我有一个包含枚举（TypeCode）和用户对象的项目集合，我需要将其展平以显示在网格中。这很难解释，所以让我举一个简单的例子。
收藏有这样的项目：
TypeCode | User
---------------
1 | Don Smith
1 | Mike Jones
1 | James Ray
2 | Tom Rizzo
2 | Alex Homes
3 | Andy Bates
我需要输出为：
1 | 2 | 3
Don Smith | Tom Rizzo | Andy Bates
Mike Jones | Alex Homes |
James Ray | |
我尝试过使用 foreach 执行此操作，但我不能这样做，因为我会将新项目插入到 foreach 中的集合中，从而导致错误。
这可以在 Linq 中以更清洁的方式完成吗？
我并不是说这是一个很好的转轴方式——但它是一个转轴......
// sample data
var data = new[] {
new { Foo = 1, Bar = "Don Smith"},
new { Foo = 1, Bar = "Mike Jones"},
new { Foo = 1, Bar = "James Ray"},
new { Foo = 2, Bar = "Tom Rizzo"},
new { Foo = 2, Bar = "Alex Homes"},
new { Foo = 3, Bar = "Andy Bates"},
};
// group into columns, and select the rows per column
var grps = from d in data
group d by d.Foo
into grp
select new {
Foo = grp.Key,
Bars = grp.Select(d2 => d2.Bar).ToArray()
};
// find the total number of (data) rows
int rows = grps.Max(grp => grp.Bars.Length);
// output columns
foreach (var grp in grps) {
Console.Write(grp.Foo + "\t");
}
Console.WriteLine();
// output data
for (int i = 0; i < rows; i++) {
foreach (var grp in grps) {
Console.Write((i < grp.Bars.Length ? grp.Bars[i] : null) + "\t");
}
Console.WriteLine();
}
Marc 的回答给出了不能直接注入 Grid 的稀疏矩阵。
我尝试从Vasu 提供的链接中扩展代码，如下所示：
public static Dictionary<TKey1, Dictionary<TKey2, TValue>> Pivot3<TSource, TKey1, TKey2, TValue>(
this IEnumerable<TSource> source
, Func<TSource, TKey1> key1Selector
, Func<TSource, TKey2> key2Selector
, Func<IEnumerable<TSource>, TValue> aggregate)
{
return source.GroupBy(key1Selector).Select(
x => new
{
X = x.Key,
Y = source.GroupBy(key2Selector).Select(
z => new
{
Z = z.Key,
V = aggregate(from item in source
where key1Selector(item).Equals(x.Key)
&& key2Selector(item).Equals(z.Key)
select item
)
}
).ToDictionary(e => e.Z, o => o.V)
}
).ToDictionary(e => e.X, o => o.Y);
}
internal class Employee
{
public string Name { get; set; }
public string Department { get; set; }
public string Function { get; set; }
public decimal Salary { get; set; }
}
public void TestLinqExtenions()
{
var l = new List<Employee>() {
new Employee() { Name = "Fons", Department = "R&D", Function = "Trainer", Salary = 2000 },
new Employee() { Name = "Jim", Department = "R&D", Function = "Trainer", Salary = 3000 },
new Employee() { Name = "Ellen", Department = "Dev", Function = "Developer", Salary = 4000 },
new Employee() { Name = "Mike", Department = "Dev", Function = "Consultant", Salary = 5000 },
new Employee() { Name = "Jack", Department = "R&D", Function = "Developer", Salary = 6000 },
new Employee() { Name = "Demy", Department = "Dev", Function = "Consultant", Salary = 2000 }};
var result5 = l.Pivot3(emp => emp.Department, emp2 => emp2.Function, lst => lst.Sum(emp => emp.Salary));
var result6 = l.Pivot3(emp => emp.Function, emp2 => emp2.Department, lst => lst.Count());
}
* 不能说任何关于性能的事情。
您可以使用 Linq 的 .ToLookup 以您正在寻找的方式进行分组。
var lookup = data.ToLookup(d => d.TypeCode, d => d.User);
然后就是把它变成一种你的消费者可以理解的形式。例如：
//Warning: untested code
var enumerators = lookup.Select(g => g.GetEnumerator()).ToList();
int columns = enumerators.Count;
while(columns > 0)
{
for(int i = 0; i < enumerators.Count; ++i)
{
var enumerator = enumerators[i];
if(enumator == null) continue;
if(!enumerator.MoveNext())
{
--columns;
enumerators[i] = null;
}
}
yield return enumerators.Select(e => (e != null) ? e.Current : null);
}
将其放入 IEnumerable<> 方法中，它将（可能）返回 User 的集合（列）的集合（行），其中将 null 放入没有数据的列中。
我想这与 Marc 的答案相似，但我会发布它，因为我花了一些时间来研究它。结果由" | "您的示例中的分隔。当使用 group by 而不是构造新的匿名类型时，它还使用IGrouping<int, string>从 LINQ 查询返回的类型。这是经过测试的工作代码。
var Items = new[] {
new { TypeCode = 1, UserName = "Don Smith"},
new { TypeCode = 1, UserName = "Mike Jones"},
new { TypeCode = 1, UserName = "James Ray"},
new { TypeCode = 2, UserName = "Tom Rizzo"},
new { TypeCode = 2, UserName = "Alex Homes"},
new { TypeCode = 3, UserName = "Andy Bates"}
};
var Columns = from i in Items
group i.UserName by i.TypeCode;
Dictionary<int, List<string>> Rows = new Dictionary<int, List<string>>();
int RowCount = Columns.Max(g => g.Count());
for (int i = 0; i <= RowCount; i++) // Row 0 is the header row.
{
Rows.Add(i, new List<string>());
}
int RowIndex;
foreach (IGrouping<int, string> c in Columns)
{
Rows[0].Add(c.Key.ToString());
RowIndex = 1;
foreach (string user in c)
{
Rows[RowIndex].Add(user);
RowIndex++;
}
for (int r = RowIndex; r <= Columns.Count(); r++)
{
Rows[r].Add(string.Empty);
}
}
foreach (List<string> row in Rows.Values)
{
Console.WriteLine(row.Aggregate((current, next) => current + " | " + next));
}
Console.ReadLine();
我还用这个输入测试了它：
var Items = new[] {
new { TypeCode = 1, UserName = "Don Smith"},
new { TypeCode = 3, UserName = "Mike Jones"},
new { TypeCode = 3, UserName = "James Ray"},
new { TypeCode = 2, UserName = "Tom Rizzo"},
new { TypeCode = 2, UserName = "Alex Homes"},
new { TypeCode = 3, UserName = "Andy Bates"}
};
这产生了以下结果，表明第一列不需要包含最长的列表。如果需要，您可以使用OrderBy获取按 TypeCode 排序的列。
1 | 3 | 2
Don Smith | Mike Jones | Tom Rizzo
| James Ray | Alex Homes
| Andy Bates |
@Sanjaya.Tio 我对您的回答很感兴趣，并创建了此改编，以最大限度地减少 keySelector 的执行。（未经测试）
public static Dictionary<TKey1, Dictionary<TKey2, TValue>> Pivot3<TSource, TKey1, TKey2, TValue>(
this IEnumerable<TSource> source
, Func<TSource, TKey1> key1Selector
, Func<TSource, TKey2> key2Selector
, Func<IEnumerable<TSource>, TValue> aggregate)
{
var lookup = source.ToLookup(x => new {Key1 = key1Selector(x), Key2 = key2Selector(x)});
List<TKey1> key1s = lookup.Select(g => g.Key.Key1).Distinct().ToList();
List<TKey2> key2s = lookup.Select(g => g.Key.Key2).Distinct().ToList();
var resultQuery =
from key1 in key1s
from key2 in key2s
let lookupKey = new {Key1 = key1, Key2 = key2}
let g = lookup[lookupKey]
let resultValue = g.Any() ? aggregate(g) : default(TValue)
select new {Key1 = key1, Key2 = key2, ResultValue = resultValue};
Dictionary<TKey1, Dictionary<TKey2, TValue>> result = new Dictionary<TKey1, Dictionary<TKey2, TValue>>();
foreach(var resultItem in resultQuery)
{
TKey1 key1 = resultItem.Key1;
TKey2 key2 = resultItem.Key2;
TValue resultValue = resultItem.ResultValue;
if (!result.ContainsKey(key1))
{
result[key1] = new Dictionary<TKey2, TValue>();
}
var subDictionary = result[key1];
subDictionary[key2] = resultValue;
}
return result;
}