Is this defined by the language? Is there a defined maximum? Is it different in different browsers?
JavaScript has two number types: Number and BigInt.
The most frequently-used number type, Number, is a 64-bit floating point IEEE 754 number.
The largest exact integral value of this type is Number.MAX_SAFE_INTEGER, which is:
253-1, or
+/- 9,007,199,254,740,991, or
nine quadrillion seven trillion one hundred ninety-nine billion two hundred fifty-four million seven hundred forty thousand nine hundred ninety-one
To put this in perspective: one quadrillion bytes is a petabyte (or one thousand terabytes).
"Safe" in this context refers to the ability to represent integers exactly and to correctly compare them.
From the spec:
Note that all the positive and negative integers whose magnitude is no
greater than 253 are representable in the Number type (indeed, the
integer 0 has two representations, +0 and -0).
To safely use integers larger than this, you need to use BigInt, which has no upper bound.
Note that the bitwise operators and shift operators operate on 32-bit integers, so in that case, the max safe integer is 231-1, or 2,147,483,647.
const log = console.log
var x = 9007199254740992
var y = -x
log(x == x + 1) // true !
log(y == y - 1) // also true !
// Arithmetic operators work, but bitwise/shifts only operate on int32:
log(x / 2) // 4503599627370496
log(x >> 1) // 0
log(x | 1) // 1
Technical note on the subject of the number 9,007,199,254,740,992: There is an exact IEEE-754 representation of this value, and you can assign and read this value from a variable, so for very carefully chosen applications in the domain of integers less than or equal to this value, you could treat this as a maximum value.
In the general case, you must treat this IEEE-754 value as inexact, because it is ambiguous whether it is encoding the logical value 9,007,199,254,740,992 or 9,007,199,254,740,993.
>= ES6:
Number.MIN_SAFE_INTEGER;
Number.MAX_SAFE_INTEGER;
<= ES5
From the reference:
Number.MAX_VALUE;
Number.MIN_VALUE;
console.log('MIN_VALUE', Number.MIN_VALUE);
console.log('MAX_VALUE', Number.MAX_VALUE);
console.log('MIN_SAFE_INTEGER', Number.MIN_SAFE_INTEGER); //ES6
console.log('MAX_SAFE_INTEGER', Number.MAX_SAFE_INTEGER); //ES6
It is 253 == 9 007 199 254 740 992. This is because Numbers are stored as floating-point in a 52-bit mantissa.
The min value is -253.
This makes some fun things happening
Math.pow(2, 53) == Math.pow(2, 53) + 1
>> true
And can also be dangerous :)
var MAX_INT = Math.pow(2, 53); // 9 007 199 254 740 992
for (var i = MAX_INT; i < MAX_INT + 2; ++i) {
// infinite loop
}
Further reading: http://blog.vjeux.com/2010/javascript/javascript-max_int-number-limits.html
In JavaScript, there is a number called Infinity.
Examples:
(Infinity>100)
=> true
// Also worth noting
Infinity - 1 == Infinity
=> true
Math.pow(2,1024) === Infinity
=> true
This may be sufficient for some questions regarding this topic.
Jimmy's answer correctly represents the continuous JavaScript integer spectrum as -9007199254740992 to 9007199254740992 inclusive (sorry 9007199254740993, you might think you are 9007199254740993, but you are wrong!
Demonstration below or in jsfiddle).
console.log(9007199254740993);
However, there is no answer that finds/proves this programatically (other than the one CoolAJ86 alluded to in his answer that would finish in 28.56 years ;), so here's a slightly more efficient way to do that (to be precise, it's more efficient by about 28.559999999968312 years :), along with a test fiddle:
/**
* Checks if adding/subtracting one to/from a number yields the correct result.
*
* #param number The number to test
* #return true if you can add/subtract 1, false otherwise.
*/
var canAddSubtractOneFromNumber = function(number) {
var numMinusOne = number - 1;
var numPlusOne = number + 1;
return ((number - numMinusOne) === 1) && ((number - numPlusOne) === -1);
}
//Find the highest number
var highestNumber = 3; //Start with an integer 1 or higher
//Get a number higher than the valid integer range
while (canAddSubtractOneFromNumber(highestNumber)) {
highestNumber *= 2;
}
//Find the lowest number you can't add/subtract 1 from
var numToSubtract = highestNumber / 4;
while (numToSubtract >= 1) {
while (!canAddSubtractOneFromNumber(highestNumber - numToSubtract)) {
highestNumber = highestNumber - numToSubtract;
}
numToSubtract /= 2;
}
//And there was much rejoicing. Yay.
console.log('HighestNumber = ' + highestNumber);
Many earlier answers have shown 9007199254740992 === 9007199254740992 + 1 is true to verify that 9,007,199,254,740,991 is the maximum and safe integer.
But what if we keep doing accumulation:
input: 9007199254740992 + 1 output: 9007199254740992 // expected: 9007199254740993
input: 9007199254740992 + 2 output: 9007199254740994 // expected: 9007199254740994
input: 9007199254740992 + 3 output: 9007199254740996 // expected: 9007199254740995
input: 9007199254740992 + 4 output: 9007199254740996 // expected: 9007199254740996
We can see that among numbers greater than 9,007,199,254,740,992, only even numbers are representable.
It's an entry to explain how the double-precision 64-bit binary format works. Let's see how 9,007,199,254,740,992 be held (represented) by using this binary format.
Using a brief version to demonstrate it from 4,503,599,627,370,496:
1 . 0000 ---- 0000 * 2^52 => 1 0000 ---- 0000.
|-- 52 bits --| |exponent part| |-- 52 bits --|
On the left side of the arrow, we have bit value 1, and an adjacent radix point. By consuming the exponent part on the left, the radix point is moved 52 steps to the right. The radix point ends up at the end, and we get 4503599627370496 in pure binary.
Now let's keep incrementing the fraction part with 1 until all the bits are set to 1, which equals 9,007,199,254,740,991 in decimal.
1 . 0000 ---- 0000 * 2^52 => 1 0000 ---- 0000.
(+1)
1 . 0000 ---- 0001 * 2^52 => 1 0000 ---- 0001.
(+1)
1 . 0000 ---- 0010 * 2^52 => 1 0000 ---- 0010.
(+1)
.
.
.
1 . 1111 ---- 1111 * 2^52 => 1 1111 ---- 1111.
Because the 64-bit double-precision format strictly allots 52 bits for the fraction part, no more bits are available if we add another 1, so what we can do is setting all bits back to 0, and manipulate the exponent part:
┏━━▶ This bit is implicit and persistent.
┃
1 . 1111 ---- 1111 * 2^52 => 1 1111 ---- 1111.
|-- 52 bits --| |-- 52 bits --|
(+1)
1 . 0000 ---- 0000 * 2^52 * 2 => 1 0000 ---- 0000. * 2
|-- 52 bits --| |-- 52 bits --|
(By consuming the 2^52, radix
point has no way to go, but
there is still one 2 left in
exponent part)
=> 1 . 0000 ---- 0000 * 2^53
|-- 52 bits --|
Now we get the 9,007,199,254,740,992, and for the numbers greater than it, the format can only handle increments of 2 because every increment of 1 on the fraction part ends up being multiplied by the left 2 in the exponent part. That's why double-precision 64-bit binary format cannot hold odd numbers when the number is greater than 9,007,199,254,740,992:
(consume 2^52 to move radix point to the end)
1 . 0000 ---- 0001 * 2^53 => 1 0000 ---- 0001. * 2
|-- 52 bits --| |-- 52 bits --|
Following this pattern, when the number gets greater than 9,007,199,254,740,992 * 2 = 18,014,398,509,481,984 only 4 times the fraction can be held:
input: 18014398509481984 + 1 output: 18014398509481984 // expected: 18014398509481985
input: 18014398509481984 + 2 output: 18014398509481984 // expected: 18014398509481986
input: 18014398509481984 + 3 output: 18014398509481984 // expected: 18014398509481987
input: 18014398509481984 + 4 output: 18014398509481988 // expected: 18014398509481988
How about numbers between [ 2 251 799 813 685 248, 4 503 599 627 370 496 )?
1 . 0000 ---- 0001 * 2^51 => 1 0000 ---- 000.1
|-- 52 bits --| |-- 52 bits --|
The value 0.1 in binary is exactly 2^-1 (=1/2) (=0.5)
So when the number is less than 4,503,599,627,370,496 (2^52), there is one bit available to represent the 1/2 times of the integer:
input: 4503599627370495.5 output: 4503599627370495.5
input: 4503599627370495.75 output: 4503599627370495.5
Less than 2,251,799,813,685,248 (2^51)
input: 2251799813685246.75 output: 2251799813685246.8 // expected: 2251799813685246.75
input: 2251799813685246.25 output: 2251799813685246.2 // expected: 2251799813685246.25
input: 2251799813685246.5 output: 2251799813685246.5
/**
Please note that if you try this yourself and, say, log
these numbers to the console, they will get rounded. JavaScript
rounds if the number of digits exceed 17. The value
is internally held correctly:
*/
input: 2251799813685246.25.toString(2)
output: "111111111111111111111111111111111111111111111111110.01"
input: 2251799813685246.75.toString(2)
output: "111111111111111111111111111111111111111111111111110.11"
input: 2251799813685246.78.toString(2)
output: "111111111111111111111111111111111111111111111111110.11"
And what is the available range of exponent part? 11 bits allotted for it by the format.
From Wikipedia (for more details, go there)
So to make the exponent part be 2^52, we exactly need to set e = 1075.
To be safe
var MAX_INT = 4294967295;
Reasoning
I thought I'd be clever and find the value at which x + 1 === x with a more pragmatic approach.
My machine can only count 10 million per second or so... so I'll post back with the definitive answer in 28.56 years.
If you can't wait that long, I'm willing to bet that
Most of your loops don't run for 28.56 years
9007199254740992 === Math.pow(2, 53) + 1 is proof enough
You should stick to 4294967295 which is Math.pow(2,32) - 1 as to avoid expected issues with bit-shifting
Finding x + 1 === x:
(function () {
"use strict";
var x = 0
, start = new Date().valueOf()
;
while (x + 1 != x) {
if (!(x % 10000000)) {
console.log(x);
}
x += 1
}
console.log(x, new Date().valueOf() - start);
}());
The short answer is “it depends.”
If you’re using bitwise operators anywhere (or if you’re referring to the length of an Array), the ranges are:
Unsigned: 0…(-1>>>0)
Signed: (-(-1>>>1)-1)…(-1>>>1)
(It so happens that the bitwise operators and the maximum length of an array are restricted to 32-bit integers.)
If you’re not using bitwise operators or working with array lengths:
Signed: (-Math.pow(2,53))…(+Math.pow(2,53))
These limitations are imposed by the internal representation of the “Number” type, which generally corresponds to IEEE 754 double-precision floating-point representation. (Note that unlike typical signed integers, the magnitude of the negative limit is the same as the magnitude of the positive limit, due to characteristics of the internal representation, which actually includes a negative 0!)
ECMAScript 6:
Number.MAX_SAFE_INTEGER = Math.pow(2, 53)-1;
Number.MIN_SAFE_INTEGER = -Number.MAX_SAFE_INTEGER;
Other may have already given the generic answer, but I thought it would be a good idea to give a fast way of determining it :
for (var x = 2; x + 1 !== x; x *= 2);
console.log(x);
Which gives me 9007199254740992 within less than a millisecond in Chrome 30.
It will test powers of 2 to find which one, when 'added' 1, equals himself.
Anything you want to use for bitwise operations must be between 0x80000000 (-2147483648 or -2^31) and 0x7fffffff (2147483647 or 2^31 - 1).
The console will tell you that 0x80000000 equals +2147483648, but 0x80000000 & 0x80000000 equals -2147483648.
JavaScript has received a new data type in ECMAScript 2020: BigInt. It introduced numerical literals having an "n" suffix and allows for arbitrary precision:
var a = 123456789012345678901012345678901n;
Precision will still be lost, of course, when such big integer is (maybe unintentionally) coerced to a number data type.
And, obviously, there will always be precision limitations due to finite memory, and a cost in terms of time in order to allocate the necessary memory and to perform arithmetic on such large numbers.
For instance, the generation of a number with a hundred thousand decimal digits, will take a noticeable delay before completion:
console.log(BigInt("1".padEnd(100000,"0")) + 1n)
...but it works.
Try:
maxInt = -1 >>> 1
In Firefox 3.6 it's 2^31 - 1.
I did a simple test with a formula, X-(X+1)=-1, and the largest value of X I can get to work on Safari, Opera and Firefox (tested on OS X) is 9e15. Here is the code I used for testing:
javascript: alert(9e15-(9e15+1));
I write it like this:
var max_int = 0x20000000000000;
var min_int = -0x20000000000000;
(max_int + 1) === 0x20000000000000; //true
(max_int - 1) < 0x20000000000000; //true
Same for int32
var max_int32 = 0x80000000;
var min_int32 = -0x80000000;
Let's get to the sources
Description
The MAX_SAFE_INTEGER constant has a value of 9007199254740991 (9,007,199,254,740,991 or ~9 quadrillion). The reasoning behind that number is that JavaScript uses double-precision floating-point format numbers as specified in IEEE 754 and can only safely represent numbers between -(2^53 - 1) and 2^53 - 1.
Safe in this context refers to the ability to represent integers exactly and to correctly compare them. For example, Number.MAX_SAFE_INTEGER + 1 === Number.MAX_SAFE_INTEGER + 2 will evaluate to true, which is mathematically incorrect. See Number.isSafeInteger() for more information.
Because MAX_SAFE_INTEGER is a static property of Number, you always use it as Number.MAX_SAFE_INTEGER, rather than as a property of a Number object you created.
Browser compatibility
In JavaScript the representation of numbers is 2^53 - 1.
However, Bitwise operation are calculated on 32 bits ( 4 bytes ), meaning if you exceed 32bits shifts you will start loosing bits.
In the Google Chrome built-in javascript, you can go to approximately 2^1024 before the number is called infinity.
Scato wrotes:
anything you want to use for bitwise operations must be between
0x80000000 (-2147483648 or -2^31) and 0x7fffffff (2147483647 or 2^31 -
1).
the console will tell you that 0x80000000 equals +2147483648, but
0x80000000 & 0x80000000 equals -2147483648
Hex-Decimals are unsigned positive values, so 0x80000000 = 2147483648 - thats mathematically correct. If you want to make it a signed value you have to right shift: 0x80000000 >> 0 = -2147483648. You can write 1 << 31 instead, too.
Firefox 3 doesn't seem to have a problem with huge numbers.
1e+200 * 1e+100 will calculate fine to 1e+300.
Safari seem to have no problem with it as well. (For the record, this is on a Mac if anyone else decides to test this.)
Unless I lost my brain at this time of day, this is way bigger than a 64-bit integer.
Node.js and Google Chrome seem to both be using 1024 bit floating point values so:
Number.MAX_VALUE = 1.7976931348623157e+308
I would like to understand how mIoU is calculated for multi-class classification. The formula for each class is
IoU formula
and then the average is done over the classes to get the mIoU. However, I don't understand what happens for the classes that are not represented. The formula becomes a division by 0, so I ignore them and the average is only computed for the classes represented.
The problem is that when a prediction is wrong, the accuracy is really lowered. It adds another class to make the average. For instance : in semantic segmentation the ground-truth of an image is made of 4 classes (0,1,2,3) and 6 classes are represented over the dataset. The prediction is also made of 4 classes (0,1,4,5) but all the items classified in 2 and 3 (in the ground-truth) are classified in 4 and 5 (in the prediction). In this case should we calculate the mIoU over 6 classes ? Even if 4 classes are totally wrong and there respective IoU is 0 ? So the problem is that if just one pixel is predicted in a class that is not in the ground_truth, we have to divide by a higher denominator and it lows a lot the score.
Is it the correct way to compute the mIoU for multi-class (and the semantic segmentation) ?
Instead of calculating the miou of each image and then calculate the "mean" miou over all the images, I calculate the miou as one big image. If a class is not in the image and is not predicited, I set there respective iou equal to 1.
From scratch :
def miou(gt,pred,nbr_mask):
intersection = np.zeros(nbr_mask) # int = (A and B)
den = np.zeros(nbr_mask) # den = A + B = (A or B) + (A and B)
for i in range(len(gt)):
for j in range(height):
for k in range(width):
if pred[i][j][k]==gt[i][j][k]:
intersection[gt[i][j][k]]+=1
den[pred[i][j][k]] += 1
den[gt[i][j][k]] += 1
mIoU = 0
for i in range(nbr_mask):
if den[i]!=0:
mIoU+=intersection[i]/(den[i]-intersection[i])
else:
mIoU+=1
mIoU=mIoU/nbr_mask
return mIoU
With gt the array of ground truth labels and pred the prediction of theassociated images (have to correspond in the array and be the same size).
Adding to the previous answer, this is a great fast and efficient pytorch GPU implementation of calculating the mIOU and classswise IOU for a batch of size (N, H, W) (both pred mask and labels), taken from the NeurIPS 2021 paper "Few-Shot Segmentation via Cycle-Consistent Transformer", github repo available here.
def intersectionAndUnionGPU(output, target, K, ignore_index=255):
# 'K' classes, output and target sizes are N or N * L or N * H * W, each value in range 0 to K - 1.
assert (output.dim() in [1, 2, 3])
assert output.shape == target.shape
output = output.view(-1)
target = target.view(-1)
output[target == ignore_index] = ignore_index
intersection = output[output == target]
area_intersection = torch.histc(intersection, bins=K, min=0, max=K-1)
area_output = torch.histc(output, bins=K, min=0, max=K-1)
area_target = torch.histc(target, bins=K, min=0, max=K-1)
area_union = area_output + area_target - area_intersection
return area_intersection, area_union, area_target
Example usage:
output = torch.rand(4, 5, 224, 224) # model output; batch size=4; channels=5, H,W=224
preds = F.softmax(output, dim=1).argmax(dim=1) # (4, 224, 224)
labels = torch.randint(0,5, (4, 224, 224))
i, u, _ = intersectionAndUnionGPU(preds, labels, 5) # 5 is num_classes
classwise_IOU = i/u # tensor of size (num_classes)
mIOU = i.sum()/u.sum() # mean IOU, taking (i/u).mean() is wrong
Hope this helps everyone!
(A non-GPU implementation is available as well in the repo!)
For example if I have 10 samples in the FFT in a system sampled at 100 Hz. Then how are the results represented in FFT bin output ?
The frequency resolution is going to be how many Hz each DFT bin represents. This is, as you have noted, given by fs/N. In this case :--
resolution = 100/10 = 10 hz
Definition of FFT states that frequency range [0,fs] is represented by N point.
But when the result comes :--
X[0] is the constant term (also refereed to as DC bias)
X[1] is the 10 hz
X[2] is the 20 hz
X[3] is the 30 hz
X[4] is the 40 hz
X[5] is the 50 hz
X[6] to X[9] are the complex conjugates of X[4] to X[1] respectively.
so my question is :--
X[6] is the 60hz signal & its value will be complex conjugate of 40 hz signal .. is it right ?
Or X[6] does not contains spectral of 60hz but it is only complex conjugate of X[4] ?
Please suggest.
The frequencies of each bin in the DFT go from $n = 0 to N-1$ where each frequency is given as $n F_s/N$.
Due to the sampling process, for any signals (either real and complex signals) the upper half of this frequency spectrum that extends from $F_s/2$ to 1 sample less than $F_s$ is equivalent to the negative half spectrum (the spectrum that extends from $-F_s/2% to 1 sample less than 0.). This applies to the DFT and all digital signals.
This is demonstrated in the graphics below, showing how an analog spectrum (top line), convolves with the sampling spectrum (second line) of a system sampled at 20 Hz to result in the digital spectrum of a sampled signal. Since the spectrum is unique over a 20 Hz range (and repeats everywhere else), we only need to show the frequencies over any 20 Hz range to represent the signal. This can be identically shown from -10 Hz to +10 Hz, or 0 to 20 Hz.
Representing the sampled spectrum of a real signal sampled at 20 Hz over the range of -10Hz to +10Hz:
The same signal can also be represented over the range of 0 to 20 Hz:
Representing the sampled spectrum of a complex signal sampled at 20 Hz over the range of -10Hz to +10Hz:
The same signal can also be represented over the range of 0 to 20 Hz:
Since the spectrum for the DFT is discrete, the samples go from $n = 0 to N-1$ where each frequency is given as $F_s/n$ as described above. There is also only N samples in the DFT of course but as you rotate the samples in the DFT, you effectively move through the expanded spectrum above. It is helpful to view the spectrum on the surface of a cylinder with a circumference that goes from 0 to $F_s$ where you see that $F_s$ is equivalent to 0, and going backwards from 0 is equivalent to going into the negative half spectrum.
So in your example specifically:
X[0] = DC
X1 = 10
X2 = 20
X3 = 30
X4 = 40
X[5] = 50
X[6] = 60
X[7] = 70
X[8] = 80
X[9] = 90
Can also be represented as
X[0] = DC
X1 = 10
X2 = 20
X3 = 30
X4 = 40
X[5] = -50
X[6] = -40
X[7] = -30
X[8] = -20
X[9] = -10
Note that the command "FFTSHIFT" in MATLAB shifts the DFT vector accordingly to produce the following order representing the range from -F_s/2 to +F_s/2 :
fftshift([X[0], X1, X2, X3, X4, X[5], X[6], X[7], X[8], X[9]]) =
[X[5], X[6], X[7], X[8], X[9], X[0], X1, X2, X3, X4]
Assuming your input signal is purely real, then X[6] is just the complex conjugate of X[4]. The top half of the spectrum is essentially redundant for real signals. See: this question and answer for further details.
Note that the input signal should be bandwidth-limited to fs / 2, otherwise aliasing will occur, so for a baseband signal there should not be any components >= fs / 2.