我有一个对象数组,我正在尝试按字母数字排序它们,请看下面的例子:
var objs = {
'obj1': {'name': 'Object21'},
'obj2': {'name': 'Object140'},
'obj3': {'name': 'Object28'},
'obj4': {'name': 'Object251'}
};
调用_.sortBy(objs, function(obj) { return obj.name; }
时输出为:
如何使用Underscore以字母顺序排序?我知道我可以只使用名称创建一个单独的数组,但有没有更好的方法使用Underscore这样做而不创建额外的变量?
答案 0 :(得分:6)
我已经设法使用谷歌找到了解决方案:-)这里是我将来用于那些需要它的人,它实际上被称为“自然分类”
通过致电_.sortByNat(objs, function(obj) { return obj.name; })
/*
* Backbone.js & Underscore.js Natural Sorting
*
* @author Kevin Jantzer <https://gist.github.com/kjantzer/7027717>
* @since 2013-10-17
*
* NOTE: make sure to include the Natural Sort algorithm by Jim Palmer (https://github.com/overset/javascript-natural-sort)
*/
// add _.sortByNat() method
_.mixin({
sortByNat: function(obj, value, context) {
var iterator = _.isFunction(value) ? value : function(obj){ return obj[value]; };
return _.pluck(_.map(obj, function(value, index, list) {
return {
value: value,
index: index,
criteria: iterator.call(context, value, index, list)
};
}).sort(function(left, right) {
var a = left.criteria;
var b = right.criteria;
return naturalSort(a, b);
}), 'value');
}
});
/*
* Natural Sort algorithm for Javascript - Version 0.7 - Released under MIT license
* Author: Jim Palmer (based on chunking idea from Dave Koelle)
* https://github.com/overset/javascript-natural-sort
*/
function naturalSort (a, b) {
var re = /(^-?[0-9]+(\.?[0-9]*)[df]?e?[0-9]?$|^0x[0-9a-f]+$|[0-9]+)/gi,
sre = /(^[ ]*|[ ]*$)/g,
dre = /(^([\w ]+,?[\w ]+)?[\w ]+,?[\w ]+\d+:\d+(:\d+)?[\w ]?|^\d{1,4}[\/\-]\d{1,4}[\/\-]\d{1,4}|^\w+, \w+ \d+, \d{4})/,
hre = /^0x[0-9a-f]+$/i,
ore = /^0/,
i = function(s) { return naturalSort.insensitive && (''+s).toLowerCase() || ''+s },
// convert all to strings strip whitespace
x = i(a).replace(sre, '') || '',
y = i(b).replace(sre, '') || '',
// chunk/tokenize
xN = x.replace(re, '\0$1\0').replace(/\0$/,'').replace(/^\0/,'').split('\0'),
yN = y.replace(re, '\0$1\0').replace(/\0$/,'').replace(/^\0/,'').split('\0'),
// numeric, hex or date detection
xD = parseInt(x.match(hre)) || (xN.length != 1 && x.match(dre) && Date.parse(x)),
yD = parseInt(y.match(hre)) || xD && y.match(dre) && Date.parse(y) || null,
oFxNcL, oFyNcL;
// first try and sort Hex codes or Dates
if (yD)
if ( xD < yD ) return -1;
else if ( xD > yD ) return 1;
// natural sorting through split numeric strings and default strings
for(var cLoc=0, numS=Math.max(xN.length, yN.length); cLoc < numS; cLoc++) {
// find floats not starting with '0', string or 0 if not defined (Clint Priest)
oFxNcL = !(xN[cLoc] || '').match(ore) && parseFloat(xN[cLoc]) || xN[cLoc] || 0;
oFyNcL = !(yN[cLoc] || '').match(ore) && parseFloat(yN[cLoc]) || yN[cLoc] || 0;
// handle numeric vs string comparison - number < string - (Kyle Adams)
if (isNaN(oFxNcL) !== isNaN(oFyNcL)) { return (isNaN(oFxNcL)) ? 1 : -1; }
// rely on string comparison if different types - i.e. '02' < 2 != '02' < '2'
else if (typeof oFxNcL !== typeof oFyNcL) {
oFxNcL += '';
oFyNcL += '';
}
if (oFxNcL < oFyNcL) return -1;
if (oFxNcL > oFyNcL) return 1;
}
return 0;
}
// extend Array to have a natural sort
Array.prototype.sortNat = function(){
return Array.prototype.sort.call(this, naturalSort)
}
答案 1 :(得分:1)
您将需要创建自己的迭代器函数然后使用它,您无法使用迭代器函数实际执行此操作,但您可以接近它:
var objs = {
'obj1': {'name': 'Object21'},
'obj2': {'name': 'Object140'},
'obj3': {'name': 'Object28'},
'obj4': {'name': 'AnObject251'}
};
_.sortBy(objs, function(obj) {
var cc = [], s = obj.name;
for(var i = 0, c; c = s.charAt(i); i++)
c == +c ? cc.push(+c) : cc.push(c.charCodeAt(0));
return +cc.join('');
});
> Object21
Object28
Object140
AnObject251
&#34; AnObject251&#34;由于它的长度,它排在最后一个位置。
答案 2 :(得分:1)
您需要Alphanum sorting algorithm和优雅的implementation in JavaScript:
function alphanum(a, b) {
function chunkify(t) {
var tz = [], x = 0, y = -1, n = 0, i, j;
while (i = (j = t.charAt(x++)).charCodeAt(0)) {
var m = (i == 46 || (i >=48 && i <= 57));
if (m !== n) {
tz[++y] = "";
n = m;
}
tz[y] += j;
}
return tz;
}
var aa = chunkify(a);
var bb = chunkify(b);
for (x = 0; aa[x] && bb[x]; x++) {
if (aa[x] !== bb[x]) {
var c = Number(aa[x]), d = Number(bb[x]);
if (c == aa[x] && d == bb[x]) {
return c - d;
} else return (aa[x] > bb[x]) ? 1 : -1;
}
}
return aa.length - bb.length;
}