Im looping over a collection of coordinate values and doing math on the coordinates to see if the calculated values are in a hashmap. if they are in the hash map then I want to run an additional function. since I had multiple cases I wanted to check for each coord in the collection, I figured a switch statement would be cool to use to replace my if statements so all my checks could be visually and logically grouped. When I replaced my if statements with a switch, my code returned bad results. When I debugged, I realized the switch statements would sometimes execute even when the case was false(I added console.logs to output the result of the same switch condition and it would print false, but should only run when true). Here is a small example:
var idm = {0:1, 3:1, 9:1, 10:1, 11:1, 12:1, 20:1, 21:1, 23:1}
var findNeighbors = function(b) {
var u,d,l,r,lRow,rRow;
var currentBuilding = parseInt(b);
var currRow = Math.floor(currentBuilding/column);
//remove value from map so we dont recount it.
delete idm[currentBuilding];
u = currentBuilding - column;
d = currentBuilding + column;
l = currentBuilding - 1;
lRow = Math.floor(l/column);
r = currentBuilding + 1;
rRow = Math.floor(r/column);
console.log("current idx:" + currentBuilding);
console.log("u:" + u + ", d:" + d + ", l:" + l + " r:" + r);
// debugger;
switch(true) {
case (idm.hasOwnProperty(u) === true):
console.log((idm.hasOwnProperty(u)));
console.log("map has " + currentBuilding + " -> u: " + u);
findNeighbors(u);
case (idm.hasOwnProperty(d) === true):
console.log((idm.hasOwnProperty(d)));
console.log("map has " + currentBuilding + " -> d: " + d);
findNeighbors(d);
case (lRow === currRow && idm.hasOwnProperty(l) === true):
console.log((lRow === currRow && idm.hasOwnProperty(l)));
console.log("map has " + currentBuilding + " -> l: " + l);
findNeighbors(l);
case (rRow === currRow && idm.hasOwnProperty(r) === true):
console.log((rRow === currRow && idm.hasOwnProperty(r)))
console.log("map has " + currentBuilding + " -> r: " + u);
findNeighbors(r);
}
console.log("---------------------------");
}
I figured a switch statement would be cool to use to replace my if statements so all my checks could be visually and logically grouped.
Well, write code that works not code that looks cool. You were forgetting break statements, so the execution flow fell through - without evaluating the other case expressions after the first one matched. Btw switching on a constant is a horrible (uncool) practice.
Use standard if/else instead.
Related
I gave an example of using .tofixed() with math, functions, and arrays, to a beginner coder friend who has been reviewing these topics in his class.
const bananaX = 9;
const bananaY = 2.9768;
bananaArray = [bananaX , bananaY];
console.log("X before array = " + bananaX);
console.log("Y before array = " + bananaY + '\n')
console.log("X,Y after array = " + bananaArray + '\n')
console.log("Value of X in array: " + bananaArray[0]+ '\n')
console.log("Value of Y in array: " + bananaArray[1]+ '\n')
function bananaDivision (bananaArray){
console.log("Value of X after function = " + bananaX);
console.log("Value of Y after function = " + bananaY + '\n')
let bananaDivided = Math.abs(bananaX/bananaY );
console.log (`X divided by Y = + ${bananaDivided}` + '\n')
let bananaFixed = bananaDivided.toFixed(2);
console.log("After using .toFixed(2) : " + bananaFixed + '\n');
};
bananaDivision();
They were understanding and following along no problem.
Then they asked me - "What if we put a decimal in the .toFixed ?"
So I ran:
const bananaX = 9;
const bananaY = 2.9768;
bananaArray = [bananaX , bananaY];
console.log("X before array = " + bananaX);
console.log("Y before array = " + bananaY + '\n')
console.log("X,Y after array = " + bananaArray + '\n')
console.log("Value of X in array: " + bananaArray[0]+ '\n')
console.log("Value of Y in array: " + bananaArray[1]+ '\n')
function bananaDivision (bananaArray){
console.log("Value of X after function = " + bananaX);
console.log("Value of Y after function = " + bananaY + '\n')
let bananaDivided = Math.abs(bananaX/bananaY );
console.log (`X divided by Y = + ${bananaDivided}` + '\n')
let bananaFixed = bananaDivided.toFixed(2);
let bananaFixed1 = bananaDivided.toFixed(.69420);
let bananaFixed2 = bananaDivided.toFixed(1.69420);
console.log("After using .toFixed(2) : " + bananaFixed + '\n');
console.log("After using .toFixed(.69420) : " + bananaFixed1 + '\n');
console.log("After using .toFixed(1.69420) : " + bananaFixed2 + '\n');
};
bananaDivision();
I explained it as that .toFixed is looking at the first number within the () and that the decimals are ignored.
Am I correct? For my own curiousity, is there a crazy way to break .toFixed() so that it actually uses decimals? I'm experimenting atm but wanted to know if someone already figured that out.
I explained it as that .toFixed is looking at the first number within the () and that the decimals are ignored.
This would be correct. That is essentially what happens.
For full correctness, the input of toFixed() will be converted to an integer. The specification states that the argument must first be converted to a number - NaN will be converted to a zero. Numbers with a fractional part will be rounded down.
Which means that if you pass any number, you essentially get the integer part of it.
It also means that non-numbers can be used:
const n = 3;
console.log(n.toFixed("1e1")); // 1e1 scientific notation for 10
You're close, since toFixed() expects an integer it will handle converting decimal numbers before doing anything else. It uses toIntegerOrInfinity() to do that, which itself uses floor() so the number is always rounded down.
Most of Javascript handles type conversion implicitly, so it's something you should really understand well if you don't want to run into problems. There's a free book series that explains that concept and a lot of other important Javascript knowledge very well, it's called You Don't Know JS Yet.
just a demo how .tofixed works !!!!!!
function roundFloat(x, digits) {
const arr = x.toString().split(".")
if (arr.length < 2) {
return x
}else if(arr[1] === ""){
return arr[0]
}else if(digits < 1){
return arr[0]
}
const st = parseInt(x.toString().split(".")[1]);
let add = false;
const rudgt = digits
const fX = parseInt(st.toString().split("")[rudgt]);
fX > 5 ? add = true : add = false
nFloat = parseInt(st.toString().split("").slice(0, rudgt).join(""))
if (add) {
nFloat += 1
}
const repeat0 = (() => {
if (rudgt - st.toString().length < 0) {
return 0
}
return rudgt - st.toString().length
})()
const output = x.toString().split(".")[0] + "." + nFloat.toString() + "0".repeat(repeat0);
return output
}
console.log(roundFloat(1.200, 2))
I am currently trying to complete an assignment for an intro2Javascript course. The question basically asks me to return a string of multiples of 2 parameters (num, numMultiple). Each time it increments the value i until i = numMultiple. For example:
5 x 1 = 5\n
5 x 2 = 10\n
5 x 3 = 15\n
5 x 4 = 20\n
This was my attempt:
function showMultiples(num, numMultiples) {
var result;
for (i = 1; i <= numMultiples; i++) {
result = num * i
multiples = "" + num + " x " + i + " = " + result + "\n"
return (multiples)
}
}
...and because the assignment comes with pre-written console logs:
console.log('showMultiples(2,8) returns: ' + showMultiples(2, 8));
console.log('showMultiples(3,2) returns: ' + showMultiples(3, 2));
console.log('showMultiples(5,4) returns: ' + showMultiples(5, 4));
console.log('\n');
This is my output:
showMultiples(2,8) returns: 2 x 1 = 2
Scratchpad/1:59:1
showMultiples(3,2) returns: 3 x 1 = 3
Scratchpad/1:60:1
showMultiples(5,4) returns: 5 x 1 = 5
UPDATE
You were doing two things incorrectly:
1) You were returning after the first iteration through your loop
2) You were assigning to multiples instead of appending to it.
Since you want to gather all the values and then show the final result first, I add all of the values to an array, and then use unshift() to add the final element (the result) to the beginning of the array. Then I use join() to return a string representation of the desired array.
function showMultiples(num, numMultiples) {
var result;
var multiples = [];
for (let i = 1; i <= numMultiples; i++) {
result = num * i
multiples.push("" + num + " x " + i + " = " + result + "\n")
}
multiples.unshift(multiples[multiples.length-1]);
return (multiples.join(''))
}
console.log('showMultiples(2,8) returns: ' + showMultiples(2, 8));
console.log('showMultiples(3,2) returns: ' + showMultiples(3, 2));
console.log('showMultiples(5,4) returns: ' + showMultiples(5, 4));
console.log('\n');
You need to declare all variables, because without you get global variables (beside that it does not work in 'strict mode').
The second point is to use multiples with an empty string for collecting all intermediate results and return that value at the end of the function.
For keeping the last result, you could use another variable and append that value at the end for return.
function showMultiples(num, numMultiples) {
var i,
result,
multiples = "",
temp = '';
for (i = 1; i <= numMultiples; i++) {
result = num * i;
temp = num + " x " + i + " = " + result + "\n";
multiples += temp;
}
return temp + multiples;
}
console.log('showMultiples(2,8) returns: ' + showMultiples(2, 8));
console.log('showMultiples(3,2) returns: ' + showMultiples(3, 2));
console.log('showMultiples(5,4) returns: ' + showMultiples(5, 4));
As other answers say, your problem is in multiple.
You are clearing multiple every iteration and storing the new value, but you do not want that, you want to add the new result, and to do so you use this code:
multiples = multiple + "" + num + " x " + i + " = " + result + "\n"
which can be compressed in what the rest of the people answered:
multiples += "" + num + " x " + i + " = " + result + "\n"
Probably you already know, but to ensure:
a += b ---> a = a + b
a -= b ---> a = a - b
a *= b ---> a = a * b
and there are even more.
I am not good at algorithm analysis. The source code is from this place: https://repl.it/KREy/4
Instead of dynamic programming, this piece of code uses a cache to optimize the BigO by sacrificing memory. However, I just don't know how to calculate the BigO mathematically after this cache mechanism is added. May anyone genius give an explanation?
To ease reading I will copy and paste them in the following space:
// using cache to optimize the solution 1 from http://www.techiedelight.com/longest-palindromic-subsequence-using-dynamic-programming/
const cache = {};
var runningtime = 0;
var runningtimeWithCache = 0;
function computeGetLP(x, start, end){
const answer = a => {
runningtime++;
return a;
}
console.log("try to compute: " + x + " " + start + " " + end + " ");
if(start > end)
return answer(0);
if(start == end)
return answer(1);
if(x[start] == x[end])
return answer(2 + computeGetLP(x, start+1, end-1));
return answer(Math.max(computeGetLP(x, start+1, end),
computeGetLP(x, start, end-1)));
}
function computeGetLPWithCache(x, start, end){
const answer = a => {
runningtimeWithCache ++;
console.log("do cache: " + x + " " + start + " " + end + " is " + a);
cache["" + x + start + end] = a;
return a;
}
console.log("try to compute: " + x + " " + start + " " + end + " ");
if(cache["" + x + start + end]){
console.log("hit cache " + x + " " + start + " " + end + " "+ ": ",cache["" + x + start + end]);
return cache["" + x + start + end];
}
if(start > end)
return answer(0);
if(start == end)
return answer(1);
if(x[start] == x[end])
return answer(2 + computeGetLPWithCache(x, start+1, end-1));
return answer(Math.max(computeGetLPWithCache(x, start+1, end),
computeGetLPWithCache(x, start, end-1)));
}
const findLongestPadlindrom1 = s => computeGetLPWithCache(s, 0, s.length-1)
const findLongestPadlindrom2 = s => computeGetLP(s, 0, s.length-1)
const log = (function(){
var lg = [];
var output = function(text){
lg.push(text);
}
output.getRecord = function(){
return lg;
}
return output;
})();
log("Now let's do it with cache")
log("result: "+findLongestPadlindrom1("ABBDCACB"))
log("running time is: " + runningtimeWithCache)
log("Now let's do it without cache")
log("result: "+findLongestPadlindrom2("ABBDCACB"))
log("running time is: " + runningtime)
log.getRecord();
I'm not an expert in algorithms either, but I remember cache techniques like this from Introduction to Algorithms, chapter 15, just beside Dynamic Programming. It has the same big O to DP, which is O(n^2) in your case.
Each call to computeGetLPWithCache() costs O(1) for it does not contain loops. Consider the worst case where x[start] != x[end] in each recursion. How many times are we going to call computeGetLPWithCache()?
Let n = length(x), [start, end] represent a call to computeGetLPWithCache(x, start, end), and F(n) equals the number of calls. In computeGetLPWithCache(x, 0, n), 2 sub calls - [0, n-1] and [1, n] - are issued. The former costs F(n), and when we're doing the latter, we discover that in each iteration, the first call's [start, end] range is a true subset of [0, n-1] whose result is already written to cache during the [0, n-1] call, thus no need for recursing. Only the second call which has the element n in it has to be calculated; there're n such calls [1,n][2,n][3,n]...[n,n] (one in each stack layer), so F(n+1) = F(n) + O(n).
F(n) = F(n-1) + O(n-1) = F(n-2) + O(n-2) + O(n-1) = ... = O(1+2+...+(n-1)) = O(n^2).
Hope I've got the meaning through. Replies are welcome.
im just started to learn JavaScript and i´m trying to simplify some code, but couldn´t get a working solution. The working part is this:
switch (v) {
case 0:
if (localStorage.FGAz0) {
localStorage.FGAz0 = Number(localStorage.FGAz0)+1;
} else {
localStorage.FGAz0 = 1;
}
document.getElementById("Ergebnis").innerHTML = "You have " + localStorage.FGAz0+ " Visitor(s).";
break;
case 1:
if (localStorage.FGAz1) {
localStorage.FGAz1 = Number(localStorage.FGAz1)+1;
} else {
localStorage.FGAz1 = 1;
}
document.getElementById("Ergebnis").innerHTML = "You have " + localStorage.FGAz1+ " Visitor(s).";
break;
case 2:
if (localStorage.FGAz2) {
localStorage.FGAz2 = Number(localStorage.FGAz2)+1;
} else {
localStorage.FGAz2 = 1;
}
document.getElementById("Ergebnis").innerHTML = "You have " + localStorage.FGAz2+ " Visitor(s).";
break;
case 3:
if (localStorage.FGAz3) {
localStorage.FGAz3 = Number(localStorage.FGAz3)+1;
} else {
localStorage.FGAz3 = 1;
}
document.getElementById("Ergebnis").innerHTML = "You have " + localStorage.FGAz3+ " Visitor(s).";
break;
case 4:
if (localStorage.FGAz4) {
localStorage.FGAz4 = Number(localStorage.FGAz4)+1;
} else {
localStorage.FGAz4 = 1;
}
document.getElementById("Ergebnis").innerHTML = "You have " + localStorage.FGAz4+ " Visitor(s).";
break;
case 5:
if (localStorage.FGAz5) {
localStorage.FGAz5 = Number(localStorage.FGAz5)+1;
} else {
localStorage.FGAz5 = 1;
}
document.getElementById("Ergebnis").innerHTML = "You have " + localStorage.FGAz5+ " Visitor(s).";
break;
default:
if (localStorage.FahrGastAnzahl) {
localStorage.FahrGastAnzahl = Number(localStorage.FahrGastAnzahl)+1;
} else {
localStorage.FahrGastAnzahl= 1;
}
document.getElementById("Ergebnis").innerHTML = "You have " + localStorage.FahrGastAnzahl+ " Visitor(s).";}
} else {
document.getElementById("Ergebnis").innerHTML = "Sorry, dein Browser unterstützt die Speicherung von lokalen Daten nicht...";
}
and i am trying to short it to the var depending on "v" which only had numbers. At the moment i have this:
if (localStorage.FGAz + "v") {
(localStorage.FGAz + "v") = Number(localStorage.FGAz + "v")+1;
} else {
(localStorage.FGAz + "v") = 1;
document.getElementById("Ergebnis").innerHTML = "You have " + (localStorage.FGAz + "v") + " Visitor(s).";}
Something isn´t right with the adding of the variable "v", but i don´t know what and didn´t found a solution on searching. Hope someone can help me. Please no jquery, i haven´t learned that yet.
Firs of all, make sure you understand the difference between v and "v":
v is a variable name, which can hold any value (string, number, etc), eg:v = 1; v = "1"; v = "xxx";
"v" (mind the brackets) is a string value itself (v = "v", where v is the variable name and "v" is the variable value). Everything inside the brackets (" or ') is a string.
If you wish to cast numerical (integer in this example) value to string, you can use v.toString() or simply append an empty string to the value v + "".
Secondly, please get some knowledge on Property Accessors.
In short: you can access properties of an object in two different ways: using dot notation (localStorage.FGAz1) or brackets notation (localStorage["FGAz1"]).
Appending string value to dot notated property accessor (document.getElementBy + "Id") will firstly evaluate the value of property (document.getElementBy - which evaluates to undefined) and THEN will concatenate it with the string (undefined + "Id" - which results in "undefinedId"). On the other hand, appending string to accessor value inside bracket notation (document["getElementBy" + "Id"]) will firstly evaluate the value of accessor ("getElementBy" + "Id" - which evaluates to "getElementById"), and then access the object's property (document["getElementById"] - which returns the function).
Based on your example:
localStorage.FGAz is not defined, so the localStorage.FGAz + "v" evaluates to undefined + "v" which results in "undefinedv" (Notice the 'v' added to the end of 'undefined'). Sentence if ("undefinedv") always evaluates to true (HERE you can find out why).
In conclusion:
Use brackets notation:
if (localStorage['FGAz' + v.toString()]) {. This will evaluate to if (localStorage['FGAz1']) { for v = 1, if (localStorage['FGAzXXX']) { for v = XXX and so on.
I'm hoping it makes sens to you, if not leave me a question in comments.
I'm looking at the code above and it looks like you're doing some sort of visitor counter. However the localstorage would only work for the one person.
If you're going to be doing vistor counter, you could use ajax to send the information to a database and then check how many people are online.
Simple visit counter:
JSFIDDLE LINK
Code from JSFIDDLE
<div id="visitCount"></div>
if (localStorage.visitCount){
localStorage.visitCount = Number(localStorage.visitCount)+1;
} else {
localStorage.visitCount = 1;
}
document.getElementById("visitCount").innerHTML = "You've been to this page "+ localStorage.visitCount + " time(s)."
Instead of using
localStorage.FGAz + "v"
Use
localStorage["FGAz" + v]
v in "" quotations makes it a string value not variable.
I'm trying to take a character and increment or decrement it. I'm attempting to use String.fromCharCode to accomplish this. While the following works in a console
'Na' + String.fromCharCode(78 + n) // n is 2 in this example, can even hard-code it
seems to work properly and gives me NaP, something else in my code is giving me Na̎ instead.
Here's the code block that is being executed
if (ifTypes(a, b, 'integer', 'NaN')) { // disregard this, inside code IS executing
console.log("a: " + JSON.stringify(a) + " b: " + JSON.stringify(b));
var n = a[1] === 'NaN' ? b[0] : a[0];
var output = 'Na' + String.fromCharCode(78 + n);
console.log("output: " + output);
return output;
}
From the console for verification:
a: [null,"NaN"] b: ["2","integer"]
output: Na̎ // <-- SO's code highlighter is messing that up
And yes, if anyone recognizes what I'm doing, I am implementing the interpreter from xkcd's 1537. If you want to see this code in action and maybe try to read through my console output, I've got it online here. Just click in the lighter bar and press enter, it's just simulating a terminal.
I suspect the problem is some strange ascii/unicode toggle. I've tried putting the 'Na' inside the String.fromCharCode but it gives the similar results. Though a should be [NaN, "NaN"], I don't think that's the issue. I do need to track down that bug too.
I think the problem is that you're adding "2" and 78, instead of 2 and 78.
function print(s) {
document.querySelector("#result").innerHTML += "<pre>" + s + "</pre>";
}
var a = [null, "NaN"];
var b = ["2", "integer"];
var n = a[1] === 'NaN' ? b[0] : a[0];
var output = 'Na' + String.fromCharCode(78 + n);
print("output 1: " + output);
output = 'Na' + String.fromCharCode(78 + (typeof n == "number" ? n : parseInt(n, 10)));
print("output 2: " + output);
<div id="result"></div>
n is a string, not an integer, so the addition isn't doing what you expect.
Change
var output = 'Na' + String.fromCharCode(78 + n);
to
var output = 'Na' + String.fromCharCode(78 + parseInt(n));