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So I have this question, which I don't quite understand.
I would like to understand the approach of the problem
Imagine the following scenario:
You are the HR manager of a company with 1000 employees numbered for 1 to 1000. Your boss told you to give a big Christmas bonus to employees, but didn’t tell you their names. Instead they gave you two indications:
1) the sum of the proper divisors (including 1 but not itself) of the employee number is greater than the employee number itself
2) no subset of those divisors sums to the employee number itself.
How many employees are eligible for the bonus and what are their number?
For example:
- Number 12: the proper divisors are 1, 2, 3, 4 and 6. The sum is 1+2+3+4+6 = 16 which is greater than 12 and matches the first condition. However, the subset 2+4+6=12 which violates the second condition.
My conclusion is:
I have to get those numbers from 1 to 1000, where the sum of the number's divisors are greater than the number itself (including 1 but not itself), but none of the divisors' subsets can be added to be equal with the number itself?
My steps would be:
Put the divisors of the numbers from 1 to 1000 into an array
Get those numbers, when the divisors' sum (including 1 but not itself) are greater than the number itself and resize the array to only those numbers.
I have to check every subset of the remaining number's divisors and remove those when a subset of the divisors can be equal with the number itself.
Could you help me if it's a good approach or do any of you know a more efficient/better way?
Any help would be appreciated!
NOTE: I don't want you to solve it, I want to understand it!
I have made this so far, which covers the first two steps that I intended to do. Last step is beyond my knowledge, but I have the solution for that also.
This is my code:
<script type="text/javascript">
function getDivisors(n){
var divisors=new Array();
for(var x=1;x<n;x++){
if(n%x==0) divisors.push(x);
}
return divisors;
}
function getNumbers(n){
var numbers=new Array(),
sum=0;
for(var x=1;x<=n;x++){
sum=getDivisors(x).reduce((a, b) => a + b, 0);
if(sum>x) numbers.push(x);
// console.log("Number: "+x+" sum:"+sum);
}
return numbers;
}
var remainingNumbers = getNumbers(1000);
console.log(remainingNumbers);
</script>
This is the answer for the question:
var out = document.getElementById('outLine');
out.innerHTML += "X\t»\tSUM\tSUM-X\tLIST\r\n";
function isSubsetSum(set, n, sum) {
if (sum == 0) { return true; }
if (n == 0 && sum != 0) { return false; }
if (set[n - 1] > sum) { return isSubsetSum(set, n - 1, sum); }
return isSubsetSum(set, n - 1, sum) ||
isSubsetSum(set, n - 1, sum - set[n - 1]);
}
function v1chNum(x) {
var r = [1];
var n = 0;
var m = x/2;
for(var i = 2; i <= m; i++ ) {
if(x%i==0) {
if(r.indexOf(i)==-1) {
r.push(i);
n += i;
}
m = x/i;
if(r.indexOf(m)==-1) {
r.push(m);
n += m;
}
}
}
if( n > x ) {
r.sort(function(a, b) {return b - a;});
if(!isSubsetSum(r,r.length,x)) {
out.innerHTML += x+"\t»\t"+n+"\t"+(n-x)+"\t"+r+"\r\n";
} else { return false; }
} else { return false; }
}
for(var x = 1; x<1000; x++) {
v1chNum(x);
}
<pre id="outLine"></pre>
PHP approach:
$empList = [];
for ($emp = 1; $emp <= 100; $emp++) {
$multiples = [];
$subset = [];
$cond1 = false;
$cond2 = false;
// Get multiples.
for ($i = 1; $i < $emp; $i++) {
if ($emp % $i == 0) $multiples[]= $i;
}
// Condition 1
if (array_sum($multiples) > $emp) $cond1 = true;
foreach ($multiples as $num) {
if ($num % 2 == 0) $subset[]= $num;
}
// Condition 2
if (array_sum($subset) > $emp) $cond2 = true;
if ($cond1 && $cond2) $empList[] = $emp;
}
echo "<pre>";
var_dump($empList);
echo "</pre>";
Output:
Array
(
[0] => 24
[1] => 36
[2] => 40
[3] => 48
[4] => 60
[5] => 72
[6] => 80
[7] => 84
[8] => 96
)
My code in python:
def getDivisors(n):
div = []
for i in range(1, n-1):
if(n%i == 0):
div.append(i)
return div
def sumDivisors(arr):
div_sum = 0
for i in arr:
div_sum += i
return div_sum
def subLists(arr):
lists = [[]]
for i in range(len(arr)):
orig = lists[:]
new = arr[i]
for j in range(len(lists)):
lists[j] = lists[j] + [new]
lists = orig + lists
return lists
def sumSublists(lists, n):
for i in range(len(lists)):
sum_list = sum(lists[i])
if (sum_list == n):
return False
return True
for num in range(100):
arr = getDivisors(int(num))
lists = subLists(arr)
if ((sumDivisors(arr) > int(num))and(sumSublists(lists, int(num)))):
print(str(num) + '/n')
I would like to know if it's possible to change the data type for a column. For instance, the json data passed to the grid are strings, but I would like slickgrid to consider it as integers or floats to be able to sort it correctly.
var data = [{"NOM": "Saguenay - Lac-Saint-Jean", "CODE": "02", "id": "0", "integer": "1"},]
I would like the 'integer' column to be an int not a string, without changing the data itself.
Thank you for your help.
As I mentioned in my comment, you are looking at the wrong place (no offense); there is no need to change datatype as actually this will not fix your problem with sort, since the SlickGrid default sort is string sort. But you could use custom sort to fix your problem.
So here is the solution: Define sort function and use them as needed. Here is a list of custom sort functions you could create:
function sorterStringCompare(a, b) {
var x = a[sortcol], y = b[sortcol];
return sortdir * (x === y ? 0 : (x > y ? 1 : -1));
}
function sorterNumeric(a, b) {
var x = (isNaN(a[sortcol]) || a[sortcol] === "" || a[sortcol] === null) ? -99e+10 : parseFloat(a[sortcol]);
var y = (isNaN(b[sortcol]) || b[sortcol] === "" || b[sortcol] === null) ? -99e+10 : parseFloat(b[sortcol]);
return sortdir * (x === y ? 0 : (x > y ? 1 : -1));
}
function sorterRating(a, b) {
var xrow = a[sortcol], yrow = b[sortcol];
var x = xrow[3], y = yrow[3];
return sortdir * (x === y ? 0 : (x > y ? 1 : -1));
}
function sorterDateIso(a, b) {
var regex_a = new RegExp("^((19[1-9][1-9])|([2][01][0-9]))\\d-([0]\\d|[1][0-2])-([0-2]\\d|[3][0-1])(\\s([0]\\d|[1][0-2])(\\:[0-5]\\d){1,2}(\\:[0-5]\\d){1,2})?$", "gi");
var regex_b = new RegExp("^((19[1-9][1-9])|([2][01][0-9]))\\d-([0]\\d|[1][0-2])-([0-2]\\d|[3][0-1])(\\s([0]\\d|[1][0-2])(\\:[0-5]\\d){1,2}(\\:[0-5]\\d){1,2})?$", "gi");
if (regex_a.test(a[sortcol]) && regex_b.test(b[sortcol])) {
var date_a = new Date(a[sortcol]);
var date_b = new Date(b[sortcol]);
var diff = date_a.getTime() - date_b.getTime();
return sortdir * (diff === 0 ? 0 : (date_a > date_b ? 1 : -1));
}
else {
var x = a[sortcol], y = b[sortcol];
return sortdir * (x === y ? 0 : (x > y ? 1 : -1));
}
}
and then in your columns definition you would use whichever custom filter you need, in your case the sorterNumeric() is what you're looking for...so your columns definition would look like the following (custom filter are at the end):
var columns = [
{id:"column1", name:"column1", field: "Column String", width:40, sortable:true, sorter:sorterStringCompare},
{id:"column2", name:"column2", field: "Column integer", width:40, sortable:true, sorter:sorterNumeric},
{id:"column3", name:"column3", field: "Column rating", width:40, sortable:true, sorter:sorterRating}
];
Saguenay...? Quebecois? :)
EDIT
I forgot to add the piece of code that attach the new sorter property to the onSort event (of course without it then it won't work), make sure you have same object name for grid and dataView, correct to whatever your variables naming are (if need be), here is the code:
grid.onSort.subscribe(function (e, args) {
var cols = args.sortCols;
dataView.sort(function (dataRow1, dataRow2) {
for (var i = 0, l = cols.length; i < l; i++) {
sortdir = cols[i].sortAsc ? 1 : -1;
sortcol = cols[i].sortCol.field;
var result = cols[i].sortCol.sorter(dataRow1, dataRow2); // sorter property from column definition comes in play here
if (result != 0) {
return result;
}
}
return 0;
});
args.grid.invalidateAllRows();
args.grid.render();
});
You could also put your code directly into the last onSort.subscribe but I suggest having the sorter into a separate function since it is cleaner (which is the code I sent).
I used this to sort the numbers correctly.
grid.onSort.subscribe(function(e, args) {
var cols = args.sortCols;
data.sort(function(dataRow1, dataRow2) {
for (var i = 0, l = cols.length; i < l; i++) {
var result = sortOnString(cols, i, dataRow1, dataRow2);
if (result != 0) {
return result;
}
}
return 0;
});
grid.invalidate();
grid.render();
});
function sortOnString(cols, i, dataRow1, dataRow2) {
var field = cols[i].sortCol.field;
var sign = cols[i].sortAsc ? 1 : -1;
console.log("name filed " + field);
if (field === 'Folio' || field === 'Orden') {
var value1 = parseInt(dataRow1[field]),
value2 = parseInt(dataRow2[field]);
console.log("name value 1 " + value1 + "name value 2 " + value2);
} else {
var value1 = dataRow1[field],
value2 = dataRow2[field];
}
var result = (value1 == value2 ? 0 : (value1 > value2 ? 1 : -1)) * sign;
return result;
}
I need to convert a Google Spreadsheet column index into its corresponding letter value, for example, given a spreadsheet:
I need to do this (this function obviously does not exist, it's an example):
getColumnLetterByIndex(4); // this should return "D"
getColumnLetterByIndex(1); // this should return "A"
getColumnLetterByIndex(6); // this should return "F"
Now, I don't recall exactly if the index starts from 0 or from 1, anyway the concept should be clear.
I didn't find anything about this on gas documentation.. am I blind? Any idea?
Thank you
I wrote these a while back for various purposes (will return the double-letter column names for column numbers > 26):
function columnToLetter(column)
{
var temp, letter = '';
while (column > 0)
{
temp = (column - 1) % 26;
letter = String.fromCharCode(temp + 65) + letter;
column = (column - temp - 1) / 26;
}
return letter;
}
function letterToColumn(letter)
{
var column = 0, length = letter.length;
for (var i = 0; i < length; i++)
{
column += (letter.charCodeAt(i) - 64) * Math.pow(26, length - i - 1);
}
return column;
}
This works good
=REGEXEXTRACT(ADDRESS(ROW(); COLUMN()); "[A-Z]+")
even for columns beyond Z.
Simply replace COLUMN() with your column number. The value of ROW() doesn't matter.
No need to reinvent the wheel here, use the GAS range instead:
var column_index = 1; // your column to resolve
var ss = SpreadsheetApp.getActiveSpreadsheet();
var sheet = ss.getSheets()[0];
var range = sheet.getRange(1, column_index, 1, 1);
Logger.log(range.getA1Notation().match(/([A-Z]+)/)[0]); // Logs "A"
=SUBSTITUTE(ADDRESS(1,COLUMN(),4), "1", "")
This takes your cell, gets it's address as e.g. C1, and removes the "1".
How it works
COLUMN() gives the number of the column of the cell.
ADDRESS(1, ..., <format>) gives an address of a cell, in format speficied by <format> parameter. 4 means the address you know - e.g. C1.
The row doesn't matter here, so we use 1.
See ADDRESS docs
Finally, SUBSTITUTE(..., "1", "") replaces the 1 in the address C1, so you're left with the column letter.
This works on ranges A-Z
formula =char(64+column())
js String.fromCharCode(64+colno)
an google spreadsheet appscript code, based on #Gardener would be:
function columnName(index) {
var cname = String.fromCharCode(65 + ((index - 1) % 26));
if (index > 26)
cname = String.fromCharCode(64 + (index - 1) / 26) + cname;
return cname;
}
In javascript:
X = (n) => (a=Math.floor(n/26)) >= 0 ? X(a-1) + String.fromCharCode(65+(n%26)) : '';
console.assert (X(0) == 'A')
console.assert (X(25) == 'Z')
console.assert (X(26) == 'AA')
console.assert (X(51) == 'AZ')
console.assert (X(52) == 'BA')
Adding to #SauloAlessandre's answer, this will work for columns up from A-ZZ.
=if(column() >26,char(64+(column()-1)/26),) & char(65 + mod(column()-1,26))
I like the answers by #wronex and #Ondra Žižka. However, I really like the simplicity of #SauloAlessandre's answer.
So, I just added the obvious code to allow #SauloAlessandre's answer to work for wider spreadsheets.
As #Dave mentioned in his comment, it does help to have a programming background, particularly one in C where we added the hex value of 'A' to a number to get the nth letter of the alphabet as a standard pattern.
Answer updated to catch the error pointed out by #Sangbok Lee. Thank you!
I was looking for a solution in PHP. Maybe this will help someone.
<?php
$numberToLetter = function(int $number)
{
if ($number <= 0) return null;
$temp; $letter = '';
while ($number > 0) {
$temp = ($number - 1) % 26;
$letter = chr($temp + 65) . $letter;
$number = ($number - $temp - 1) / 26;
}
return $letter;
};
$letterToNumber = function(string $letters) {
$letters = strtoupper($letters);
$letters = preg_replace("/[^A-Z]/", '', $letters);
$column = 0;
$length = strlen($letters);
for ($i = 0; $i < $length; $i++) {
$column += (ord($letters[$i]) - 64) * pow(26, $length - $i - 1);
}
return $column;
};
var_dump($numberToLetter(-1));
var_dump($numberToLetter(26));
var_dump($numberToLetter(27));
var_dump($numberToLetter(30));
var_dump($letterToNumber('-1A!'));
var_dump($letterToNumber('A'));
var_dump($letterToNumber('B'));
var_dump($letterToNumber('Y'));
var_dump($letterToNumber('Z'));
var_dump($letterToNumber('AA'));
var_dump($letterToNumber('AB'));
Output:
NULL
string(1) "Z"
string(2) "AA"
string(2) "AD"
int(1)
int(1)
int(2)
int(25)
int(26)
int(27)
int(28)
Simple way through Google Sheet functions, A to Z.
=column(B2) : value is 2
=address(1, column(B2)) : value is $B$1
=mid(address(1, column(B2)),2,1) : value is B
It's a complicated way through Google Sheet functions, but it's also more than AA.
=mid(address(1, column(AB3)),2,len(address(1, column(AB3)))-3) : value is AB
I also was looking for a Python version here is mine which was tested on Python 3.6
def columnToLetter(column):
character = chr(ord('A') + column % 26)
remainder = column // 26
if column >= 26:
return columnToLetter(remainder-1) + character
else:
return character
A comment on my answer says you wanted a script function for it. All right, here we go:
function excelize(colNum) {
var order = 1, sub = 0, divTmp = colNum;
do {
divTmp -= order; sub += order; order *= 26;
divTmp = (divTmp - (divTmp % 26)) / 26;
} while(divTmp > 0);
var symbols = "0123456789abcdefghijklmnopqrstuvwxyz";
var tr = c => symbols[symbols.indexOf(c)+10];
return Number(colNum-sub).toString(26).split('').map(c=>tr(c)).join('');
}
This can handle any number JS can handle, I think.
Explanation:
Since this is not base26, we need to substract the base times order for each additional symbol ("digit"). So first we count the order of the resulting number, and at the same time count the number to substract. And then we convert it to base 26 and substract that, and then shift the symbols to A-Z instead of 0-P.
Anyway, this question is turning into a code golf :)
Java Apache POI
String columnLetter = CellReference.convertNumToColString(columnNumber);
This will cover you out as far as column AZ:
=iferror(if(match(A2,$A$1:$AZ$1,0)<27,char(64+(match(A2,$A$1:$AZ$1,0))),concatenate("A",char(38+(match(A2,$A$1:$AZ$1,0))))),"No match")
A function to convert a column index to letter combinations, recursively:
function lettersFromIndex(index, curResult, i) {
if (i == undefined) i = 11; //enough for Number.MAX_SAFE_INTEGER
if (curResult == undefined) curResult = "";
var factor = Math.floor(index / Math.pow(26, i)); //for the order of magnitude 26^i
if (factor > 0 && i > 0) {
curResult += String.fromCharCode(64 + factor);
curResult = lettersFromIndex(index - Math.pow(26, i) * factor, curResult, i - 1);
} else if (factor == 0 && i > 0) {
curResult = lettersFromIndex(index, curResult, i - 1);
} else {
curResult += String.fromCharCode(64 + index % 26);
}
return curResult;
}
function lettersFromIndex(index, curResult, i) {
if (i == undefined) i = 11; //enough for Number.MAX_SAFE_INTEGER
if (curResult == undefined) curResult = "";
var factor = Math.floor(index / Math.pow(26, i));
if (factor > 0 && i > 0) {
curResult += String.fromCharCode(64 + factor);
curResult = lettersFromIndex(index - Math.pow(26, i) * factor, curResult, i - 1);
} else if (factor == 0 && i > 0) {
curResult = lettersFromIndex(index, curResult, i - 1);
} else {
curResult += String.fromCharCode(64 + index % 26);
}
return curResult;
}
document.getElementById("result1").innerHTML = lettersFromIndex(32);
document.getElementById("result2").innerHTML = lettersFromIndex(6800);
document.getElementById("result3").innerHTML = lettersFromIndex(9007199254740991);
32 --> <span id="result1"></span><br> 6800 --> <span id="result2"></span><br> 9007199254740991 --> <span id="result3"></span>
In python, there is the gspread library
import gspread
column_letter = gspread.utils.rowcol_to_a1(1, <put your col number here>)[:-1]
If you cannot use python, I suggest looking the source code of rowcol_to_a1() in https://github.com/burnash/gspread/blob/master/gspread/utils.py
Here's a two liner which works beyond ZZ using recursion:
Python
def col_to_letter(n):
l = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
return col_to_letter((n-1)//26) + col_to_letter(n%26) if n > 26 else l[n-1]
Javascript
function colToLetter(n) {
l = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
return n > 26 ? colToLetter(Math.floor((n-1)/26)) + colToLetter(n%26) : l[n-1]
}
If you need a version directly in the sheet, here a solution:
For the colonne 4, we can use :
=Address(1,4)
I keep the row number to 1 for simplicty.
The above formula returns $D$1 which is not what you want.
By modifying the formula a little bit we can remove the dollar signs in the cell reference.
=Address(1,4,4)
Adding four as the third argument tells the formula that we are not looking for absolute cell reference.
Now the returns is : D1
So you only need to remove the 1 to get the colonne lettre if you need, for example with :
=Substitute(Address(1,4,4),"1","")
That returns D.
This is a way to convert column letters to column numbers.
=mmult(ArrayFormula(ifna(vlookup(substitute(mid(rept(" ",3-len(filter(A:A,A:A<>"")))&filter(A:A,A:A<>""),sequence(1,3),1)," ",""),{char(64+sequence(26)),sequence(26)},2,0),0)*{676,26,1}),sequence(3,1,1,0))
Screenshot of the Google Sheet
Don't use 26 radix. Like below.
const n2c = n => {
if (!n) return '';
// Column number to 26 radix. From 0 to p.
// Column number starts from 1. Subtract 1.
return [...(n-1).toString(26)]
// to ascii number
.map(c=>c.charCodeAt())
.map((c,i,arr)=> {
// last digit
if (i===arr.length-1) return c;
// 10 -> p
else if (arr.length - i > 2 && arr[i+1]===48) return c===49 ? null : c-2;
// 0 -> p
else if (c===48) return 112;
// a-1 -> 9
else if (c===97) return 57;
// Subtract 1 except last digit.
// Look at 10. This should be AA not BA.
else return c-1;
})
.filter(c=>c!==null)
// Convert with the ascii table. [0-9]->[A-J] and [a-p]->[K-Z]
.map(a=>a>96?a-22:a+17)
// to char
.map(a=>String.fromCharCode(a))
.join('');
};
const table = document.createElement('table');
table.border = 1;
table.cellPadding = 3;
for(let i=0, row; i<1380; i++) {
if (i%5===0) row = table.insertRow();
row.insertCell().textContent = i;
row.insertCell().textContent = n2c(i);
}
document.body.append(table);
td:nth-child(odd) { background: gray; color: white; }
td:nth-child(even) { background: silver; }
Simple typescript functional approach
const integerToColumn = (integer: number): string => {
const base26 = (x: number): string =>
x < 26
? String.fromCharCode(65 + x)
: base26((x / 26) - 1) + String.fromCharCode(65 + x % 26)
return base26(integer)
}
console.log(integerToColumn(0)) // "A"
console.log(integerToColumn(1)) // "B"
console.log(integerToColumn(2)) // "C"
Here is a general version written in Scala. It's for a column index start at 0 (it's simple to modify for an index start at 1):
def indexToColumnBase(n: Int, base: Int): String = {
require(n >= 0, s"Index is non-negative, n = $n")
require(2 <= base && base <= 26, s"Base in range 2...26, base = $base")
def digitFromZeroToLetter(n: BigInt): String =
('A' + n.toInt).toChar.toString
def digitFromOneToLetter(n: BigInt): String =
('A' - 1 + n.toInt).toChar.toString
def lhsConvert(n: Int): String = {
val q0: Int = n / base
val r0: Int = n % base
val q1 = if (r0 == 0) (n - base) / base else q0
val r1 = if (r0 == 0) base else r0
if (q1 == 0)
digitFromOneToLetter(r1)
else
lhsConvert(q1) + digitFromOneToLetter(r1)
}
val q: Int = n / base
val r: Int = n % base
if (q == 0)
digitFromZeroToLetter(r)
else
lhsConvert(q) + digitFromZeroToLetter(r)
}
def indexToColumnAtoZ(n: Int): String = {
val AtoZBase = 26
indexToColumnBase(n, AtoZBase)
}
In PowerShell:
function convert-IndexToColumn
{
Param
(
[Parameter(Mandatory)]
[int]$col
)
"$(if($col -gt 26){[char][int][math]::Floor(64+($col-1)/26)})$([char](65 + (($col-1) % 26)))"
}
Here is a 0-indexed JavaScript function without a maximum value, as it uses a while-loop:
function indexesToA1Notation(row, col) {
const letterCount = 'Z'.charCodeAt() - 'A'.charCodeAt() + 1;
row += 1
let colName = ''
while (col >= 0) {
let rem = col % letterCount
colName = String.fromCharCode('A'.charCodeAt() + rem)
col -= rem
col /= letterCount
}
return `${colName}${row}`
}
//Test runs:
console.log(indexesToA1Notation(0,0)) //A1
console.log(indexesToA1Notation(37,9)) //J38
console.log(indexesToA1Notation(5,747)) //ABT6
I wrote it for a web-app, so I'm not 100% sure it works in Google Apps Script, but it is normal JavaScript, so I assume it will.
For some reason I cant get the snippet to show its output, but you can copy the code to some online playground if you like
Here's a zero-indexed version (in Python):
letters = []
while column >= 0:
letters.append(string.ascii_uppercase[column % 26])
column = column // 26 - 1
return ''.join(reversed(letters))
I would like to sort an array of strings (in JavaScript) such that groups of digits within the strings are compared as integers not strings. I am not worried about signed or floating point numbers.
For example, the result should be ["a1b3","a9b2","a10b2","a10b11"] not ["a1b3","a10b11","a10b2","a9b2"]
The easiest way to do this seems to be splitting each string on boundaries around groups of digits. Is there a pattern I can pass to String.split to split on character boundaries without removing any characters?
"abc11def22ghi".split(/?/) = ["abc","11","def","22","ghi"];
Or is there another way to compare strings that does not involve splitting them up, perhaps by padding all groups of digits with leading zeros so they are the same length?
"aa1bb" => "aa00000001bb", "aa10bb" => "aa00000010bb"
I am working with arbitrary strings, not strings that have a specific arrangement of digit groups.
I like the /(\d+)/ one liner from Gaby to split the array. How backwards compatible is that?
The solutions that parse the strings once in a way that can be used to rebuild the originals are much more efficient that this compare function. None of the answers handle some strings starting with digits and others not, but that would be easy enough to remedy and was not explicit in the original question.
["a100", "a20", "a3", "a3b", "a3b100", "a3b20", "a3b3", "!!", "~~", "9", "10", "9.5"].sort(function (inA, inB) {
var result = 0;
var a, b, pattern = /(\d+)/;
var as = inA.split(pattern);
var bs = inB.split(pattern);
var index, count = as.length;
if (('' === as[0]) === ('' === bs[0])) {
if (count > bs.length)
count = bs.length;
for (index = 0; index < count && 0 === result; ++index) {
a = as[index]; b = bs[index];
if (index & 1) {
result = a - b;
} else {
result = !(a < b) ? (a > b) ? 1 : 0 : -1;
}
}
if (0 === result)
result = as.length - bs.length;
} else {
result = !(inA < inB) ? (inA > inB) ? 1 : 0 : -1;
}
return result;
}).toString();
Result: "!!,9,9.5,10,a3,a3b,a3b3,a3b20,a3b100,a20,a100,~~"
Another variant is to use an instance of Intl.Collator with the numeric option:
var array = ["a100", "a20", "a3", "a3b", "a3b100", "a3b20", "a3b3", "!!", "~~", "9", "10", "9.5"];
var collator = new Intl.Collator([], {numeric: true});
array.sort((a, b) => collator.compare(a, b));
console.log(array);
I think this does what you want
function sortArray(arr) {
var tempArr = [], n;
for (var i in arr) {
tempArr[i] = arr[i].match(/([^0-9]+)|([0-9]+)/g);
for (var j in tempArr[i]) {
if( ! isNaN(n = parseInt(tempArr[i][j])) ){
tempArr[i][j] = n;
}
}
}
tempArr.sort(function (x, y) {
for (var i in x) {
if (y.length < i || x[i] < y[i]) {
return -1; // x is longer
}
if (x[i] > y[i]) {
return 1;
}
}
return 0;
});
for (var i in tempArr) {
arr[i] = tempArr[i].join('');
}
return arr;
}
alert(
sortArray(["a1b3", "a10b11", "a10b2", "a9b2"]).join(",")
);
Assuming you want to just do a numeric sort by the digits in each array entry (ignoring the non-digits), you can use this:
function sortByDigits(array) {
var re = /\D/g;
array.sort(function(a, b) {
return(parseInt(a.replace(re, ""), 10) - parseInt(b.replace(re, ""), 10));
});
return(array);
}
It uses a custom sort function that removes the digits and converts to a number each time it's asked to do a comparison. You can see it work here: http://jsfiddle.net/jfriend00/t87m2/.
Use this compare function for sorting...
function compareLists(a, b) {
var alist = a.split(/(\d+)/), // Split text on change from anything
// to digit and digit to anything
blist = b.split(/(\d+)/); // Split text on change from anything
// to digit and digit to anything
alist.slice(-1) == '' ? alist.pop() : null; // Remove the last element if empty
blist.slice(-1) == '' ? blist.pop() : null; // Remove the last element if empty
for (var i = 0, len = alist.length; i < len; i++) {
if (alist[i] != blist[i]){ // Find the first non-equal part
if (alist[i].match(/\d/)) // If numeric
{
return +alist[i] - +blist[i]; // Compare as number
} else {
return alist[i].localeCompare(blist[i]); // Compare as string
}
}
}
return true;
}
Syntax
var data = ["a1b3", "a10b11", "b10b2", "a9b2", "a1b20", "a1c4"];
data.sort(compareLists);
alert(data);
There is a demo at http://jsfiddle.net/h9Rqr/7/.
Here's a more complete solution that sorts according to both letters and numbers in the strings
function sort(list) {
var i, l, mi, ml, x;
// copy the original array
list = list.slice(0);
// split the strings, converting numeric (integer) parts to integers
// and leaving letters as strings
for( i = 0, l = list.length; i < l; i++ ) {
list[i] = list[i].match(/(\d+|[a-z]+)/g);
for( mi = 0, ml = list[i].length; mi < ml ; mi++ ) {
x = parseInt(list[i][mi], 10);
list[i][mi] = !!x || x === 0 ? x : list[i][mi];
}
}
// sort deeply, without comparing integers as strings
list = list.sort(function(a, b) {
var i = 0, l = a.length, res = 0;
while( res === 0 && i < l) {
if( a[i] !== b[i] ) {
res = a[i] < b[i] ? -1 : 1;
break;
}
// If you want to ignore the letters, and only sort by numbers
// use this instead:
//
// if( typeof a[i] === "number" && a[i] !== b[i] ) {
// res = a[i] < b[i] ? -1 : 1;
// break;
// }
i++;
}
return res;
});
// glue it together again
for( i = 0, l = list.length; i < l; i++ ) {
list[i] = list[i].join("");
}
return list;
}
I needed a way to take a mixed string and create a string that could be sorted elsewhere, so that numbers sorted numerically and letters alphabetically. Based on answers above I created the following, which pads out all numbers in a way I can understand, wherever they appear in the string.
function padAllNumbers(strIn) {
// Used to create mixed strings that sort numerically as well as non-numerically
var patternDigits = /(\d+)/g; // This recognises digit/non-digit boundaries
var astrIn = strIn.split( patternDigits ); // we create an array of alternating digit/non-digit groups
var result = "";
for (var i=0;i<astrIn.length; i++) {
if (astrIn[i] != "") { // first and last elements can be "" and we don't want these padded out
if (isNaN(astrIn[i])) {
result += astrIn[i];
} else {
result += padOneNumberString("000000000",astrIn[i]);
}
}
}
return result;
}
function padOneNumberString(pad,strNum,left) {
// Pad out a string at left (or right)
if (typeof strNum === "undefined") return pad;
if (typeof left === "undefined") left = true;
var padLen = pad.length - (""+ strNum).length;
var padding = pad.substr(0,padLen);
return left? padding + strNum : strNum + padding;
}
Sorting occurs from left to right unless you create a custom algorithm. Letters or digits are compared digits first and then letters.
However, what you want to accomplish as per your own example (a1, a9, a10) won’t ever happen. That would require you knowing the data beforehand and splitting the string in every possible way before applying the sorting.
One final alternative would be:
a) break each and every string from left to right whenever there is a change from letter to digit and vice versa; &
b) then start the sorting on those groups from right-to-left. That will be a very demanding algorithm. Can be done!
Finally, if you are the generator of the original "text", you should consider NORMALIZING the output where a1 a9 a10 could be outputted as a01 a09 a10. This way you could have full control of the final version of the algorithm.
I use jQuery to get the browser version like this:
var x = $.browser.version;
I get a string like this: 1.9.1.1
Now, I want to do an evaluation so if x is >= 1.9.1 then do some stuff. Unfortunately, with multiple decimal points, I cannot do a parseFloat() because it converts 1.9.1.1 to simply 1.9, and the if evaluation would match a 1.9.0 version (which I do not want).
Has someone figured out a way to accomplish turning a version number (with multiple decimals) into something that can be used as a number for evaluation (or some other way to accomplish what I am trying to do here)?
Thanks -
You could do something with string.split and then do a digit by digit comparison
// arr[0] = 1
// arr[1] = 9
// arr[2] = 1
// arr[3] = 1
var arr = ($.browser.version).split('.');
The following is taken from this post
This is a function that will parse your version string and give you back a JSON object
function parseVersionString (str) {
if (typeof(str) != 'string') { return false; }
var x = str.split('.');
// parse from string or default to 0 if can't parse
var maj = parseInt(x[0]) || 0;
var min = parseInt(x[1]) || 0;
var bld = parseInt(x[2]) || 0;
var rev = parseInt(x[3]) || 0;
return {
major: maj,
minor: min,
build: bld,
revision: rev
}
}
Then you could use the following syntax
var version = parseVersionString($.browser.version);
// version.major == 1
// version.minor == 9
// version.build == 1
// version.revision == 1
Here's another version of versionCmp():
function versionCmp(v1, v2) {
v1 = String(v1).split('.');
v2 = String(v2).split('.');
var diff = 0;
while((v1.length || v2.length) && !diff)
diff = (+v1.shift() || 0) - (+v2.shift() || 0);
return (diff > 0) - (diff < 0);
}
Another possibility would be to assign a numeric value to each version number:
function valueOfVersion(ver) {
ver = String(ver).split('.');
var value = 0;
for(var i = ver.length; i--;)
value += ver[i] / Math.pow(2, i * 8) || 0;
return value;
}
This only works if each digit is less than 256 (because of the hard-coded divisor) and has a limited precision (ie the version strings can't get arbitrarily long).
You need to treat each portion of the string as a seperate integer, so split and iterate, and cmp:
// perform cmp(a, b)
// -1 = a is smaller
// 0 = equal
// 1 = a is bigger
function versionCmp(a, b) {
a = a.split(".");
b = b.split(".");
for(var i=0; i < a.length; i++) {
av = parseInt(a[i]);
bv = parseInt(b[i]);
if (av < bv) {
return -1;
} else if (av > bv) {
return 1;
}
}
return 0;
}
console.log(versionCmp("1.1.2.3", "1.2.1.0")); // should be -1
console.log(versionCmp("1.19.0.1", "1.2.0.4")); // should be 1
console.log(versionCmp("1.2.3.4", "1.2.3.4")); // should be 0
You could remove all dots and then parse it as an integer.
Take note tho, this solution doesn't work in the long term.