What’s the problem?
When a programmer is given the task of sorting file names in a list, it might be tempting to sort using something like std::sort(). The problem with that is: std::sort() sorts alphabetically. Suppose we have a list of file like this:
chapter1.txt
chapter10.txt
chapter2.txt
chapter3p1.txt
chapter3p2.txt
chapter3p10.txt
chapter11.txt
chapter20.txt
If we sort it alphabetically, aka, using std::sort() (or Array.prototype.sort() for JavaScript), what would happen?
chapter1.txt
chapter10.txt
chapter11.txt
chapter2.txt # we expect chapter 2 to be in front of chapter10
chapter20.txt
chapter3p1.txt
chapter3p10.txt
chapter3p2.txt
We see that chapter10 and chapter11 have been placed before chapter2. This make sense for an alphabetically sorted list. Both 10 and 11 is smaller than 2 since the first digit 1 is smaller than 2. However, it does not follow our intuition. As a human, we expect 2 to be bigger than 10, not the other way around.
Natural sort
This is where the algorithm of natural sort (sometimes called alphanumerical sort) comes in handy.
Natural sort order is an ordering of strings in alphabetical order, except that multi-digit numbers are treated atomically, i.e., as if they were a single character. Natural sort order has been promoted as being more human-friendly (“natural”) than the machine-oriented pure alphabetical order.
Pseudo-code for natural sort:
代码语言：javascript复制# Preprocessing
For every file name in list:
processedName = []
Scan for every character c in file name:
If consecutive digits are found:
Merge consecutive numerical characters into a single number
processedName.Append(number)
Else:
# non-numerical character
processedName.Append(c)
list[current] = processedName # use the new name as name for comparison
Sort list:
Comparing file name x and y:
For every cooresponding [xi in x] and [yi in y]:
If xi == yi:
Continue
If both xi and yi are numbers:
If xi < yi: x is smaller
Else: y is smaller
Else If both xi and yi are strings:
If xi < yi: x is smaller
Else: y is smaller
Else: # One of xi or yi is a number and the other is string
If xi is number: x is smaller
Else: y is smaller
# the list is now sorted naturally (or alphanumerically)!
After applying the algorithm, the list should look like this:
代码语言：javascript复制chapter1.txt
chapter2.txt
chapter3p1.txt
chapter3p2.txt
chapter3p10.txt
chapter10.txt
chapter11.txt
chapter20.txt
The number rigit after chapter is sorted correctly as well as the second number after chapter3p.
External resources
This blog only described a simplified version of natural sort. Some file manager might ignore leading and trailing spaces when sorting. Multiple consecutive spaces might be considered as one single space as well. However the overall algorithm did not change and all of these features are quite trivial to implement.
Checkout these resources: https://blog.codinghorror.com/sorting-for-humans-natural-sort-order/ https://rosettacode.org/wiki/Natural_sorting
Implementation in JavaScript
代码语言：javascript复制// natural sort algorithm in JavaScript by Miigon.
// 2021-03-30
//
// GitHub: https://github.com/miigon/
function natSort(arr){
return arr.map(v=>{ // split string into number/ascii substrings
let processedName = []
let str = v
for(let i=0;i<str.length;i ) {
let isNum = Number.isInteger(Number(str[i]));
let j;
for(j=i 1;j<str.length;j ) {
if(Number.isInteger(Number(str[j]))!=isNum) {
break;
}
}
processedName.push(isNum ? Number(str.slice(i,j)) : str.slice(i,j));
i=j-1;
}
// console.log(processedName);
return processedName;
}).sort((a,b) => {
let len = Math.min(a.length,b.length);
for(let i=0;i<len;i ) {
if(a[i]!=b[i]) {
let isNumA = Number.isInteger(a[i]);
let isNumB = Number.isInteger(b[i]);
if(isNumA && isNumB) {
return a[i]-b[i];
} else if(isNumA) {
return -1;
} else if(isNumB) {
return 1;
} else {
return a[i]<b[i] ? -1 : 1 ;
}
}
}
// in case of one string being a prefix of the other
return a.length - b.length;
}).map(v => v.join(''));
}
let a = ['a2','a1b10z','b1a2','a1b10','a33','7','a3','a22','a1b2','abbbb','a1b1','aaaaa','a10','a1','10'];
console.log(natSort(a).join('n'))
/*
7
10
a1
a1b1
a1b2
a1b10
a1b10z
a2
a3
a10
a22
a33
aaaaa
abbbb
b1a2
*/
