we are loading an aggrid on ui and when user right clicks on it and downloads an excel, a column which has more than 15 digits is loosing the last four digits and replaced with zeros.(It is shown as scientific format in excel and when we check the value on the cell it is different from what is shown on the grid)
Please advice
You can have a look at this Big Number package which allows you to store and modify numbers larger than Number.MAX_SAFE_INTEGER
A number that contains more than 15 digits in Excel, it changes any digits past the fifteenth place to zeros. You can cast the column values to string on export.
//Use customizeData
exportOptions: {
//--
},
customizeData: function (data) {
var ind = data.header.indexOf("ColumnName"); // This code is to find the column name's index which you want to cast.
for (var i = 0; i < data.body.length; i++) {
data.body[i][ind] = '\u200C' + data.body[i][ind]; //will cast the number to string.
}
}
Error_Image below:
https://i.stack.imgur.com/A39hf.png
I'm having trouble even beginning to think of how to do this.
I need to find a missing number in a string of random numbers that don't have separators.
Here's an example: 14036587109. In this example, the missing number is 2, the range is 0-10 (inclusive).
How can I write a JavaScript/Node.JS program to solve this?
The part I can't figure out is how the program separates the numbers; In the example above, how would the program know that the number 10 (before the last number) isn't the numbers 1 and 0.
There are two things we know about the missing integer: the total number of digits in the input tells us the number of digits in the missing integer, and (as #samgak mentioned in a comment) counting the occurrences of each digit in the input tells us which digits the missing integer is made of. This may give us a quick path to the solution, if one of the permutations of those digits is missing from the input. If it doesn't, then:
Find the integers from highest to lowest number of digits; if the range is e.g. 0-999, then search the 3-digit integers first, then 2, then 1.
If an integer is only present at one location in the input, mark it as found, and remove it from the input.
Then, start again with the longest integers that haven't been found yet, and look at the ones that are present at two locations; try both options, and then check whether all other integers that rely on the digits we're using are also present; e.g. if 357 is present at two locations:
... 1235789 ... 2435768 ...
357 357
23 43
123 243
235 435
578 576
78 76
789 768
When trying the first location for the 357, check whether there is another possibility for 23, 123, 235, 578, 78, and 789. For the second location, check 43, 243, 435, 576, 76 and 768.
If these checks show that only one of the options is possible, mark the number as found and remove it from the input.
Go on to do this for shorter integers, and for integers that are present at 3, 4, ... locations. If, after doing this to a certain point, there is still no result, you may have to recursively try several options, which will quickly lead to a huge number of options. (With especially crafted large input, it is probably possible to thwart this method and make it unusably slow.) But the average complexity with random input may be decent.
Actually, when you find an integer that is only present in one location in the input, but it is a permutation of the missing digits, you should not remove it, because it could be the missing integer. So the algorithm might be: remove all integers you can unequivocally locate in the input, then try removing all possible missing integers one by one, and look for inconsistencies, i.e. other missing numbers that don't have the correct length or digits.
It's all a question of heuristics, of course. You try something simple, if that doesn't work you try something more complicated, if that doesn't work, you try something even more complicated... and at each step there are several options, and each one could be optimal for some input strings but not for others.
E.g. if the range is 0-5000, you'd start by marking the 4-digit integers that are only present at one location. But after that, you could do the same thing again (because integers that were present twice could have had one of their options removed) until there's no more improvement, or you could check integers that are present twice, or integers that are present up to x times, or move on to 3-digit integers... I don't think there's a straightforward way to know which of these options will give the best result.
This solution should work for any input string and any start/end range:
We can think about the numbers in the string as a pool of digits that we can choose from. We start at startRange and go through to endRange, looking for each number along the way in our pool of digits.
When we find a number that can be composed from our pool of digits, we delete those digits from our pool of digits, as those digits are already being used to form a number in our range.
As soon as we come across a number that cannot be composed from our pool of digits, that must be the missing number.
const str = "14036587109"; // input
const numsLeft = str.split("").map(num => parseInt(num)); // array of numbers
const startRange = 0;
const endRange = 10;
for(let i = startRange; i <= endRange ; i++) {
// check if number can be formed given the numbers left in numsLeft
const numFound = findNum(numsLeft, i);
if(!numFound) {
console.log("MISSING: " + i); // prints 2
break;
}
}
function findNum(numsLeft, i) {
// array of digits
const numsToFind = String(i).split("").map(num => parseInt(num));
// default is true, if all digits are found in numsLeft
let found = true;
numsToFind.forEach(num => {
// find digit in numsLeft
const numFoundIndex = numsLeft.indexOf(num);
if(numFoundIndex < 0) {
// digit was not found in numsLeft
found = false;
return;
} else {
// digit was found; delete digit from numsLeft
numsLeft.splice(numFoundIndex, 1);
}
});
return found;
}
var input = '10436587109';
var range = [10,9,8,7,6,5,4,3,2,1,0];
var expr1 = new RegExp(range.join('|'),'g');
var expr2 = new RegExp('[0-9]','g');
var a = input.match(expr1).map(Number).concat(input.match(expr2).map(Number));
var x = range.filter(function(i){ return a.indexOf(i)===-1; });
Okay so I am making a mario inspired game with randomly generating terrain, It is all working fine however the array of random numbers that randomises the terrain must be the same each time so that the user can enter a seed which is then merged with the larger list to provide a set of random numbers based off of the seed however I cannot think of any way to make this array the same each time without writing it out, and even then making an array of 1000 numbers will be timely. Can anyone suggest a fast way (number generators online dont format it in one single line of numbers separated by numbers so cannot use them)
or could someone provide me with a list that is on a single line separated by numbers that i can easily copy and paste into an array thanks! :)
The following code in Javascript will generate 1000 random numbers separated by commas.
var string = "";
var numberOfRandomNumbers = 1000;
for (var i = 0; i < numberOfRandomNumbers; i++) {
var randomNumber = Math.floor((Math.random() * 1000) + 1); //Will generate random number between 1 and 1000
string += randomNumber+",";
}
console.log(string.substring(0, string.length - 1)); //Print string to console and remove last comma
I'm working with Youtube videos which have ids like 8DnKOc6FISU, rNsrl86inpo, 5qcmCUsw4EQ (i.e. 11 characters in the set A-Za-z0-9_-)
The goal is to convert each id to a colour (represented by the range 0-1), so they can be reliably charted.
According to this question these are 64 bit numbers. Given that:
I want to make full use of the colour space for any given set of videos
Colour perception isn't that accurate anyway
...it seems sensible to base this off the last 2-3 characters of the id.
My lead approach is a function I borrowed from here, which converts each character into a binary number like so:
function toBin(str){
var st,i,j,d;
var arr = [];
var len = str.length;
for (i = 1; i<=len; i++){
d = str.charCodeAt(len-i);
for (j = 0; j < 8; j++) {
st = d%2 == '0' ? "class='zero'" : ""
arr.push(d%2);
d = Math.floor(d/2);
}
}
}
But this leaves the question of how to convert this back to a float.
Any ideas for an elegant solution?
Many thanks for your help!
Knowing that system is base 64 (26+26+10+2), just read each symbol, add it to total and multiply result by 64 on each iteration. In the end you will have an integer number. Divide it by maximum value to normalize it to 0-1 range.
You'll start losing precision when your IDs approach 253 though, as that's maximum what JS can represent in integer precisely.
Alternatively you can count in floats from the very start by adding (letter_index / 64letter_position) for each position in line, but you'd be losing more precision in process.
I know nothing about javascript.
Assuming the string "3005600008000", I need to find a way to multiply all the digits in the odd numbered positions by 2 and the digits in the even numbered positions by 1.
This pseudo code I wrote outputs (I think) TRUE for the odd numbers (i.e. "0"),
var camid;
var LN= camid.length;
var mychar = camid.charAt(LN%2);
var arr = new Array(camid);
for(var i=0; i<arr.length; i++) {
var value = arr[i]%2;
Alert(i =" "+value);
}
I am not sure this is right: I don't believe it's chunking/splitting the string at odd (And later even) positions.
How do I that? Can you please provide some hints?
/=================================================/
My goal is to implement in a web page a validation routine for a smartcard id number.
The logic I am trying to implement is as follows:
· 1) Starting from the left, multiply all the digits in the odd numbered positions by 2 and the digits in the even numbered positions by 1.
· 2) If the result of a multiplication of a single digit by 2 results in a two-digit number (say "7 x 2 = 14"), add the digits of the result together to produce a new single-digit result ("1+4=5").
· 3) Add all single-digit results together.
· 4) The check digit is the amount you must add to this result in order to reach the next highest multiple of ten. For instance, if the sum in step #3 is 22, to reach the next highest multiple of 10 (which is 30) you must add 8 to 22. Thus the check digit is 8.
That is the whole idea. Google searches on smartcard id validation returned nothing and I am beginning to think this is overkill to do this in Javascript...
Any input welcome.
var theArray = camid.split(''); // create an array entry for each digit in camid
var size = theArray.length, i, eachValue;
for(i = 0; i < size; i++) { // iterate over each digit
eachValue = parseInt(theArray[i], 10); // test each string digit for an integer
if(!isNaN(eachValue)) {
alert((eachValue % 2) ? eachValue * 2 : eachValue); // if mod outputs 1 / true (due to odd number) multiply the value by 2. If mod outputs 0 / false output value
}
}
I discovered that what I am trying to do is called a Luhn validation.
I found an algorithm right here.
http://sites.google.com/site/abapexamples/javascript/luhn-validation
Thanks for taking the time to help me out. Much appreciated.
It looks like you might be building to a Luhn validation. If so, notice that you need to count odd/even from the RIGHT not the left of the string.