如何通過Lua的「分數」和「索引」對內部表進行排序？

Question

我已經存儲在變量T以下的Lua表：

{ 
    ["mn"] = { ["index"] = 7, ["key"] = "mn", ["score"] = 0 }, 
    ["kl"] = { ["index"] = 6, ["key"] = "kl", ["score"] = .4 }, 
    ["ef"] = { ["index"] = 3, ["key"] = "ef", ["score"] = .3 }, 
    ["ab"] = { ["index"] = 1, ["key"] = "ab", ["score"] = 0 }, 
    ["cd"] = { ["index"] = 2, ["key"] = "cd", ["score"] = .1 }, 
    ["gh"] = { ["index"] = 4, ["key"] = "gh", ["score"] = 0 }, 
    ["ij"] = { ["index"] = 5, ["key"] = "ij", ["score"] = .2 } 
} 

我要排序的所有T表以下列方式內部表：
1.表具有較高score都放在頂端。
2.等於score的表格按其index排序。

{ 
    [1] = { ["index"] = 6, ["key"] = "kl", ["score"] = .4 }, -- highest "score" 
    [2] = { ["index"] = 3, ["key"] = "ef", ["score"] = .3 }, 
    [3] = { ["index"] = 5, ["key"] = "ij", ["score"] = .2 }, 
    [4] = { ["index"] = 2, ["key"] = "cd", ["score"] = .1 }, 
    [5] = { ["index"] = 1, ["key"] = "ab", ["score"] = 0 }, -- lowest "score", lowest "index" 
    [6] = { ["index"] = 4, ["key"] = "gh", ["score"] = 0 }, -- when scores are the same, sort by their "index" instead 
    [7] = { ["index"] = 7, ["key"] = "mn", ["score"] = 0 } -- lowest "score", highest "index" 
} 

如何做到這一點的Lua表排序：

所以，整理後，下面的順序表應在輸出產生的？

Answer 1

您需要先將表中的哈希值轉換爲表格，然後使用自定義排序函數對該表格的元素進行排序，該函數按score（降序）排序，然後按index（升序）排序相同的得分。

像這樣的東西應該工作：

local hash = { 
    ["mn"] = { ["index"] = 7, ["key"] = "mn", ["score"] = 0 }, 
    ["kl"] = { ["index"] = 6, ["key"] = "kl", ["score"] = .4 }, 
    ["ef"] = { ["index"] = 3, ["key"] = "ef", ["score"] = .3 }, 
    ["ab"] = { ["index"] = 1, ["key"] = "ab", ["score"] = 0 }, 
    ["cd"] = { ["index"] = 2, ["key"] = "cd", ["score"] = .1 }, 
    ["gh"] = { ["index"] = 4, ["key"] = "gh", ["score"] = 0 }, 
    ["ij"] = { ["index"] = 5, ["key"] = "ij", ["score"] = .2 } 
} 
local tbl = {} 
for _,v in pairs(hash) do 
    table.insert(tbl, v) 
end 
table.sort(tbl, function(a,b) 
    return a.score > b.score or a.score == b.score and a.index < b.index 
    end) 

Answer 2

在lua中，表格包含兩個數據結構：數組和字典。

排序裝置排序陣列，其中每個元件與數字索引和索引相關聯的是連續的是：1,2,3 ...

您初始表實際上是一個字典 - 每個條目有一個任意關鍵的關鍵（在你的情況下，那些是字符串）。

因此，你所描述的實際上並不是一個排序任務，你最終需要一個不同類型的表。

table.sort適用於lua表的數組部分，也就是那些索引從1開始到第一個零結束的元素。

a={s=3,['r']=3, 5,3,2, nil,21} 
       |these| 
       |ones | 

所以，首先要創建一個數組和排序一句：

local sorted={} 
for k,v in pairs(T) do 
    table.insert(sorted,v) 
end 
table.sort(sorted,function(a,b) 
    --your code here 
    --function should return true if element `a` from the array `sorted` 
    --must be higher (to the left of) `b` 
end) 

或者，你可以在條目存儲在無論是在字典和陣列部分相同的表中，table.sort功能會忽略字典。但使用pairs循環表並不明智，並且同時添加新元素。因此，慣用的方式仍然涉及中間複製。