我想对数组进行排序。
数组中的项目具有关系。
例如。 list[5]
应为before list[9]
,after list[3]
样本中的预期值仅用于测试。它确实不存在。
这是一个包含关系和预期索引的示例数组。
var list = [{
id: '0001',
before: '0002',
expected: 0
}, {
id: '0002',
before: '0007',
after: '0001',
expected: 4
}, {
id: '0003',
before: '0006',
after: '0001',
expected: 2
}, {
id: '0004',
after: '0007',
expected: 11
}, {
id: '0005',
before: '0003',
after: '0001',
expected: 1
}, {
id: '0006',
before: '0002',
after: '0001',
expected: 3
}, {
id: '0007',
before: '00010',
after: '0002',
expected: 5
}, {
id: '0008',
before: '00012',
after: '0007',
expected: 9
}, {
id: '0009',
before: '0011',
after: '0001',
expected: 7
}, {
id: '0010',
before: '0009',
after: '0007',
expected: 6
}, {
id: '0011',
before: '0008',
after: '0001',
expected: 8
}, {
id: '0012',
before: '0004',
after: '0010',
expected: 10
}];
答案 0 :(得分:3)
JanTengň扩展:这是一个例子,说明你的问题(在有向无环图中找到一个总订单)是如何无法解决的:
var list = [{ id:A, before:B }, // "First" in total order
{ id:B, after:A }, // "Last" in total order
{ id:C, after:A, before:B },
{ id:D, after:A, before:B }];
C
和D
之间没有总排序:您可以将它们称为相等,但是如果不是D
,您有一个列表D0 -> D1 -> D2
会怎么样?
根据您的问题类型,这可以通过预处理来解决:将2级节点的路径减少到单个节点,并通过2级相同的节点调用并行路径(也可以简化为单个节点)。在这样的预处理结束时,你会留下一棵树 - 在你的情况下应该是一个列表/路径(或单个节点,因为你减少了2级顶点的路径)。
请注意,“之前”和“之后”的信息是多余的:您只需要其中一个。声明“A在B之前”等同于“B在A之后”并且您的非循环图只需要反映“之前”或“之后”的方向。您正在寻找的是包含所有节点的图形路径(因为它们是在“之前”或“之后”指示您自动从头到尾按顺序获取路径 - 如果存在这样的路径):
// First build the adjacency list for "X before Y"
var befores = { };
for(var i = 0; i < count; ++i)
befores[list[i].id] = null;
function insert(before, after) {
if(!before || !after)
return;
befores[before] = { next:befores[before], id:after };
}
for(var i = 0; i < count; ++i) {
var item = list[i];
insert(item.after, item.id); // "X after Y" -> "Y before X"
insert(item.id, item.before);
}
// build complete the graph as a lookup table
// id is "before" all elements in lookup[id]
var lookup = { };
for(var i = 0; i < count; ++i) {
var id = list[i].id;
var beforeList = [id];
var beforeSet = { };
beforeSet[id] = 1;
// use "A before B" and "B before C" to gain "A before C"
for(var j = 0; j < beforeList.length; ++j) {
for(var item = befores[beforeList[j]]; item != null; item = item.next) {
if(!beforeSet[item.id]) {
beforeList.push(item.id);
beforeSet[item.id] = 1;
}
}
}
// for our comparison we don't care if id is in beforeSet
lookup[id] = beforeSet;
// slice beforeList to get only the elements after id here:
//beforeList = beforeList.slice(1, beforeList.length);
}
// Now sort using the following
// a) if rhs is present in "before"-set of lhs then lhs < rhs
// b) if rhs is not present then rhs < lhs
// c) there is information missing from our graph if a) and b) for lhs analogous lead to a different conclusion!
list.sort(function(lhs, rhs) {
if(!lhs.after || !rhs.before) return -1;
if(!lhs.before || !rhs.after) return 1;
if(lhs.id === rhs.id) return 0;
// different ids guaranteed, doesn't matter if lookup[id] contains id itself
var result = lookup[lhs.id][rhs.id] ? -1 : 1;
// expect reversing lhs and rhs to get inverse result
var expected = lookup[rhs.id][lhs.id] ? 1 : -1;
if(result != expected) {
// alert: there is no adjacency information between lhs and rhs!
}
return result;
});
自己测试一下:
var list = [{
id: '0001', before: '0002', expected: 0}, {
id: '0002', before: '0007', after: '0001', expected: 4}, {
id: '0003', before: '0006', after: '0001', expected: 2}, {
id: '0004', after: '0007', expected: 11}, {
id: '0005', before: '0003', after: '0001', expected: 1}, {
id: '0006', before: '0002', after: '0001', expected: 3}, {
id: '0007', before: '0010', after: '0002', expected: 5}, {
id: '0008', before: '0012', after: '0007', expected: 9}, {
id: '0009', before: '0011', after: '0001', expected: 7}, {
id: '0010', before: '0009', after: '0007', expected: 6}, {
id: '0011', before: '0008', after: '0001', expected: 8}, {
id: '0012', before: '0004', after: '0010', expected: 10
}];
// re-used variable
var count = list.length;
var out = document.getElementById("out");
function toHTMLItem(item) {
var result = item.expected + " (";
if(item.after) result += item.after + " < ";
result += "<b>" + item.id + "</b>";
if(item.before) result += " < " + item.before;
result += ")";
return result;
}
function toHTMLList(list) {
var result = "<p>";
for(var i = 0; i < count; ++i) {
result += toHTMLItem(list[i]) + "<br>";
}
result += "</p>";
return result;
}
// out.innerHTML += toHTMLList(list);
var befores = { };
for(var i = 0; i < count; ++i)
befores[list[i].id] = null;
function insert(before, after) {
if(!before || !after)
return;
befores[before] = { next:befores[before], id:after };
}
for(var i = 0; i < count; ++i) {
var item = list[i];
insert(item.after, item.id);
insert(item.id, item.before);
}
function toHTMLTable(table, list) {
var result = "<p>";
var count = list.length;
for(var i = 0; i < count; ++i) {
var id = list[i].id;
result += id + " < ";
for(var item = table[id]; item != null; item = item.next) {
result += item.id + ", ";
}
result += "o<br>";
}
result += "</p>";
return result;
}
// out.innerHTML += toHTMLTable(befores, list);
// next build a lookup-table of a completed adjacency list
var lookup = { };
for(var i = 0; i < count; ++i) {
var id = list[i].id;
var beforeList = [id];
var beforeSet = { };
beforeSet[id] = 1;
// use "A before B" and "B before C" to gain "A before C"
for(var j = 0; j < beforeList.length; ++j) {
for(var item = befores[beforeList[j]]; item != null; item = item.next) {
if(!beforeSet[item.id]) {
beforeList.push(item.id);
beforeSet[item.id] = 1;
}
}
}
beforeList = beforeList.slice(1, beforeList.length);
beforeList.sort();
lookup[id] = beforeSet;
}
function toHTMLLookup(lookup, list) {
var result = "<p>";
for(var i = 0, imax = list.length; i < imax; ++i) {
var id = list[i].id;
var bs = lookup[id];
result += id + " < ";
for(var j = 0, jmax = imax; j < jmax; ++j) {
if(j == i) continue;
if(bs[list[j].id]) result += list[j].id + ", ";
}
result += "o<br>";
}
result += "</p>";
return result;
}
// out.innerHTML += toHTMLLookup(lookup, list);
// Search function in befores:
// a) if rhs is present in union of befores set lhs < rhs
// b) if rhs is not present in union of befores set rhs < lhs
list.sort((function() {
var enableAlert = true;
return function(lhs, rhs) {
if(!lhs.after || !rhs.before) return -1;
if(!lhs.before || !rhs.after) return 1;
if(lhs.id === rhs.id) return 0;
// different ids guaranteed, doesn't matter if lookup[id] contains id itself
var result = lookup[lhs.id][rhs.id] ? -1 : 1;
// expect reversing lhs and rhs to get inverse result
var expected = lookup[rhs.id][lhs.id] ? 1 : -1;
if(enableAlert && result != expected) {
// restrict to a single alert per execution
enableAlert = false;
out.innerHTML += "<p><b>ALERT</b> unresolved adjacency between " + lhs.id + " and " + rhs.id + "!</p>";
}
return result;
};
})());
// out.innerHTML += toHTMLList(list);
var error = count;
for(var i = 0; i < error; ++i) {
if(list[i].expected != i)
error = i;
}
if(error < count) {
out.innerHTML += "<h2>error!</h2><p>list[" + error + "] contains " + toHTMLItem(list[i]) + "</p>";
} else {
out.innerHTML += "<h2>success!</h2>";
}
// Finally print the output
out.innerHTML += toHTMLList(list);
<div id="out"/>
对于这个问题的一般主题,请考虑topological sorting。
答案 1 :(得分:1)
部分答案,评论太久了。您可能已经想到了这一点,我希望它可以帮助某人找到解决方案。
此功能在就地排序中非常有用:在数组中移动项目:
Object.defineProperty(Array.prototype,"move",{
value: function(from,to) {
var x = this.splice(from,1);
this.splice(to,0,x[0]);
}
});
var list = ['a','b','c','d','e'];
// move the index 1 (b) to position 3 (after d)
list.move(1,3); // acdbe
祝你好运。
答案 2 :(得分:0)
如果我理解这一点,expected
是你喜欢该元素的索引吗?在这种情况下,您可以将.sort()
与自定义排序功能结合使用。
list.sort(function (a, b) {
var ae = a.expected,
be = b.expected;
if (ae > be) return 1;
if (ae < be) return -1;
if (ae === be) return 0;
});
(如果您因牺牲可读性而牺牲时,可以缩短该块。)
答案 3 :(得分:0)
你想要的是一个自定义排序功能,正如Mike所说。我想你真正想要的是:
list.sort(function(a, b) {
if (/* a before b, b after a */) {
return -1;
} else if (/* b before a, a after b */) {
return 1;
}
return 0;
});
基本上,您在自定义排序函数中返回的内容决定了两个参数之间的关系。返回-1表示第一个项目在之前,+1表示第二个项目在之前,0表示没有可用的关系。
对于您的“a b之后,b之前的b”标准,也许可以尝试a.id === b.before|| b.id === a.after
,而a后面的反向为。
根据您的确切列表,这可能会或可能不会按预期返回 - 如果您有大量丢失或冲突的数据,排序可能最终无法完全按预期返回。
一般来说,我会避免同时拥有“之前”和“之后”信息 - 你应该只需要一个或另一个,并且两者都可以引入更多的冲突,而不仅仅是单一的关系。