应用排列以生成表格
我有以下键值对，我需要根据键找到所有组合。在图中，第一个表显示了所有键值对，第二个表显示了预期输出的提取。所有可能的组合都应该成为表格。
public ActionResult ViewSystemValues ()
{
var model = new List<ViewSystemValuesVM>();//Each item in Model consists of the Key Value pair mentioned in the question
var tclist = new List<TcSet>(); //Each item is the link to the Property that holds the key
var tcid = new List<int>(); //List of distinct keys in tclist
//For simplicity I have eliminated the code that would populate the Lists and the model with the description as comments
string[] myArray = new string[tclist.Count()];
var model1 = new List<String[]>();
int tclistcount = tclist.Count();
foreach(var x in model)
{
foreach (var item in model)
for(int i=0;i<tclistcount;i++)
{
if (tcid[i] == item.TcSetID)
myArray[i] = item.Value;
}
model1.Add(myArray);
}
}
方法
myArray对应于我的置换表中的一行。 model包含有效的键值对。我检查model和tcid中的密钥是否相同，可以在myArray中添加与tcid索引相同的值，以便获取相应的值tcid。
但我只得到最后一个可能的组合
colour fruit vegatable
yellow banana beans
yellow banana beans
yellow banana beans
yellow banana beans
yellow banana beans
yellow banana beans
2 个答案:
答案 0 :(得分：1)
这样做你想要的吗？
s.Length == 1此查询中的此基本案例，new [] { new [] { "onion", "potato", "tomato", "beans" } }基本上将new [] { new [] { "onion", }, new [] { "potato", }, new [] { "tomato", }, new [] { "beans", }, }变为build。非基本情况递归调用Func<string[][], string[][]> build = null;
build = s =>
(
s.Length == 1
? from x in s[0] select new [] { x }
: from xs in build(s.Skip(1).ToArray())
from x in s[0] select new [] { x }.Concat(xs).ToArray()
).ToArray();
，然后将每个递归数组的开头追加到第一个数组的值。很高兴你问过？
或者我可以写这样的查询：
var source = new []
{
new { Key = "colour", Value = "red", },
new { Key = "colour", Value = "blue", },
new { Key = "colour", Value = "violet", },
new { Key = "colour", Value = "black", },
new { Key = "colour", Value = "yellow", },
new { Key = "fruit", Value = "apple", },
new { Key = "fruit", Value = "mango", },
new { Key = "fruit", Value = "banana", },
new { Key = "vegetable", Value = "onion", },
new { Key = "vegetable", Value = "potato", },
new { Key = "vegetable", Value = "tomato", },
new { Key = "vegetable", Value = "beans", },
};
string[][] data =
source
.ToLookup(x => x.Key, x => x.Value)
.Select(x => x.ToArray())
.ToArray();
这可能会帮助您更轻松地分解它。
如果我从这些数据开始：
new []
{
new [] { "red", "apple", "onion" }
new [] { "blue", "apple", "onion" }
new [] { "violet", "apple", "onion" }
new [] { "black", "apple", "onion" }
new [] { "yellow", "apple", "onion" }
new [] { "red", "mango", "onion" }
new [] { "blue", "mango", "onion" }
new [] { "violet", "mango", "onion" }
new [] { "black", "mango", "onion" }
new [] { "yellow", "mango", "onion" }
new [] { "red", "banana", "onion" }
new [] { "blue", "banana", "onion" }
new [] { "violet", "banana", "onion" }
new [] { "black", "banana", "onion" }
new [] { "yellow", "banana", "onion" }
new [] { "red", "apple", "potato" }
new [] { "blue", "apple", "potato" }
new [] { "violet", "apple", "potato" }
new [] { "black", "apple", "potato" }
new [] { "yellow", "apple", "potato" }
new [] { "red", "mango", "potato" }
new [] { "blue", "mango", "potato" }
new [] { "violet", "mango", "potato" }
new [] { "black", "mango", "potato" }
new [] { "yellow", "mango", "potato" }
new [] { "red", "banana", "potato" }
new [] { "blue", "banana", "potato" }
new [] { "violet", "banana", "potato" }
new [] { "black", "banana", "potato" }
new [] { "yellow", "banana", "potato" }
new [] { "red", "apple", "tomato" }
new [] { "blue", "apple", "tomato" }
new [] { "violet", "apple", "tomato" }
new [] { "black", "apple", "tomato" }
new [] { "yellow", "apple", "tomato" }
new [] { "red", "mango", "tomato" }
new [] { "blue", "mango", "tomato" }
new [] { "violet", "mango", "tomato" }
new [] { "black", "mango", "tomato" }
new [] { "yellow", "mango", "tomato" }
new [] { "red", "banana", "tomato" }
new [] { "blue", "banana", "tomato" }
new [] { "violet", "banana", "tomato" }
new [] { "black", "banana", "tomato" }
new [] { "yellow", "banana", "tomato" }
new [] { "red", "apple", "beans" }
new [] { "blue", "apple", "beans" }
new [] { "violet", "apple", "beans" }
new [] { "black", "apple", "beans" }
new [] { "yellow", "apple", "beans" }
new [] { "red", "mango", "beans" }
new [] { "blue", "mango", "beans" }
new [] { "violet", "mango", "beans" }
new [] { "black", "mango", "beans" }
new [] { "yellow", "mango", "beans" }
new [] { "red", "banana", "beans" }
new [] { "blue", "banana", "beans" }
new [] { "violet", "banana", "beans" }
new [] { "black", "banana", "beans" }
new [] { "yellow", "banana", "beans" }
}
我得到了这个结果：
var source = new []
{
new { Key = "colour", Value = "red", },
new { Key = "fruit", Value = "apple", },
new { Key = "vegetable", Value = "onion", },
new { Key = "nation", Value = "france", },
new { Key = "nation", Value = "australia", },
};
这适用于任意数量的键。
试试这个数据：
new []
{
new [] { "red", "apple", "onion", "france" }
new [] { "red", "apple", "onion", "australia" }
}
我得到了这个结果：
int如果您的密钥是var source = new[]
{
new { Key = 1, Value = "red", },
new { Key = 1, Value = "apple", },
new { Key = 2, Value = "onion", },
new { Key = 3, Value = "france", },
new { Key = 3, Value = "australia", },
};
string[][] data =
source
.ToLookup(x => x.Key, x => x.Value)
.Select(x => x.ToArray())
.ToArray();
Func<string[][], string[][]> build = null;
build = s =>
(
s.Length == 1
? from x in s[0] select new[] { x }
: from xs in build(s.Skip(1).ToArray())
from x in s[0]
select new[] { x }.Concat(xs).ToArray()
).ToArray();
，则代码的工作原理相同：
build这是public string[][] Build(string[][] source)
{
return
(
source.Length == 1
? from x in source[0] select new[] { x }
: from xs in Build(source.Skip(1).ToArray())
from x in source[0]
select new[] { x }.Concat(xs).ToArray()
).ToArray();
}
的非匿名版本：
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134 0.00292 high dangerous reassure full
135 0.00048
136 0.0039 experience darren
137 0.00097 _ _param_ui_control_scripts_save _param_pattern_value
138 0.00292 find director
答案 1 :(得分：1)
这不是问题的真正答案。但是，当变量的数量固定时，可以采用一种方法。这种方法可以扩展到可变输入
public ActionResult tablecreate()
{
string[] key = { "colour", "fruit", "vegetable" };
var kvpair = new List<Tuple<String,String>>();
kvpair.Add(new Tuple<String,String>("colour","red"));
kvpair.Add(new Tuple<String, String>("colour", "blue"));
kvpair.Add(new Tuple<String, String>("colour", "yellow"));
kvpair.Add(new Tuple<String, String>("colour", "black"));
kvpair.Add(new Tuple<String, String>("fruit", "mango"));
kvpair.Add(new Tuple<String, String>("fruit", "apple"));
kvpair.Add(new Tuple<String, String>("vegetable", "potato"));
kvpair.Add(new Tuple<String, String>("vegetable", "tomato"));
kvpair.Add(new Tuple<String, String>("vegetable", "beans"));
var combi = new List<Tuple<String,String,String>>();
int i = 0, j = 0, k = 0;
var colorval = new List<String>();
var fruitval = new List<String>();
var vegval =new List<String>();
foreach (var item in kvpair)
{
if (item.Item1 == "colour")
colorval.Add(item.Item2);
else if (item.Item1 == "fruit")
fruitval.Add(item.Item2);
else if (item.Item1 == "vegetable")
vegval.Add(item.Item2);
}
foreach(var col in colorval)
{
foreach(var fru in fruitval)
{
foreach(var veg in vegval)
{
combi.Add(new Tuple<string, string, string>(col, fru, veg));
}
}
}
输出将生成问题中提到的表
我可以使用以下代码获取属于每个键的值，但仍然缺少组合
var serialize = new List<Tuple<int, List<string>>>();
for(int i=0;i<tcid.Count();i++)
{
var val = new List<string>();
foreach(var item in model)
{
if(tcid[i]==item.TcSetID)
val.Add(item.Value);
}
serialize.Add(new Tuple<int,List<string>>(tcid[i],val));
}
现在serailize我们拥有每个键的所有可能值。
为了使其更清晰serailize仅用作变量名。 serailise列表与上面代码中的colorval，fruitval和vegval类似，具有恒定数量的变量。
所以serailize将保持以下值：
serialize [item1 :colour, item2 : {blue,red,black,yellow}]
serialize [item1 :fruit, item2 : {apple,mango}]
serialize [item1 :vegetable, item2 : {tomato,potato,beans}]
我有一个函数可以在serialize.item2
public List<String> Cartesian (List<String> A, List<String> B)
{
int Alenght = A.Count();
int Blength = B.Count();
var S = new List<String>();
for (int i = 0; i < Alenght;i++ )
{
for(int j=0;j<Blength;j++)
{
S.Add(A[i] + "delim" + B[j]);
}
}
return S;
}
该函数调用如下
var A = new List<String>();
foreach(var item in serialize)
{
if(j==0)
{
A = item.Item2;
j = 1;
}
else
A = Cartesian(A, item.Item2);
}
然后将列表A中的字符串作为字符串数组
//distinctlist contains all the distinct strings in A
foreach(var item in distinctlist)
{
var row = new ViewsVM();
row.row = item.Split(new string[] { "delim"},StringSplitOptions.None);
ViewsVM.Add(row);
}
<强> ViewsVM
public class ViewsVM
{
public string[] row { get; set; }
}
希望它有所帮助。但我应该警告这不是最好的方法。
Enigmativity 可以更好地解决问题。我发布这个答案只是因为我还找到了替代方法。
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