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How to get the first two digits of a large float number in Javascript

I want to get the first two digits of a large float number e.g from 1115487.2644548 i want to get returned 11.154872644548.
Which Javascript native function to use to achieve such operation.
PS : toPrecision(2) isn't what i want.
Thanks.

Simple division
const num = 1115487.2644548;
console.log(num/100000);
console.log(parseInt(num/100000));
If the length of the integer is unknown, try this - other answers shows log versions
const num = 1115487.2644548;
const len = String(parseInt(num)).length
const two = len-2;
console.log(num/10**two); // without testing the len > 2

You could adjust by using the logarithm of ten.
function getTwo(v) {
return v / 10 ** (Math.floor(Math.log10(Math.floor(v))) - 1);
}
console.log(getTwo(1115487.2644548));
console.log(getTwo(111.54872644548));
console.log(getTwo(11154.872644548));
console.log(getTwo(99999));
console.log(getTwo(999.9999999999998863131622783839702606201171875));

How to simulate the result of a Google Sheet multiplication formula in Javascript?

If the value of f5 cell in a Google Sheet is 1.1000 (a number formatted to 4 decimal places) and the value of f6 is = f5 * 1.073, how can I ensure I get the same result multiplying those values in Javascript, eg:
var original_value = 1.1000;
var derivative_value = original_value * 1.073;
Specifically, my question is - will the result of the Javascript multiplication (derivative_value) be the same as the result of the Google formula (f6)? And if not, how can I make it so that it is?
Context / What I've Tried
For context, this question is part of a larger question I am trying to resolve for which I have set up this JSFiddle.
The JSFiddle has an input for the original_value and an input for the multiplier.
It outputs the result to four decimal places and adds trailing zeros where required (this is the required format for the result).
It is an attempt to check that the Javascript code I am writing will produce the same result as the Google Sheet formula.
[ The JSFiddle has been updated to also log decimal.js results to the console for comparison ]
Edit
There was a suggestion to use decimal.js but I'm not sure how it would be applied - something like the following?
var original_value = new Decimal(1.1000);
// some different multipliers for testing
var multiplier_a = new Decimal(1.073);
var multiplier_b = new Decimal(1.1);
// some different results for testing
var derivative_value_a = original_value.times(multiplier_a).toString();
var derivative_value_b = original_value.times(multiplier_b).toString();
console.log(derivative_value_a); // 1.1803
console.log(derivative_value_b); // 1.21
Is that any more accurate than plain Javascript original_value * multiplier? More importantly for this question, will it always simulate the same result that a Google Sheet formula produces?

JavaScript is using so called double precision float format (64 bit)- https://tc39.github.io/ecma262/#sec-terms-and-definitions-number-value
Google Sheets seem to use the same format, you can test it by =f6*1E13 - round(f6*1E13) to see that f6 is not STORED as a fixed number format, only FORMATTED
see Number.toFixed how to FORMAT numbers in Javascript
to generate some test data:
[...Array(10)].forEach(() => {
const f5 = 1.1
const x = Math.random() / 100
const f6 = f5 * x
console.log(x, f6.toFixed(4))
})
and compare in Google Sheet:
https://docs.google.com/spreadsheets/d/1jKBwzM41nwIEyatLUHEUwteK8ImJg334hzJ8nKkUZ5M/view
=> all rounded numbers are equal.
P.S.: you need to copy the console output, paste into the Sheet, use the menu item Data > Split text into columns... > Space, then multiply by 1.1 in 3rd column and finally format all numbers

After revisiting this I have updated the jsFiddle.
The main components of what I believe are a satisfactory solution are:
Convert both original_value and multiplier to decimal.js objects.
Do the multiplication using the decimal.js times method.
Do the rounding using the decimal.js toDecimalPlaces method.
Use the argument values (4,7) to define 4 decimal places with ROUND_HALF_CEIL rounding, equivalent to Math.round (reference)
For example:
var my_decimal_js_value = new Decimal(original_value).times(new Decimal(multiplier)).toDecimalPlaces(4, 7);
In order to add any necessary trailing zeros to the result, I use:
function trailingZeros(my_decimal_js_value) {
var result = my_decimal_js_value;
// add zeros if required:
var split_result = result.toString().split(".");
// if there are decimals present
if (split_result[1] != undefined) {
// declare trailing_zeros;
var trailing_zeros;
// get the amount of decimal numbers
decimals_present = split_result[1].length;
// if one decimal number, add three trailing zeros
if (decimals_present === 1) {
trailing_zeros = "000";
result += trailing_zeros;
}
// if two decimal numbers, add two trailing zeros
else if (decimals_present === 2) {
trailing_zeros = "00";
result += trailing_zeros;
}
// if three decimal numbers, add one trailing zero
else if (decimals_present === 3) {
trailing_zeros = "0";
result += trailing_zeros;
}
// if four decimal numbers, just convert result to string
else if (decimals_present === 4) {
result = result.toString();
}
}
// if there are no decimals present, add a decimal place and four zeros
else if (split_result[1] === undefined) {
trailing_zeros = ".0000";
result += trailing_zeros;
}
return result;
}
I am still not absolutely certain that this mimics the Google Sheet multiplication formula, however using decimal.js, or another dedicated decimal library, seems to be the preferred method over plain JavaScript (to avoid possible rounding errors), based on posts such as these:
http://www.jacklmoore.com/notes/rounding-in-javascript
Is floating point math broken?
https://spin.atomicobject.com/2016/01/04/javascript-math-precision-decimals

toFixed function not working properly ( please give a reason not an alternative)

toFixed() function responding differently for float values.
For Example:
var a = 2.555;
var b = 5.555;
console.log(a.toFixed(2)); /* output is 2.56 */
console.log(b.toFixed(2)); /* output is 5.55 */
For 2.555/3.555 results are (2.56/3.56)
and
For other values(not sure for all values) it is showing #.55 (# refers to any number)
I am confused can any one help me out.
Thanks in advance.

Javascript uses a binary floating point representation for numbers (IEEE754).
Using this representation the only numbers that can be represented exactly are in the form n/2m where both n and m are integers.
Any number that is not a rational where the denominator is an integral power of two is impossible to represent exactly because in binary it is a periodic number (it has infinite binary digits after the point).
The number 0.5 (i.e. 1/2) is fine, (in binary is just 0.1₂) but for example 0.55 (i.e. 11/20) cannot be represented exactly (in binary it's 0.100011001100110011₂… i.e. 0.10(0011)₂ with the last part 0011₂ repeating infinite times).
If you need to do any computation in which the result depends on exact decimal numbers you need to use an exact decimal representation. A simple solution if the number of decimals is fixed (e.g. 3) is to keep all values as integers by multiplying them by 1000...
2.555 --> 2555
5.555 --> 5555
3.7 --> 3700
and adjusting your computation when doing multiplications and divisions accordingly (e.g. after multiplying two numbers you need to divide the result by 1000).
The IEEE754 double-precision format is accurate with integers up to 9,007,199,254,740,992 and this is often enough for prices/values (where the rounding is most often an issue).

Try this Demo Here
function roundToTwo(num) {
alert(+(Math.round(num + "e+2") + "e-2"));
}
roundToTwo(2.555);
roundToTwo(5.555);

toFixed() method depending on Browser rounds down or retain.
Here is the solution for this problem, check for "5" at the end
var num = 5.555;
var temp = num.toString();
if(temp .charAt(temp .length-1)==="5"){
temp = temp .slice(0,temp .length-1) + '6';
}
num = Number(temp);
Final = num.toFixed(2);
Or reusable function would be like
function toFixedCustom(num,upto){
var temp = num.toString();
if(temp .charAt(temp .length-1)==="5"){
temp = temp .slice(0,temp .length-1) + '6';
}
num = Number(temp);
Final = num.toFixed(upto);
return Final;
}
var a = 2.555;
var b = 5.555;
console.log(toFixedCustom(a,2));
console.log(toFixedCustom(b,2));

JavaScript factorial prevent infinity

I have been using this function for calculating factorial numbers in JavaScript:
var f = [];
function factorial (n) {
if (n == 0 || n == 1)
return 1;
if (f[n] > 0)
return f[n];
return f[n] = factorial(n-1) * n;
}
All seemed to be going well until I tried the number 500. It returned infinity.
Is there a way that I can prevent infinity as an answer?
Thank you.

You indeed need to use bignumbers. With math.js you can do:
// configure math.js to work with enough precision to do our calculation
math.config({precision: 2000});
// evaluate the factorial using a bignumber value
var value = math.bignumber(500);
var result = math.factorial(value);
// output the results
console.log(math.format(result, {notation: 'fixed'}));
This will output:
1220136825991110068701238785423046926253574342803192842192413588385845373153881997605496447502203281863013616477148203584163378722078177200480785205159329285477907571939330603772960859086270429174547882424912726344305670173270769461062802310452644218878789465754777149863494367781037644274033827365397471386477878495438489595537537990423241061271326984327745715546309977202781014561081188373709531016356324432987029563896628911658974769572087926928871281780070265174507768410719624390394322536422605234945850129918571501248706961568141625359056693423813008856249246891564126775654481886506593847951775360894005745238940335798476363944905313062323749066445048824665075946735862074637925184200459369692981022263971952597190945217823331756934581508552332820762820023402626907898342451712006207714640979456116127629145951237229913340169552363850942885592018727433795173014586357570828355780158735432768888680120399882384702151467605445407663535984174430480128938313896881639487469658817504506926365338175055478128640000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

500! is, for lack of a better term, "[bleep]ing huge".
It is far, far beyond what can be stored in a double-precision float, which is what JavaScript uses for numbers.
There's no way to prevent this, other than use numbers that are reasonable :p
EDIT: To show you just how huge it is, here's the answer:
500! = 1220136825991110068701238785423046926253574342803192842192413588385845373153881997605496447502203281863013616477148203584163378722078177200480785205159329285477907571939330603772960859086270429174547882424912726344305670173270769461062802310452644218878789465754777149863494367781037644274033827365397471386477878495438489595537537990423241061271326984327745715546309977202781014561081188373709531016356324432987029563896628911658974769572087926928871281780070265174507768410719624390394322536422605234945850129918571501248706961568141625359056693423813008856249246891564126775654481886506593847951775360894005745238940335798476363944905313062323749066445048824665075946735862074637925184200459369692981022263971952597190945217823331756934581508552332820762820023402626907898342451712006207714640979456116127629145951237229913340169552363850942885592018727433795173014586357570828355780158735432768888680120399882384702151467605445407663535984174430480128938313896881639487469658817504506926365338175055478128640000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
That right there is a 1,135-digit number. For comparison, double-precision floats can handle about 15 digits of precision.

You could consider using an arbitrary precision numeric library. This is a question of its own, though. Here's one related question: https://stackoverflow.com/questions/744099/is-there-a-good-javascript-bigdecimal-library.

I dont know if anyone has solved this elsewise...
I'm a novice beginner in coding and dont know all the aspects. But after I faced this factorial problem myself, i came here when searching for the answer. I solved the 'infinity' display problem in another way. I dont know if its very efficient or not. But it does show the results of even verry high intergers.
Sorry for any redundancy or untidiness in the code.
<!DOCTYPE html>
<html>
<head>
<title>Factorial</title>
<script src='http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js'></script>
</head>
<body>
<input type='text' id='number' />
<input type='button' value='!Factorial!' id='btn' />
<script>
var reslt=1;
var counter=0;
var mantissa=0; //stores the seperated matissa
var exponent=0; //stores the seperated exponent
$(document).ready(function (){
$('#btn').click(function (){
var num=parseFloat($('#number').val()); //number input by user
for(i=1;i<=num;i++){
reslt=reslt*i;
//when the result becomes so high that the exponent reaches 306, the number is divided by 1e300
if((parseFloat(reslt.toExponential().toString().split("e")[1]))>=300){
reslt=reslt/1e300; //the result becomes small again to be able to be iterated without becoming infinity
counter+=1; //the number of times that the number is divided in such manner is recorded by counter
}
}
//the mantissa of the final result is seperated first
mantissa=parseFloat(reslt.toExponential().toString().split("e")[0]);
//the exponent of the final result is obtained by adding the remaining exponent with the previously dropped exponents (1e300)
exponent=parseFloat(reslt.toExponential().toString().split("e")[1])+300*counter;
alert(mantissa+"e+"+exponent); //displays the result as a string by concatenating
//resets the variables and fields for the next input if any
$('#number').val('');
reslt=1;
mantissa=0;
exponent=0;
counter=0;
});
});
</script>
</body>
</html>

Javascript numbers can only get so big before they just become "Infinity". If you want to support bigger numbers, you'll have to use BigInt.
Examples:
// Without BigInt
console.log(100 ** 1000) // Infinity
// With BigInt
// (stackOverflow doesn't seem to print the result,
// unless I turn it into a string first)
console.log(String(100n ** 1000n)) // A really big number
So, for your specific bit of code, all you need to do is turn your numeric literals into BigInt literals, like this:
var f = [];
function factorial (n) {
if (n == 0n || n == 1n)
return 1n;
if (f[n] > 0n)
return f[n];
return f[n] = factorial(n-1n) * n;
}
console.log(String(factorial(500n)));
You'll find that you computer can run that piece of code in a snap.

Hi this is due to the nature of java script as it can't represents number above 253-1 reference so to solve this either wrap the number with BigInt(n) or add to the number >> 3n
const factorial = (n) => {
n = BigInt(n)
if ( n < 1 ) return 1n
return factorial(n - 1n) * n
}

Converting a double to an int in Javascript without rounding

In C# the following code returns 2:
double d = 2.9;
int i = (int)d;
Debug.WriteLine(i);
In Javascript, however, the only way of converting a "double" to an "int" that I'm aware of is by using Math.round/floor/toFixed etc. Is there a way of converting to an int in Javascript without rounding? I'm aware of the performance implications of Number() so I'd rather avoid converting it to a string if at all possible.

Use parseInt().
var num = 2.9
console.log(parseInt(num, 10)); // 2
You can also use |.
var num = 2.9
console.log(num | 0); // 2

I find the "parseInt" suggestions to be pretty curious, because "parseInt" operates on strings by design. That's why its name has the word "parse" in it.
A trick that avoids a function call entirely is
var truncated = ~~number;
The double application of the "~" unary operator will leave you with a truncated version of a double-precision value. However, the value is limited to 32 bit precision, as with all the other JavaScript operations that implicitly involve considering numbers to be integers (like array indexing and the bitwise operators).
edit — In an update quite a while later, another alternative to the ~~ trick is to bitwise-OR the value with zero:
var truncated = number|0;

Similar to C# casting to (int) with just using standard lib:
Math.trunc(1.6) // 1
Math.trunc(-1.6) // -1

Just use parseInt() and be sure to include the radix so you get predictable results:
parseInt(d, 10);

There is no such thing as an int in Javascript. All Numbers are actually doubles behind the scenes* so you can't rely on the type system to issue a rounding order for you as you can in C or C#.
You don't need to worry about precision issues (since doubles correctly represent any integer up to 2^53) but you really are stuck with using Math.floor (or other equivalent tricks) if you want to round to the nearest integer.
*Most JS engines use native ints when they can but all in all JS numbers must still have double semantics.

A trick to truncate that avoids a function call entirely is
var number = 2.9
var truncated = number - number % 1;
console.log(truncated); // 2
To round a floating-point number to the nearest integer, use the addition/subtraction trick. This works for numbers with absolute value < 2 ^ 51.
var number = 2.9
var rounded = number + 6755399441055744.0 - 6755399441055744.0; // (2^52 + 2^51)
console.log(rounded); // 3
Note:
Halfway values are rounded to the nearest even using "round half to even" as the tie-breaking rule. Thus, for example, +23.5 becomes +24, as does +24.5. This variant of the round-to-nearest mode is also called bankers' rounding.
The magic number 6755399441055744.0 is explained in the stackoverflow post "A fast method to round a double to a 32-bit int explained".
// Round to whole integers using arithmetic operators
let trunc = (v) => v - v % 1;
let ceil = (v) => trunc(v % 1 > 0 ? v + 1 : v);
let floor = (v) => trunc(v % 1 < 0 ? v - 1 : v);
let round = (v) => trunc(v < 0 ? v - 0.5 : v + 0.5);
let roundHalfEven = (v) => v + 6755399441055744.0 - 6755399441055744.0; // (2^52 + 2^51)
console.log("number floor ceil round trunc");
var array = [1.5, 1.4, 1.0, -1.0, -1.4, -1.5];
array.forEach(x => {
let f = x => (x).toString().padStart(6," ");
console.log(`${f(x)} ${f(floor(x))} ${f(ceil(x))} ${f(round(x))} ${f(trunc(x))}`);
});

As #Quentin and #Pointy pointed out in their comments, it's not a good idea to use parseInt() because it is designed to convert a string to an integer. When you pass a decimal number to it, it first converts the number to a string, then casts it to an integer. I suggest you use Math.trunc(), Math.floor(), ~~num, ~~v , num | 0, num << 0, or num >> 0 depending on your needs.
This performance test demonstrates the difference in parseInt() and Math.floor() performance.
Also, this post explains the difference between the proposed methods.

What about this:
if (stringToSearch.IndexOfAny( ".,;:?!".ToCharArray() ) == -1) { ... }

I think that the easiest solution is using the bitwise not operator twice:
const myDouble = -66.7;
console.log(myDouble); //-66.7
const myInt = ~~myDouble;
console.log(myInt); //-66
const myInt = ~~-myDouble;
console.log(myInt); //66