How to find difference between two values? [closed] - javascript

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I want return the difference between 2 values how to do that?
0.0.0.1.0 and 0.0.0.1.12
so the difference between these two values is 12
so how to calculate that I tried with Math.abs() but it is fine with single digits.

Assuming that they are strings (since you can't have more than 1 full stop in a valid JS number), you could split it by . character and calculate the difference of individual components:
function numStringDiff(a, b) {
// split them on dot characters
const aParts = a.split('.');
const bParts = b.split('.');
// loop using longest array length as a container for mapped results
return Array(Math.max(aParts.length, bParts.length)).fill(undefined).map((_, i) => {
const i1 = parseInt(aParts[i] || 0); // fetch aParts[i] or default to 0
const i2 = parseInt(bParts[i] || 0); // fetch bParts[i] or default to 0
// return compared value after defaulting the values.
return i2 - i1;
});
}
console.log(numStringDiff('0.0.0.1.0', '0.0.0.1.12'));
The problem here is that, as you stated in the comments, they can be of different length. To make it work in this scenario, we must iterate an amount of times equal to the length of the longest array and ensure that any missing items in the shorter one are defaulted to some non-breaking value like 0 so that we can safely subtract every digit present in the longest list with something or 0.
Note that 0 is a value I only used to ensure you can calculate a difference between different-length arrays, choose any (numeric or float) value that fits your needs.
If in this case the second argument has less dots than the first, negative difference will be returned, otherwise if first is longer than last, positive difference will be returned.
Some examples:
numStringDiff('1.1.1', '1.1') // => [0, 0, -1]
numStringDiff('1.1', '1.1.1') // => [0, 0, 1]
numStringDiff('1.1.1', '1.1.1') // => [0, 0, 0]
For the absolute distance between two values, one can simply .map over this array:
numStringDiff('1.1.1', '1.1').map(num => Math.abs(num));
// OR, using short form:
numStringDiff('1.1.1', '1.1').map(Math.abs);
And finally, should you need the result as a string, simply .join it back together with '.':
numStringDiff('1.1.1', '1.1').map(Math.abs).join('.');
Do know what you are trying to achieve though. If you're trying to manually bisect version numbers (like semver versions) I'd recommend against it since there will always be scenario's uncovered by this function such as pre-releases that wouldn't include only digits but rather 0.0.0-pre-foo-version or something. Since I don't know what it is you're trying to do exactly I'll leave that a responsibility for you to figure out :)

Related

what does '&' do in this solution? [duplicate]

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How does '&' work in relation to odd and even? In JS
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I'm working through a problem on CodeSignal and trying to understand some of the solutions that other people have submitted. One of the solutions was as follows, and I don't understand what the ampersand is doing.
(a) => a.reduce((p,v,i) => (p[i&1]+=v,p), [0,0])
The problem is:
Several people are standing in a row and need to be divided into two teams. The first person goes into team 1, the second goes into team 2, the third goes into team 1 again, the fourth into team 2, and so on.
You are given an array of positive integers - the weights of the people. Return an array of two integers, where the first element is the total weight of team 1, and the second element is the total weight of team 2 after the division is complete.
Example
For a = [50, 60, 60, 45, 70], the output should be
solution(a) = [180, 105].
In this solution, the & operator is used to perform a bitwise AND operation. In JavaScript, the & operator compares each bit of the first operand to the corresponding bit of the second operand. If both bits are 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.
In the given solution, the & operator is used to determine whether the index i of the current element in the array is even or odd. If i is even, the result of i & 1 will be 0. If i is odd, the result of i & 1 will be 1.

Finding a pattern in an array that is not always consistant

I have an ordered data set of decimal numbers. This data is always similar - but not always the same. The expected data is a few, 0 - 5 large numbers, followed by several (10 - 90) average numbers then follow by smaller numbers. There are cases where a large number may be mixed into the average numbers' See the following arrays.
let expectedData = [35.267,9.267,9.332,9.186,9.220,9.141,9.107,9.114,9.098,9.181,9.220,4.012,0.132];
let expectedData = [35.267,32.267,9.267,9.332,9.186,9.220,9.141,9.107,30.267,9.114,9.098,9.181,9.220,4.012,0.132];
I am trying to analyze the data by getting the average without high numbers on front and low numbers on back. The middle high/low are fine to keep in the average. I have a partial solution below. Right now I am sort of brute forcing it but the solution isn't perfect. On smaller datasets the first average calculation is influenced by the large number.
My question is: Is there a way to handle this type of problem, which is identifying patterns in an array of numbers?
My algorithm is:
Get an average of the array
Calculate an above/below average value
Remove front (n) elements that are above average
remove end elements that are below average
Recalculate average
In JavaScript I have: (this is partial leaving out below average)
let total= expectedData.reduce((rt,cur)=> {return rt+cur;}, 0);
let avg = total/expectedData.length;
let aboveAvg = avg*0.1+avg;
let remove = -1;
for(let k=0;k<expectedData.length;k++) {
if(expectedData[k] > aboveAvg) {
remove=k;
} else {
if(k==0) {
remove = -1;//no need to remove
}
//break because we don't want large values from middle removed.
break;
}
}
if(remove >= 0 ) {
//remove front above average
expectedData.splice(0,remove+1);
}
//remove belows
//recalculate average
I believe you are looking for some outlier detection Algorithm. There are already a bunch of questions related to this on Stack overflow.
However, each outlier detection algorithm has its own merits.
Here are a few of them
https://mathworld.wolfram.com/Outlier.html
High outliers are anything beyond the 3rd quartile + 1.5 * the inter-quartile range (IQR)
Low outliers are anything beneath the 1st quartile - 1.5 * IQR
Grubbs's test
You can check how it works for your expectations here
Apart from these 2, the is a comparison calculator here . You can visit this to use other Algorithms per your need.
I would have tried to get a sliding window coupled with an hysteresis / band filter in order to detect the high value peaks, first.
Then, when your sliding windows advance, you can add the previous first value (which is now the last of analyzed values) to the global sum, and add 1 to the number of total values.
When you encounter a peak (=something that causes the hysteresis to move or overflow the band filter), you either remove the values (may be costly), or better, you set the value to NaN so you can safely ignore it.
You should keep computing a sliding average within your sliding window in order to be able to auto-correct the hysteresis/band filter, so it will reject only the start values of a peak (the end values are the start values of the next one), but once values are stabilized to a new level, values will be kept again.
The size of the sliding window will set how much consecutive "stable" values are needed to be kept, or in other words how much UNstable values are rejected when you reach a new level.
For that, you can check the mode of the values (rounded) and then take all the numbers in a certain range around the mode. That range can be taken from the data itself, for example by taking the 10% of the max - min value. That helps you to filter your data. You can select the percent that fits your needs. Something like this:
let expectedData = [35.267,9.267,9.332,9.186,9.220,9.141,9.107,9.114,9.098,9.181,9.220,4.012,0.132];
expectedData.sort((a, b) => a - b);
/// Get the range of the data
const RANGE = expectedData[ expectedData.length - 1 ] - expectedData[0];
const WINDOW = 0.1; /// Window of selection 10% from left and right
/// Frequency of each number
let dist = expectedData.reduce((acc, e) => (acc[ Math.floor(e) ] = (acc[ Math.floor(e) ] || 0) + 1, acc), {});
let mode = +Object.entries(dist).sort((a, b) => b[1] - a[1])[0][0];
let newData = expectedData.filter(e => mode - RANGE * WINDOW <= e && e <= mode + RANGE * WINDOW);
console.log(newData);

Algorithm to obtain the item(s) at a precise number [closed]

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I find that algorithm/functionality in two games already, but I always wanted to know what was the logic behind it.
Basically, there is a list of items and each of them has an id.
For example:
item_1 has id: 1
item_2 has id: 2
item_3 has id: 4
item_4 has id: 8
item_5 has id: 16
etc.
The id is multiplied by two every new item.
There is then a number, let's say 4, that indicate what the current item is. Is this case that would be item_3, but the tricky part is that number could also select multiple items at once like 7 which is 4 + 2 + 1 (item_3, item_2, item_1) or 17 which is 16 + 1 (item_5, item_1). It can go really high like 16384 if you have a long list and still be perfectly accurate for the multiple selections.
How do I solve this problem?
The algorithm you described is basically outputting where the 1's are in the binary representation of the number.
For 7, its binary representation is 111. There are three 1's: in the first, second, and third position from the left respectively, so it's item 1, 2 and 3. Note that we are counting from the left.
Another example:
For 10, its binary representation is 1010. There are two 1's: in the second and fourth position from the left, so the output would be items 2 and 4.
Here is an implementation in C#.
public static List<int> FindOnes(int number) {
var list = new List<int>();
var binaryString = Convert.ToString(number, 2);
for (int i = 0 ; i < binaryString.Length ; i++) {
if (binaryString[binaryString.Length - i - 1] == '1') {
list.Add(i + 1);
}
}
return list;
}
// usage:
FindOnes(7) // [1,2,3]
No idea how the games you're talking about implement it, but if this was me I would do it using bits in the binary expression of the number (example code in java).
public boolean isItemSelected(final int number, final int itemId) {
return (number & (1 << (itemId - 1))) != 0;
}
The trick here being that the binary representation of a number (from right to left) already denotes whether 1, 2, 4, 8, 16, etc. is required additively to make the number using only powers of two. The left shift simply makes a number which (in binary) is all 0's except a 1 in the 'itemId - 1'th slot. The & will match if that bit is 1 in the given number. And then checking that the result is not 0 simply turns it into a boolean.
Obviously you can combine this with some looping or anything else if you want to build the array/List of all the 'itemIds' which match.
In Javascript, you could take the number, convert it to a binary value, take the bits, reverse it and take the values (index plus one) or zero for a filtering of truthy values.
var value = 13,
items = [...value.toString(2)].reverse().map((v, i) => +v && (i + 1)).filter(Boolean);
console.log(items);

compare values in an array to a number in javascript [closed]

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I am writing a program that determines if a number is between two values in an array.
Here is an example of the array I am using.
var attackArray = new Array (2);
attackArray[0] = new Array("0","1","2","2","2","3","4");
attackArray[1] = new Array("2","3","2","3","2","3","4");
I am using the following code to compare the number against the first two values in the array. I then loop through the array until I find a line that meets the requirements. The number must be >= to the first number and <= the second number.
Here is the code that I am using.
function leveltest ( number)
{
var attack = attackArray.length;
for ( var count = 0 ; count < attack; count ++)
{
if ((number >= Number(attackArray [count][0])) && (number <= Number(attackArray [count][1])))
{
do something ;
}
}
}
If someone can look at my code and explain what I am doing wrong.
I believe you are trying to compare a number to each range of numbers defined by the item values with the same index in element 0 and element 1 of attackArray. If that is right, then the following applies.
The problems present in your code snippet were:
You have the index wrong on line 3. Your third line, attackArray[2] = new Array("1","3","2","3","2","3","4"); is creating a new third element in the attackArray created on the first line. Instead, I think you are wanting to populate the second element of attackArray which should be attackArray[1] = new Array("1","3","2","3","2","3","4"); Or you could use different array syntax as shown below.
In the function, you were using the length of attackArray var attack = attackArray.length;, to control the for loop following. Instead, you will want, var attack = attackArray[0].length; so long as attackArray[0] and attackArray[1] are the same length. You can think of it like this, you were getting your length along the wrong dimension of your array. You were getting the length "down" your array or list of objects, ran than "across" the horizontal dim of your array.
In the function, you are confused on how to loop through the array, and you have this attackArray [count][0] and attackArray [count][1] backwards. Instead they should be attackArray[0][count] and attackArray[1][count]. This will allow you to properly compare your number with each item in element 0 and the item of the same index in element 1.
The following code should be a concise, correct working piece of code to accomplish your goal. You can take and plug this in to jsfiddle.net and it should work in Chrome with the Javascript console used to view the results in the log. Here it is:
var attackArray = [];
attackArray[0] = ["0","0","2","2","2","3","4"];
attackArray[1] = ["1","3","2","3","2","3","4"];
function leveltest (number){
var attack = attackArray[0].length;
for (var count = 0;count < attack;count ++){
if ((number >= Number(attackArray [0][count])) &&
(number <= Number(attackArray [1][count]))) {
console.log(number + " matches at index " + count);
}
}
}
leveltest(2);
Looks like your second element in attackArray has the wrong index.
attackArray[2] = new Array("1","3","2","3","2","3","4");
attackArray.length == 2
you "count" can go up to 1, attackArray[1] is not defined by you.
The second comparation inside the if is wrong. At the second loop it will be attackArray[1][1] and you created an attackArray[0] and attackArray[2].

Is there a better way of writing v = (v == 0 ? 1 : 0); [closed]

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I want to toggle a variable between 0 and 1. If it's 0 I want to set it to 1, else if it's 1 I want to set it to 0.
This is such a fundamental operation that I write so often I'd like to investigate the shortest, clearest possible way of doing it. Here's my best so far:
v = (v == 0 ? 1 : 0);
Can you improve on this?
Edit: the question is asking how to write the above statement in the fewest characters while retaining clarity - how is this 'not a real question'? This wasn't intended to be a code-golf exercise, though some interesting answers have come out of people approaching it as golf - it's nice to see golf being used in a constructive and thought-provoking manner.
You can simply use:
v = 1 - v;
This of course assumes that the variable is initialised properly, i.e. that it only has the value 0 or 1.
Another method that is shorter but uses a less common operator:
v ^= 1;
Edit:
To be clear; I never approached this question as code golf, just to find a short way of doing the task without using any obscuring tricks like side effects of operators.
Since 0 is a false value and 1 is a true value.
v = (v ? 0 : 1);
If you are happy to use true and false instead of numbers
v = !v;
or if they must be numbers:
v = +!v; /* Boolean invert v then cast back to a Number */
v = (v + 1) % 2 and if you need to cycle through more values just change 2 for (n + 1). Say you need to cycle 0,1,2 just do v = (v + 1) % 3.
You could write a function for it and use it like:
v = inv(v)
If you don't care about any possibility other than 1:
v = v ? 0 : 1;
In the above case, v will end up being 1 if v is 0, false, undefined or null. Take care using this kind of approach - v will be 0 even if v is "hello world".
Lines like v = 1 - v, or v ^= 1 or v= +!v will all get the job done, but they constitute what I would refer to as hacks. These are not beautiful lines of code, but cheap tricks to have the intended effect. 1 - v does not communicate "toggle the value between 0 and 1". This makes your code less expressive and introduces a place (albeit a small one) where another developer will have to parse your code.
Having instead a function like v = toggle(v) communicates the intent at the quickest glance.
(Honesty and mathematical integrity - given the number of votes on this "answer" - have led me to edit this answer. I held off as long as possible because it was intended as a short quip and not as anything "deep" so putting in any explanation seemed counter to the purpose. However, the comments are making it clear that I should be clear to avoid misunderstanding.)
My original answer:
The wording of this part of the specification:
If it's 0, I want to set it to 1, else set it to 0.
implies that the most accurate solution is:
v = dirac_delta(0,v)
First, the confession: I did get my delta functions confused. The Kronecker delta would have been slightly more appropriate, but not by much as I wanted something that was domain-independent (the Kronecker delta is mainly used just for integers). But I really shouldn't have used delta functions at all, I should have said:
v = characteristic_function({0},v)
Let me clarify. Recall that a function is a triple, (X,Y,f), where X and Y are sets (called the domain and codomain respectively) and f is a rule that assigns an element of Y to each element of X. We often write the triple (X,Y,f) as f: X &rightarrow; Y. Given a subset of X, say A, there is a characteristic function which is a function χA: X &rightarrow; {0,1} (it can also be thought of as a function to a larger codomain such as &Nopf; or &Ropf;). This function is defined by the rule:
χA(x) = 1 if x &in; A and χA(x) = 0 if x ∉ A.
If you like truth tables, it's the truth table for the question "Is the element x of X an element of the subset A?".
So from this definition, it's clear that the characteristic function is what is needed here, with X some big set containing 0 and A = {0}. That's what I should have written.
And so to delta functions. For this, we need to know about integration. Either you already know it, or you don't. If you don't, nothing I can say here will tell you about the intricacies of the theory, but I can give a one sentence summary. A measure on a set X is in essence "that which is needed to make averages work". That is to say that if we have a set X and a measure μ on that set then there is a class of functions X &rightarrow; &Ropf;, called measurable functions for which the expression ∫X f dμ makes sense and is, in some vague sense, the "average" of f over X.
Given a measure on a set, one can define a "measure" for subsets of that set. This is done by assigning to a subset the integral of its characteristic function (assuming that this is a measurable function). This can be infinite, or undefined (the two are subtly different).
There are lots of measures around, but there are two that are important here. One is the standard measure on the real line, &Ropf;. For this measure, then ∫&Ropf; f dμ is pretty much what you get taught in school (is calculus still taught in schools?): sum up little rectangles and take smaller and smaller widths. In this measure, the measure of an interval is its width. The measure of a point is 0.
Another important measure, which works on any set, is called the point measure. It is defined so that the integral of a function is the sum of its values:
∫X f dμ = ∑x &in;X f(x)
This measure assigns to each singleton set the measure 1. This means that a subset has finite measure if and only if it is itself finite. And very few functions have finite integral. If a function has a finite integral, it must be non-zero only on a countable number of points. So the vast majority of functions that you probably know do not have finite integral under this measure.
And now to delta functions. Let's take a very broad definition. We have a measurable space (X,μ) (so that's a set with a measure on it) and an element a &in; X. We "define" the delta function (depending on a) to be the "function" δa: X &rightarrow; &Ropf; with the property that δa(x) = 0 if x ≠ a and ∫X δa dμ = 1.
The most important fact about this to get a-hold of is this: The delta function need not be a function. It is not properly defined: I have not said what δa(a) is.
What you do at this point depends on who you are. The world here divides in to two categories. If you are a mathematician, you say the following:
Okay, so the delta function might not be defined. Let's look at its hypothetical properties and see if we can find a proper home for it where it is defined. We can do that, and we end up with distributions. These are not (necessarily) functions, but are things that behave a little like functions, and often we can work with them as if they were functions; but there are certain things that they don't have (such as "values") so we need to be careful.
If you are not a mathematician, you say the following:
Okay, so the delta function might not be properly defined. Who says so? A bunch of mathematicians? Ignore them! What do they know?
Having now offended my audience, I shall continue.
The dirac delta is usually taken to be the delta function of a point (often 0) in the real line with its standard measure. So those who are complaining in the comments about me not knowing my deltas are doing so because they are using this definition. To them, I apologise: although I can wriggle out of that by using the Mathematician's defence (as popularised by Humpty Dumpty: simply redefine everything so that it is correct), it is bad form to use a standard term to mean something different.
But there is a delta function which does do what I want it to do and it is that which I need here. If I take a point measure on a set X then there is a genuine function δa : X &rightarrow; &Ropf; which satisfies the criteria for a delta function. This is because we are looking for a function X &rightarrow; &Ropf; which is zero except at a and such that the sum of all of its values is 1. Such a function is simple: the only missing piece of information is its value at a, and to get the sum to be 1 we just assign it the value 1. This is none other than the characteristic function on {a}. Then:
∫X δa dμ = ∑x &in; X δa(x) = δa(a) = 1.
So in this case, for a singleton set, the characteristic function and the delta function agree.
In conclusion, there are three families of "functions" here:
The characteristic functions of singleton sets,
The delta functions,
The Kronecker delta functions.
The second of these is the most general as any of the others is an example of it when using the point measure. But the first and third have the advantage that they are always genuine functions. The third is actually a special case of the first, for a particular family of domains (integers, or some subset thereof).
So, finally, when I originally wrote the answer I wasn't thinking properly (I wouldn't go so far as to say that I was confused, as I hope I've just demonstrated I do know what I'm talking about when I actually think first, I just didn't think very much). The usual meaning of the dirac delta is not what is wanted here, but one of the points of my answer was that the input domain was not defined so the Kronecker delta would also not have been right. Thus the best mathematical answer (which I was aiming for) would have been the characteristic function.
I hope that that is all clear; and I also hope that I never have to write a mathematical piece again using HTML entities instead of TeX macros!
in general whenever you need to toggle between two values , you can just subtract the current value from the sum of the two toggle values :
0,1 -> v = 1 - v
1,2 -> v = 3 - v
4,5 -> v = 9 - v
You could do
v = Math.abs(--v);
The decrement sets the value to 0 or -1, and then the Math.abs converts -1 to +1.
If it must be the integer 1 or 0, then the way you're doing it is fine, though parentheses aren't needed. If these a are to be used as booleans, then you can just do:
v = !v;
v = v == 0 ? 1 : 0;
Is enough !
List of solutions
There are three solutions I would like to propose. All of them convert any value to 0 (if 1, true etc.) or 1 (if 0, false, null etc.):
v = 1*!v
v = +!v
v = ~~!v
and one additional, already mentioned, but clever and fast (although works only for 0s and 1s):
v = 1-v
Solution 1
You can use the following solution:
v = 1*!v
This will first convert the integer to the opposite boolean (0 to True and any other value to False), then will treat it as integer when multiplying by 1. As a result 0 will be converted to 1 and any other value to 0.
As a proof see this jsfiddle and provide any values you wish to test: jsfiddle.net/rH3g5/
The results are as follows:
-123 will convert to integer 0,
-10 will convert to integer 0,
-1 will convert to integer 0,
0 will convert to integer 1,
1 will convert to integer 0,
2 will convert to integer 0,
60 will convert to integer 0,
Solution 2
As mblase75 noted, jAndy had some other solution that works as mine:
v = +!v
It also first makes boolean from the original value, but uses + instead of 1* to convert it into integer. The result is exactly the same, but the notation is shorter.
Solution 3
The another approach is to use ~~ operator:
v = ~~!v
It is pretty uncommon and always converts to integer from boolean.
To sum up another answer, a comment and my own opinion, I suggest combining two things:
Use a function for the toggle
Inside this function use a more readable implementation
Here is the function which you could place in a library or maybe wrap it in a Plugin for another Javascript Framework.
function inv(i) {
if (i == 0) {
return 1
} else {
return 0;
}
}
And the usage is simply:
v = inv(v);
The advantages are:
No code Duplication
If you or anybody read this again in the future, you will understand your code in a minimum of time.
This is missing:
v = [1, 0][v];
It works as round robin as well:
v = [2, 0, 1][v]; // 0 2 1 0 ...
v = [1, 2, 0][v]; // 0 1 2 0 ...
v = [1, 2, 3, 4, 5, 0][v]; // 0 1 2 3 4 5 ...
v = [5, 0, 1, 2, 3, 4][v]; // 0 5 4 3 2 1 0 ...
Or
v = {0: 1, 1: 0}[v];
The charme of the last solution, it works with all other values as well.
v = {777: 'seven', 'seven': 777}[v];
For a very special case, like to get a (changing) value and undefined, this pattern may be helpful:
v = { undefined: someValue }[v]; // undefined someValue undefined someValue undefined ...
I don't know why you want to build your own booleans? I like the funky syntaxes, but why not write understandable code?
This is not the shortest/fastest, but the most clearest (and readable for everyone) is using the well-known if/else state:
if (v === 0)
{
v = 1;
}
else
{
v = 0;
}
If you want to be really clear, you should use booleans instead of numbers for this. They are fast enough for most cases. With booleans, you could just use this syntax, which will win in shortness:
v = !v;
Another form of your original solution:
v = Number(v == 0);
EDIT: Thanks TehShrike and Guffa for pointing out the error in my original solution.
I would make it more explicit.
What does v mean?
For example when v is some state. Create an object Status. In DDD an value object.
Implement the logic in this value object. Then you can write your code in a more functional way which is more readable. Switching status can be done by creating a new Status based on the current status. Your if statement / logic is then encapsulated in your object, which you can unittest. An valueObject is always immutable, so it has no identity. So for changing it's value you have to create a new one.
Example:
public class Status
{
private readonly int _actualValue;
public Status(int value)
{
_actualValue = value;
}
public Status(Status status)
{
_actualValue = status._actualValue == 0 ? 1 : 0;
}
//some equals method to compare two Status objects
}
var status = new Status(0);
Status = new Status(status);
Since this is JavaScript, we can use the unary + to convert to int:
v = +!v;
This will logical NOT the value of v (giving true if v == 0 or false if v == 1). Then we convert the returned boolean value into its corresponding integer representation.
Another way to do it:
v = ~(v|-v) >>> 31;
One more:
v=++v%2
(in C it would be simple ++v%=2)
ps. Yeah, I know it's double assignment, but this is just raw rewrite of C's method (which doesn't work as is, cause JS pre-increment operator dosen't return lvalue.
If you're guaranteed your input is either a 1 or a 0, then you could use:
v = 2+~v;
Just for kicks: v = Math.pow(v-1,v) also toggles between 1 and 0.
define an array{1,0}, set v to v[v], therefore v with a value of 0 becomes 1, and vica versa.
Another creative way of doing it, with v being equal to any value, will always return 0 or 1
v = !!v^1;
If possible values for v are only 0 and 1, then for any integer x, the expression:
v = Math.pow((Math.pow(x, v) - x), v);
will toggle the value.
I know this is an ugly solution and the OP was not looking for this...but I was thinking about just another solution when I was in the loo :P
Untested, but if you're after a boolean I think var v = !v will work.
Reference: http://www.jackfranklin.co.uk/blog/2011/05/a-better-way-to-reverse-variables
v=!v;
will work for v=0 and v=1; and toggle the state;
If there are just two values, as in this case(0, 1), i believe it's wasteful to use int. Rather go for boolean and work in bits. I know I'm assuming but in case of toggle between two states boolean seems to be ideal choice.
v = Number(!v)
It will type cast the Inverted Boolean value to Number, which is the desired output.
Well, As we know that in javascript only that Boolean comparison will also give you expected result.
i.e. v = v == 0 is enough for that.
Below is the code for that:
var v = 0;
alert("if v is 0 output: " + (v == 0));
setTimeout(function() {
v = 1;
alert("if v is 1 Output: " + (v == 0));
}, 1000);
JSFiddle: https://jsfiddle.net/vikash2402/83zf2zz0/
Hoping this will help you :)

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