, 2) + '=' + '1.23e45' + '<br>');
document.write(test(1.2350e45, 3, 2) + '=' + '1.24e45' + '<br>');
document.write(test(1.0000, 3, 2) + '=' + '1' + '<br>');
document.write(test(1.0100, 3, 2) + '=' + '1.01' + '<br>');
document.write(test(1.2340, 4, 2) + '=' + '1.23' + '<br>');
document.write(test(1.2350, 4, 2) + '=' + '1.24' + '<br>');

countDecimals = value => {
if (Math.floor(value) === value) return 0;
let stringValue = value.toString().split(".")[1];
if (stringValue) {
return value.toString().split(".")[1].length
? value.toString().split(".")[1].length
: 0;
} else {
return 0;
}
};
formatNumber=(ans)=>{
let decimalPlaces = this.countDecimals(ans);
ans = 1 * ans;
if (decimalPlaces !== 0) {
let onePlusAns = ans + 1;
let decimalOnePlus = this.countDecimals(onePlusAns);
if (decimalOnePlus < decimalPlaces) {
ans = ans.toFixed(decimalPlaces - 1).replace(/\.?0*$/, "");
} else {
let tenMulAns = ans * 10;
let decimalTenMul = this.countDecimals(tenMulAns);
if (decimalTenMul + 1 < decimalPlaces) {
ans = ans.toFixed(decimalPlaces - 1).replace(/\.?0*$/, "");
}
}
}
}
I just add 1 to the value and count the decimal digits present in the original value and the added value. If I find the decimal digits after adding one less than the original decimal digits, I just call the toFixed() with (original decimals - 1). I also check by multiplying the original value by 10 and follow the same logic in case adding one doesn't reduce redundant decimal places.
A simple workaround to handle floating-point number rounding in JS. Works in most cases I tried.