above is result of below snippet of code
var total_points = 0
for(var i = 0; i < req.body.requisites.length; i++){
total_points = total_points + req.body.requisites[i].points
console.log(req.body.requisites[i].points , total_points)
}
console.log('total points :' + total_points)
req.body.points = total_points
I am not getting why one time it is concatenating the results (see the last values before 'total points') and next time it calculates correctly.
Appreciated if you can help.
Thanks in advance!
As per my earlier comment, it seems like some of your input must be a string instead of number and because of Javascript's coercion rules when adding a string and a number you are getting string concatenation instead of math addition.
You can force all the input to numbers so you always get addition like this:
var total_points = 0
for (var i = 0; i < req.body.requisites.length; i++) {
total_points = total_points + (+req.body.requisites[i].points);
console.log(req.body.requisites[i].points , total_points)
}
console.log('total points :' + total_points)
req.body.points = total_points
And, it might be easier to use .reduce():
req.body.points = req.body.requisites.reduce((total, val) => total + (+val)), 0);
The (+req.body.requisites[i].points) or (+val) converts it to a number if it was a numeric string.
You code seems to be okay.
This might be happening because you are sending the value for one of req.body.requisites[points] as string and that is why it gets concatenated instead on addition.
Check your input or update the question with the input you are passing ie, req.body
Hope this helps you know the reason behind the mess!
Related
I'm currently making a program with many functions that utilise Math.rand(). I'm trying to generate a string with a given keyword (in this case, lathe). I want the program to log a string that has "lathe" (or any version of it, with capitals or not), but everything I've tried has the program hit its call stack size limit (I understand exactly why, I want the program to generate a string with the word without it hitting its call stack size).
What I have tried:
function generateStringWithKeyword(randNum: number) {
const chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789+/";
let result = "";
for(let i = 0; i < randNum; i++) {
result += chars[Math.floor(Math.random() * chars.length)];
if(result.includes("lathe")) {
continue;
} else {
generateStringWithKeyword(randNum);
}
}
console.log(result);
}
This is what I have now, after doing brief research on stackoverflow I learned that it might have been better to add the if/else block with a continue, rather than using
if(!result.includes("lathe")) return generateStringWithKeyword(randNum);
But both ways I had hit the call stack size limit.
A "correct" version of your algorithm, written as an iterative function instead of as a recursive one so as not to exceed stack depth, would look something like this:
function generateStringWithKeyword(randNum: number) {
const chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789+/";
let result = "";
let attemptCnt = 0;
while (!result.toLowerCase().includes("lathe")) {
attemptCnt++;
result = "";
for (let i = 0; i < randNum; i++) {
result += chars[Math.floor(Math.random() * chars.length)];
}
if (attemptCnt > 1e6) {
console.log("I GIVE UP");
return;
}
}
console.log(result);
return result;
}
I don't like when my browser hangs because of a script that won't finish, so I put a maximum attempt count in there. A million chances seems reasonable. When you try it out, this happens:
generateStringWithKeyword(10); // I GIVE UP
Which makes sense; let's perform a rough back-of-the-envelope probability calculation to see how long we might expect this to take. The chance that "lathe" will appear in some case at position 1 of the word is (2/64)×(2/64)×(2×64)×(2/64)×(2/64) ("L" or "l" appears first, followed by "A" or "a", etc) which is approximately 3×10-8. For a word of length 10, "lathe" can appear starting at positions 1, 2, 3, 4, 5, or 6. While this isn't exactly correct, let's think of this as multiplying your chances by 6 of getting the word somewhere, so the actual chance of getting a valid result is somewhere around 1.8×10-7. So we can expect that you'd need to make approximately 1 ÷ 1.8×10-7 = 5.6 million chances to succeed.
Oh, darn, I only gave it a million. Let's up that to 10 million and try again:
generateStringWithKeyword(10); // "lATHELEYSc"
Great! Although, it does sometimes still give up. And really, an algorithm which needs millions of tries before it succeeds is very, very inefficient. You might want to read about bogosort, a sorting algorithm which works by randomly shuffling things and checking to see if they are sorted, and it keeps trying until it works. It's used for educational purposes to highlight how such techniques don't really perform well enough to be practical. Nobody would ever want to use such an algorithm for real.
So how would you do this "the right" way? Well, my suggestion here is to just build your result correctly the first time. If you have 10 characters and 5 of them need to be "lathe" in some case, then you will need 5 truly random characters. So randomly decide how many of those letters should be before "lathe". If you pick 2, for example, then put 2 random characters, plus "lathe" in a random case, plus 3 more random characters.
It could be something like this, where I mostly use your same style of for-loops and += string concatenation:
function generateStringWithKeyword(randNum: number) {
const keyword = "lathe";
if (randNum < keyword.length) throw new Error(
"This is not possible; \"" + keyword + "\" doesn't fit in " + randNum + " characters"
);
const actuallyRandNum = randNum - keyword.length;
const chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789+/";
let result = "";
const kwInsertionPoint = Math.floor(Math.random() * (actuallyRandNum + 1));
for (let i = 0; i < kwInsertionPoint; i++) {
result += chars[Math.floor(Math.random() * chars.length)];
}
for (let i = 0; i < keyword.length; i++) {
result += Math.random() < 0.5 ? keyword[i].toLowerCase() : keyword[i].toUpperCase();
}
for (let i = kwInsertionPoint; i < actuallyRandNum; i++) {
result += chars[Math.floor(Math.random() * chars.length)];
}
return result;
}
If you run this, you will see that it is very efficient, and never gives up:
console.log(Array.from({ length: 4 }, () => generateStringWithKeyword(5)).join(" "));
// "lathE LaThe lATHe LatHe"
console.log(Array.from({ length: 4 }, () => generateStringWithKeyword(7)).join(" "));
// "p6lAtHe laThE01 nlaTheK lATHeRJ"
console.log(Array.from({ length: 4 }, () => generateStringWithKeyword(10)).join(" "));
// "giMqzLaTHe 5klAthegBo oVdLatHe0q twNlATheCr"
Playground link to code
On the webpage
http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463
It lists prices for a particular item in a game, I wanted to grab the "Current guide price:" of said item, and store it as a variable so I could output it in a google spreadsheet. I only want the number, currently it is "643.8k", but I am not sure how to grab specific text like that.
Since the number is in "k" form, that means I can't graph it, It would have to be something like 643,800 to make it graphable. I have a formula for it, and my second question would be to know if it's possible to use a formula on the number pulled, then store that as the final output?
-EDIT-
This is what I have so far and it's not working not sure why.
function pullRuneScape() {
var page = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var number = page.match(/Current guide price:<\/th>\n(\d*)/)[1];
SpreadsheetApp.getActive().getSheetByName('RuneScape').appendRow([new Date(), number]);
}
Your regex is wrong. I tested this one successfully:
var number = page.match(/Current guide price:<\/th>\s*<td>([^<]*)<\/td>/m)[1];
What it does:
Current guide price:<\/th> find Current guide price: and closing td tag
\s*<td> allow whitespace between tags, find opening td tag
([^<]*) build a group and match everything except this char <
<\/td> match the closing td tag
/m match multiline
Use UrlFetch to get the page [1]. That'll return an HTTPResponse that you can read with GetBlob [2]. Once you have the text you can use regular expressions. In this case just search for 'Current guide price:' and then read the next row. As to remove the 'k' you can just replace with reg ex like this:
'123k'.replace(/k/g,'')
Will return just '123'.
https://developers.google.com/apps-script/reference/url-fetch/
https://developers.google.com/apps-script/reference/url-fetch/http-response
Obviously, you are not getting anything because the regexp is wrong. I'm no regexp expert but I was able to extract the number using basic string manipulation
var page = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var TD = "<td>";
var start = page.indexOf('Current guide price');
start = page.indexOf(TD, start);
var end = page.indexOf('</td>',start);
var number = page.substring (start + TD.length , end);
Logger.log(number);
Then, I wrote a function to convert k,m etc. to the corresponding multiplying factors.
function getMultiplyingFactor(symbol){
switch(symbol){
case 'k':
case 'K':
return 1000;
case 'm':
case 'M':
return 1000 * 1000;
case 'g':
case 'G':
return 1000 * 1000 * 1000;
default:
return 1;
}
}
Finally, tie the two together
function pullRuneScape() {
var page = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var TD = "<td>";
var start = page.indexOf('Current guide price');
start = page.indexOf(TD, start);
var end = page.indexOf('</td>',start);
var number = page.substring (start + TD.length , end);
Logger.log(number);
var numericPart = number.substring(0, number.length -1);
var multiplierSymbol = number.substring(number.length -1 , number.length);
var multiplier = getMultiplyingFactor(multiplierSymbol);
var fullNumber = multiplier == 1 ? number : numericPart * multiplier;
Logger.log(fullNumber);
}
Certainly, not the optimal way of doing things but it works.
Basically I parse the html page as you did (with corrected regex) and split the string into number part and multiplicator (k = 1000). Finally I return the extracted number. This function can be used in Google Docs.
function pullRuneScape() {
var pageContent = UrlFetchApp.fetch("http://services.runescape.com/m=itemdb_rs/Armadyl_chaps/viewitem.ws?obj=19463").getContentText();
var matched = pageContent.match(/Current guide price:<.th>\n<td>(\d+\.*\d*)([k]{0,1})/);
var numberAsString = matched[1];
var multiplier = "";
if (matched.length == 3) {
multiplier = matched[2];
}
number = convertNumber(numberAsString, multiplier);
return number;
}
function convertNumber(numberAsString, multiplier) {
var number = Number(numberAsString);
if (multiplier == 'k') {
number *= 1000;
}
return number;
}
I am using a node.js server for a multiplayer synchronized dice, but i am having some strange problems with variables changing that are not referenced or used...
var origonalRolls = rolls;
//identify epic fails
var epicFails = [];
for(var r = 0; r < rolls.length; r++)
if(rolls[r] === 1)
epicFails.push(r);
console.log("TEST 1 :: " + JSON.stringify(rolls));
console.log("TEST 2 :: " + JSON.stringify(origonalRolls));
//remove epic fails and the corresponding heighest
if(epicFails.length > 0){
for(var i = epicFails.length-1; i >= 0; i--){
rolls.splice(epicFails[i], 1);
if(rolls[rolls.length-1] >= success)
rolls.splice(rolls.length-1, 1);
}
}
console.log("TEST 3 :: " + JSON.stringify(rolls));
console.log("TEST 4 :: " + JSON.stringify(origonalRolls));
the above should find any element in the rolls array which is 1 and then add it to epicFails. it should then remove it from rolls as well as the heighest remaining roll. (note, rolls is sorted numerically)
for some reason the output of this segment of code is as follows:
TEST 1 :: [1,1,2,3,3,6,7,7,9,9]
TEST 2 :: [1,1,2,3,3,6,7,7,9,9]
TEST 3 :: [2,3,3,6,7,7]
TEST 4 :: [2,3,3,6,7,7]
I am unsure why rolls and origonalRolls start the same and end the same. I am only using rolls.
Any help and/or explanation to this problem is welcome, it's been troubling me for a long time now...
In Javascript Arrays and Objects are only shallow copied - which means that an array (rolls) copied from another array (originalRolls) is only a reference to originalRolls - it is not an entirely new array, and modifying values in one will affect values in the other.
You will need to implement a deep copy function to create an entirely new array based off another. There are numerous imlementations of deep copying arrays/objects both here and elsewhere on the net - here is one of them from a quick Google.
Replace var origonalRolls = rolls; with:
var origonalRolls = [];
for (var i = 0, len = rolls.length; i < len; i++) {
origonalRolls[i] = rolls[i];
}
Let me explain.
I have to do some fuzzy matching for a company, so ATM I use a levenshtein distance calculator, and then calculate the percentage of similarity between the two terms. If the terms are more than 80% similar, Fuzzymatch returns "TRUE".
My problem is that I'm on an internship, and leaving soon. The people who will continue doing this do not know how to use excel with macros, and want me to implement what I did as best I can.
So my question is : however inefficient the function may be, is there ANY way to make a standard function in Excel that will calculate what I did before, without resorting to macros ?
Thanks.
If you came about this googling something like
levenshtein distance google sheets
I threw this together, with the code comment from milot-midia on this gist (https://gist.github.com/andrei-m/982927 - code under MIT license)
From Sheets in the header menu, Tools -> Script Editor
Name the project
The name of the function (not the project) will let you use the func
Paste the following code
function Levenshtein(a, b) {
if(a.length == 0) return b.length;
if(b.length == 0) return a.length;
// swap to save some memory O(min(a,b)) instead of O(a)
if(a.length > b.length) {
var tmp = a;
a = b;
b = tmp;
}
var row = [];
// init the row
for(var i = 0; i <= a.length; i++){
row[i] = i;
}
// fill in the rest
for(var i = 1; i <= b.length; i++){
var prev = i;
for(var j = 1; j <= a.length; j++){
var val;
if(b.charAt(i-1) == a.charAt(j-1)){
val = row[j-1]; // match
} else {
val = Math.min(row[j-1] + 1, // substitution
prev + 1, // insertion
row[j] + 1); // deletion
}
row[j - 1] = prev;
prev = val;
}
row[a.length] = prev;
}
return row[a.length];
}
You should be able to run it from a spreadsheet with
=Levenshtein(cell_1,cell_2)
While it can't be done in a single formula for any reasonably-sized strings, you can use formulas alone to compute the Levenshtein Distance between strings using a worksheet.
Here is an example that can handle strings up to 15 characters, it could be easily expanded for more:
https://docs.google.com/spreadsheet/ccc?key=0AkZy12yffb5YdFNybkNJaE5hTG9VYkNpdW5ZOWowSFE&usp=sharing
This isn't practical for anything other than ad-hoc comparisons, but it does do a decent job of showing how the algorithm works.
looking at the previous answers to calculating Levenshtein distance, I think it would be impossible to create it as a formula.
Take a look at the code here
Actually, I think I just found a workaround. I was adding it in the wrong part of the code...
Adding this line
} else if(b.charAt(i-1)==a.charAt(j) && b.charAt(i)==a.charAt(j-1)){
val = row[j-1]-0.33; //transposition
so it now reads
if(b.charAt(i-1) == a.charAt(j-1)){
val = row[j-1]; // match
} else if(b.charAt(i-1)==a.charAt(j) && b.charAt(i)==a.charAt(j-1)){
val = row[j-1]-0.33; //transposition
} else {
val = Math.min(row[j-1] + 1, // substitution
prev + 1, // insertion
row[j] + 1); // deletion
}
Seems to fix the problem. Now 'biulding' is 92% accurate and 'bilding' is 88%. (whereas with the original formula 'biulding' was only 75%... despite being closer to the correct spelling of building)
I am using Actionscript 2.0
In a Brand new Scene. My only bit of code is:
trace(int('04755'));
trace(int('04812'));
Results in:
2541
4812
Any idea what I am doing wrong/silly?
By the way, I am getting this source number from XML, where it already has the leading 0. Also, this works perfect in Actionscript 3.
In AS3, you can try:
parseInt('04755', 10)
10 above is the radix.
parseInt(yourString);
...is the correct answer. .parseInt() is a top-level function.
Converting a string with a leading 0 to a Number in ActionScript 2 assumes that the number you want is octal. Give this function I've made for you a try:
var val:String = '00010';
function parse(str:String):Number
{
for(var i = 0; i < str.length; i++)
{
var c:String = str.charAt(i);
if(c != "0") break;
}
return Number(str.substr(i));
}
trace(parse(val)); // 10
trace(parse(val) + 10); // 20
Basically what you want to do now is just wrap your string in the above parse() function, instead of int() or Number() as you would typically.
Bit of a simple one...
try this -
temp="120";
temp2="140";
temp3=int ( temp );
temp4=int ( temp2 );
temp5=temp4+temp3;
trace(temp5);
so, all you need is...
int("190");