For my new project on World of Raids I have to implement a table sorting. The browser not stable sorting and the faster sorting trick add difficulty to the task.

### String Comparison

As mentionned in the Speed Up Javascript Sort() article, using a string as a key to represent each element is faster than using a custom sort function in most browsers.

Tables are commonly made of two data types: Numbers and Strings. Strings are the final representation so no work is involved there so our work will be focused on Numbers. What we want is to fit a number into a string. A string is a succession of characters, each one able to hold 256 values. So, we can see the problem as encoding the number into a base 256 which is trivial.

#### Padding

String comparison does not work exactly like we want it. It reads all the characters one by one and if they mismatch then it returns who is the highest. So for example `"10" < "5"`

because `'1' < '5'`

. We have to pad every numbers with `0`

at the beginning and become `"10" > "05"`

.

In this case we are adding one `0`

because 10 is 2 digits and 5 is only one. We won't know the values of other elements when making a comparison, so two solutions apply: either we iterate over all the elements and retrieve the highest, or we set an arbitrary maximum.

Characters | Maximum |
---|---|

1 | 256 |

2 | 65 536 |

3 | 16 777 216 |

4 | 4 294 967 296 |

With that table in mind, it is obvious that it would be a waste of time to inspect every element just to find the highest value. 1 or 2 characters will probably fit most uses but if you want to be safe, you will always be able to store your number into 4 characters which is acceptable.

Here is the code to do the conversion:

function compareAsc(num, digits) { var s = ""; // Encode num in base 256 while (num !== 0) { s = String.fromCharCode((num % 256) | 0) + s; num = (num / 256) | 0; digits -= 1; } // Fill with 0 while (digits > 0) { s = String.fromCharCode(0) + s; digits -= 1; } return s; } |

For the sake of simplicity, this code is using string concatenation which requires to build up new strings and make copy several times. Since this is for small strings (up to 4 characters!) this is probably alright but one looking for performances could bench this with the use of an `Array`

with a `.concat()`

at the end to build the string.

Also note that numbers are stored as float in Javascript, so we are rounding (flooring to be exact) all the values using the `|0`

trick.

#### Descending Order

We have made the code for an ascending sort but we have to handle the other case. We cannot just revert the order of the sorting or adding a minus sign before the number. What we have to do is to do a 256-complement of the final result, in other terms: digit = 255 - digit. Let see an example for a base 10.

00 -> 99 01 -> 98 ... 54 -> 45 55 -> 44 56 -> 43 ... 98 -> 01 99 -> 00

You can see that all the numbers are now sorted in the opposite order.

Here is the associated code :

function compareDesc(num, digits) { var s = ""; // Encode num in base 256 and do the 256-complement while (num !== 0) { s = String.fromCharCode(255 - (num % 256) | 0) + s; num = (num / 256) | 0; digits -= 1; } while (digits >= 0) { s = String.fromCharCode(255) + s; digits -= 1; } return s; } |

#### Floats and Negatives

We have handled all the positive integers but they are not alone. In order to deal with the negative numbers you can apply the 2-complement to the number. This is how it is done in computer arithmetic.

To handle floats, it is possible to treat them as integers by multiplying by 10^(number of decimal displayed). So `12.34`

is translated to `1234`

and `11`

to `1100`

and it's working just fine. You will probably need more than 4 characters if you are using huge numbers though.

### Stable Sorting

Implementing a table sorting requires the sorting algorithm to be stable (values that have the same key keep their original order). This property allows people to sort by multiple columns, the last one having the highest weight.

However, browser implementations of the `Array.sort()`

are not stable. From there we have two solutions, implementing a stable sort in javascript or find a way to emulate that behaviour. Since Javascript is slow in many browsers, implementing such a computation heavy algorithm in Javascript is going to slow down things and requires more efforts than I am willing to spend on this project.

The solution instead is to understand what the stable property means and find a way to code it.

Stable sorting algorithms maintain the relative order of records with equal keys. [...] Whenever there are two records (let's say R and S) with the same key, and R appears before S in the original list, then R will always appear before S in the sorted list.

We can easily transpose that sentence into the sorting function:

if (R.key == S.key) return compare(R.position, S.position) |

We need to know one more thing to do the computation: the position of each element in the list before sorting. This is straightforward to do, you have to iterate over all the elements and update their position field. This requires *n* steps which is acceptable.

Here is the full Javascript code

var sort = function (a, b) { if (a.key === b.key) return a.position - b.position; if (a.key < b.key) return -1; return 1; }; |

#### String comparison

To apply this concept to the string comparison, you can append the position at the end of the string. `"Key"`

now being `"Key1"`

, `"Key2"`

and so on. When the keys are equal, the position is used for the comparison instead. We can reuse the `compareAsc()`

function to encode the position using the minimal size.

This works well for fixed-size strings like the representation of numbers. However common strings do not share that property and a problem arise when two strings share a common base.

Take "Method" and "Methodology". It is obvious that `"Method" < "Methodology"`

. Now append the position: `"Method" + chr(250) > "Method`

the comparison changed because **ology**" + chr(251)`chr(250) > 'o'`

. (Where `chr = String.fromCharCode'`

).

The way to fix this problem is to add a `chr(0)`

before the position. Considering that a normal string never contains a `chr(0)`

, the `chr(0)`

will always be lesser than any character of the string and therefore stop the comparison. The longer string always wins!

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