Sort Array Elements (string with numbers), natural sort
JavascriptJquerySortingJavascript Problem Overview
I have an array like;
["IL0 Foo", "PI0 Bar", "IL10 Baz", "IL3 Bob says hello"]
And need to sort it so it appears like;
["IL0 Foo", "IL3 Bob says hello", "IL10 Baz", "PI0 Bar"]
I have tried a sort function;
function compare(a,b) {
if (a < b)
return -1;
if (a > b)
return 1;
return 0;
}
but this gives the order
["IL0 Foo", "IL10 Baz", "IL3 Bob says hello", "PI0 Bar"]
I have tried to think of a regex that will work but can't get my head around it.
If it helps the format will always be 2 letters, x amount of numbers, then any number of characters.
Javascript Solutions
Solution 1 - Javascript
This is called "natural sort" and can be implemented in JS like this:
function naturalCompare(a, b) {
var ax = [], bx = [];
a.replace(/(\d+)|(\D+)/g, function(_, $1, $2) { ax.push([$1 || Infinity, $2 || ""]) });
b.replace(/(\d+)|(\D+)/g, function(_, $1, $2) { bx.push([$1 || Infinity, $2 || ""]) });
while(ax.length && bx.length) {
var an = ax.shift();
var bn = bx.shift();
var nn = (an[0] - bn[0]) || an[1].localeCompare(bn[1]);
if(nn) return nn;
}
return ax.length - bx.length;
}
/////////////////////////
test = [
"img12.png",
"img10.png",
"img2.png",
"img1.png",
"img101.png",
"img101a.png",
"abc10.jpg",
"abc10",
"abc2.jpg",
"20.jpg",
"20",
"abc",
"abc2",
""
];
test.sort(naturalCompare)
document.write("<pre>" + JSON.stringify(test,0,3));
To sort in reverse order, just swap the arguments:
test.sort(function(a, b) { return naturalCompare(b, a) })
or simply
test = test.sort(naturalCompare).reverse();
Solution 2 - Javascript
You could use String#localeCompare
with options
>sensitivity
>
>Which differences in the strings should lead to non-zero result values. Possible values are:
>
>- "base"
: Only strings that differ in base letters compare as unequal. Examples: a ≠ b
, a = á
, a = A
.
>- "accent"
: Only strings that differ in base letters or accents and other diacritic marks compare as unequal. Examples: a ≠ b
, a ≠ á
, a = A
.
>- "case"
: Only strings that differ in base letters or case compare as unequal. Examples: a ≠ b
, a = á
, a ≠ A
.
>- "variant"
: Strings that differ in base letters, accents and other diacritic marks, or case compare as unequal. Other differences may also be taken into consideration. Examples: a ≠ b
, a ≠ á
, a ≠ A
.
>
> The default is "variant" for usage "sort"; it's locale dependent for usage "search".
>
>numeric
>
>Whether numeric collation should be used, such that "1" < "2" < "10". Possible values are true
and false
; the default is false
. This option can be set through an options property or through a Unicode extension key; if both are provided, the options
property takes precedence. Implementations are not required to support this property.
var array = ["IL0 Foo", "PI0 Bar", "IL10 Baz", "IL3 Bob says hello"];
array.sort(function (a,b) {
return a.localeCompare(b, undefined, { numeric: true, sensitivity: 'base' });
});
console.log(array);
Solution 3 - Javascript
var re = /([a-z]+)(\d+)(.+)/i;
var arr = ["IL0 Foo", "PI0 Bar", "IL10 Baz", "IL3 Bob says hello"];
var order = arr.sort( function(a,b){
var ma = a.match(re),
mb = b.match(re),
a_str = ma[1],
b_str = mb[1],
a_num = parseInt(ma[2],10),
b_num = parseInt(mb[2],10),
a_rem = ma[3],
b_rem = mb[3];
return a_str > b_str ? 1 : a_str < b_str ? -1 : a_num > b_num ? 1 : a_num < b_num ? -1 : a_rem > b_rem;
});
Solution 4 - Javascript
I liked georg's solution a lot, but I needed underscores ("_") to sort before numbers. Here's how I modified his code:
var chunkRgx = /(_+)|([0-9]+)|([^0-9_]+)/g;
function naturalCompare(a, b) {
var ax = [], bx = [];
a.replace(chunkRgx, function(_, $1, $2, $3) {
ax.push([$1 || "0", $2 || Infinity, $3 || ""])
});
b.replace(chunkRgx, function(_, $1, $2, $3) {
bx.push([$1 || "0", $2 || Infinity, $3 || ""])
});
while(ax.length && bx.length) {
var an = ax.shift();
var bn = bx.shift();
var nn = an[0].localeCompare(bn[0]) ||
(an[1] - bn[1]) ||
an[2].localeCompare(bn[2]);
if(nn) return nn;
}
return ax.length - bx.length;
}
/////////////////////////
test = [
"img12.png",
"img10.png",
"img2.png",
"img1.png",
"img101.png",
"img101a.png",
"abc10.jpg",
"abc10",
"abc2.jpg",
"20.jpg",
"20",
"abc",
"abc2",
"_abc",
"_ab_c",
"_ab__c",
"_abc_d",
"ab_",
"abc_",
"_ab_cd",
""
];
test.sort(naturalCompare)
document.write("<pre>" + JSON.stringify(test,0,3));
Solution 5 - Javascript
Pad numbers in string with leading zeros, then sort normally.
var naturalSort = function (a, b) {
a = ('' + a).replace(/(\d+)/g, function (n) { return ('0000' + n).slice(-5) });
b = ('' + b).replace(/(\d+)/g, function (n) { return ('0000' + n).slice(-5) });
return a.localeCompare(b);
}
var naturalSortModern = function (a, b) {
return ('' + a).localeCompare(('' + b), 'en', { numeric: true });
}
console.dir((["IL0 Foo", "PI0 Bar", "IL10 Baz", "IL3 Bob says hello"].sort(naturalSort)));
console.dir((["IL0 Foo", "PI0 Bar", "IL10 Baz", "IL3 Bob says hello"].sort(naturalSortModern)));
Solution 6 - Javascript
You could do a regex like this to get non-numeric and numeric parts of the string:
var s = "foo124bar23";
s.match(/[^\d]+|\d+/g)
returns: ["foo", "124" , "bar" , "23"]
Then in your compare function you can iterate through the parts of the two strings comparing them part-by-part. The first non-matching part determines the result of the overall comparison. For each part, check if the part starts with a digit and if so parse it as a number before doing the comparison.
Solution 7 - Javascript
Add one more alternative (why not):
var ary = ["IL0 Foo", "PI0 Bar", "IL10 Hello", "IL10 Baz", "IL3 Bob says hello"];
// break out the three components in to an array
// "IL10 Bar" => ['IL', 10, 'Bar']
function getParts(i){
i = i || '';
var parts = i.match(/^([a-z]+)([0-9]+)(\s.*)$/i);
if (parts){
return [
parts[1],
parseInt(parts[2], 10),
parts[3]
];
}
return []; // erroneous
}
ary.sort(function(a,b){
// grab the parts
var _a = getParts(a),
_b = getParts(b);
// trouble parsing (both fail = no shift, otherwise
// move the troubles element to end of the array)
if(_a.length == 0 && _b.length == 0) return 0;
if(_a.length == 0) return -1;
if(_b.length == 0) return 1;
// Compare letter portion
if (_a[0] < _b[0]) return -1;
if (_a[0] > _b[0]) return 1;
// letters are equal, continue...
// compare number portion
if (_a[1] < _b[1]) return -1;
if (_a[1] > _b[1]) return 1;
// numbers are equal, continue...
// compare remaining string
if (_a[2] < _b[2]) return -1;
if (_a[2] > _b[2]) return 1;
// strings are equal, continue...
// exact match
return 0;
});
Solution 8 - Javascript
Not pretty, but check the first two char codes. If all equal parse and compare the numbers:
var arr = ["IL0 Foo", "IL10 Baz", "IL3 Bob says hello", "PI0 Bar"];
arr.sort(function (a1, b1) {
var a = parseInt(a1.match(/\d+/g)[0], 10),
b = parseInt(b1.match(/\d+/g)[0], 10),
letterA = a1.charCodeAt(0),
letterB = b1.charCodeAt(0),
letterA1 = a1.charCodeAt(1),
letterB1 = b1.charCodeAt(1);
if (letterA > letterB) {
return 1;
} else if (letterB > letterA) {
return -1;
} else {
if (letterA1 > letterB1) {
return 1;
} else if (letterB1 > letterA1) {
return -1;
}
if (a < b) return -1;
if (a > b) return 1;
return 0;
}
});