我有两个具有此类型的IQueryable集合
public class Property
{
public string Name {get; set;}
}
集合1，具有以下名称值：
A
A
A
B
集合2，具有以下名称值：
A
B
B
我想得到的是第三个集合，其中集合1和2的名称值匹配，如果没有匹配，则为null (empty)，如下所示：
Result Collection:
A A
A null
A null
B B
null B
如何用C＃，LINQ实现这一点？
答案 0 :(得分：4)
using System;
using System.Collections.Generic;
using System.Linq;
namespace Testing
{
public class Property
{
public string Name { get; set; }
public override bool Equals(object obj)
{
var item = obj as Property;
if (item == null)
{
return false;
}
return item.Name == Name;
}
public override int GetHashCode()
{
return Name.GetHashCode();
}
}
public class JoinedProperty
{
public Property Name1 { get; set; }
public Property Name2 { get; set; }
public override string ToString()
{
return (Name1 == null ? "" : Name1.Name)
+ (Name2 == null ? "" : Name2.Name);
}
}
class Program
{
static void Main(string[] args)
{
var list1 = new List<Property>
{
new Property{ Name = "A" },
new Property{ Name = "A" },
new Property{ Name = "A" },
new Property{ Name = "B" }
};
var list2 = new List<Property>
{
new Property{ Name = "A" },
new Property{ Name = "B" },
new Property{ Name = "B" }
};
var allLetters = list1.Union(list2).Distinct().ToList();
var result = new List<JoinedProperty>();
foreach (var letter in allLetters)
{
var list1Count = list1.Count(l => l.Name == letter.Name);
var list2Count = list2.Count(l => l.Name == letter.Name);
var matchCount = Math.Min(list1Count, list2Count);
addValuesToResult(result, letter, letter, matchCount);
var difference = list1Count - list2Count;
if(difference > 0)
{
addValuesToResult(result, letter, null, difference);
}
else
{
difference = difference * -1;
addValuesToResult(result,null, letter, difference);
}
}
foreach(var res in result)
{
Console.WriteLine(res.ToString());
}
Console.ReadLine();
}
private static void addValuesToResult(List<JoinedProperty> result, Property letter1, Property letter2, int count)
{
for (int i = 0; i < count; i++)
{
result.Add(new JoinedProperty
{
Name1 = letter1,
Name2 = letter2
});
}
}
}
}
运行此命令即可获得结果
AA
A
A
BB
B
并且结果列表的内容就是你所追求的。
编辑：更新了我的答案，以使用指定的属性。
答案 1 :(得分：1)
对于这个问题似乎很感兴趣所以我试图提出一个更通用的解决方案。我从这个链接中获取了灵感https://www.codeproject.com/Articles/488643/LinQ-Extended-Joins。
我已经创建了一个fullouterjoin扩展方法，它可以执行操作所要求的操作。不确定fullouterjoin是否是正确的名称。
我使用了我的扩展方法来解决操作问题。
using System;
using System.Collections.Generic;
using System.Linq;
namespace Testing
{
public class Property
{
public string Name { get; set; }
}
public class JoinedProperty
{
public Property Name1 { get; set; }
public Property Name2 { get; set; }
public override string ToString()
{
return (Name1 == null ? "" : Name1.Name)
+ (Name2 == null ? "" : Name2.Name);
}
}
class Program
{
static void Main(string[] args)
{
var list1 = new List<Property>
{
new Property{ Name = "A" },
new Property{ Name = "A" },
new Property{ Name = "A" },
new Property{ Name = "B" }
};
var list2 = new List<Property>
{
new Property{ Name = "A" },
new Property{ Name = "B" },
new Property{ Name = "B" }
};
var result = list1.FullOuterJoin(
list2,
p1 => p1.Name,
p2 => p2.Name,
(p1, p2) => new JoinedProperty
{
Name1 = p1,
Name2 = p2
}).ToList();
foreach (var res in result)
{
Console.WriteLine(res.ToString());
}
Console.ReadLine();
}
}
public static class MyExtensions
{
public static IEnumerable<TResult>
FullOuterJoin<TSource, TInner, TKey, TResult>(this IEnumerable<TSource> source,
IEnumerable<TInner> inner,
Func<TSource, TKey> pk,
Func<TInner, TKey> fk,
Func<TSource, TInner, TResult> result)
where TSource : class where TInner : class
{
var fullList = source.Select(s => new Tuple<TSource, TInner>(s, null))
.Concat(inner.Select(i => new Tuple<TSource, TInner>(null, i)));
var joinedList = new List<Tuple<TSource, TInner>>();
foreach (var item in fullList)
{
var matchingItem = joinedList.FirstOrDefault
(
i => matches(i, item, pk, fk)
);
if(matchingItem != null)
{
joinedList.Remove(matchingItem);
joinedList.Add(combinedMatchingItems(item, matchingItem));
}
else
{
joinedList.Add(item);
}
}
return joinedList.Select(jl => result(jl.Item1, jl.Item2)).ToList();
}
private static Tuple<TSource, TInner> combinedMatchingItems<TSource, TInner>(Tuple<TSource, TInner> item1, Tuple<TSource, TInner> item2)
where TSource : class
where TInner : class
{
if(item1.Item1 == null && item2.Item2 == null && item1.Item2 != null && item2.Item1 !=null)
{
return new Tuple<TSource, TInner>(item2.Item1, item1.Item2);
}
if(item1.Item2 == null && item2.Item1 == null && item1.Item1 != null && item2.Item2 != null)
{
return new Tuple<TSource, TInner>(item1.Item1, item2.Item2);
}
throw new InvalidOperationException("2 items cannot be combined");
}
public static bool matches<TSource, TInner, TKey>(Tuple<TSource, TInner> item1, Tuple<TSource, TInner> item2, Func<TSource, TKey> pk, Func<TInner, TKey> fk)
where TSource : class
where TInner : class
{
if (item1.Item1 != null && item1.Item2 == null && item2.Item2 != null && item2.Item1 == null && pk(item1.Item1).Equals(fk(item2.Item2)))
{
return true;
}
if (item1.Item2 != null && item1.Item1 == null && item2.Item1 != null && item2.Item2 == null && fk(item1.Item2).Equals(pk(item2.Item1)))
{
return true;
}
return false;
}
}
}
答案 2 :(得分：0)
我认为，最好的选择就是使用loop;
var listA = new List<Property>
{
new Property{ Name = "A" },
new Property{ Name = "A" },
new Property{ Name = "A" },
new Property{ Name = "B" }
};
var listB = new List<Property>
{
new Property{ Name = "A" },
new Property{ Name = "B" },
new Property{ Name = "B" }
};
var joinedList = new List<JoinedProperty>();
for (int i = 0; i < listA.Count; i++)
{
var property = new JoinedProperty
{
AName = listA[i].Name,
BName = null
};
if (listB.Count < i + 1)
{
continue;
}
if (listA[i].Name == listB[i].Name)
{
property.BName = listA[i].Name;
}
joinedList.Add(property);
}
for (int i = 0; i < listB.Count; i++)
{
var property = new JoinedProperty
{
AName = null,
BName = listB[i].Name
};
if (listA.Count < i + 1)
{
continue;
}
if (listB[i].Name == listA[i].Name)
{
property.AName = listB[i].Name;
}
joinedList.Add(property);
}
public class JoinedProperty
{
public string AName { get; set; }
public string BName { get; set; }
}
另外，我认为，您的输出示例缺少一个元素;
null B
<强>输出
A A
A null
A null
B B
null B
null B
答案 3 :(得分：0)
public class Property
{
public string Name { get; set; }
}
var list1 = new List<Property>
{
new Property { Name ="A" },
new Property { Name ="A" },
new Property { Name ="A" },
new Property { Name ="B" }
};
var list2 = new List<Property>
{
new Property { Name ="A" },
new Property { Name ="B" },
new Property { Name ="B" }
};
var r = new List<string>();
int x1 = 0, x2 = 0;
int count1 = list1.Count, count2 = list2.Count;
while (true)
{
if (x1 == count1 && x2 == count2) break;
if (x1 < count1 && x2 == count2)
{
r.Add($"{list1[x1].Name}\tNULL");
++x1;
}
else if (x1 == count1 && x2 < count2)
{
r.Add($"NULL\t{list2[x2].Name}");
++x2;
}
else
{
if (list1[x1].Name == list2[x2].Name)
{
r.Add($"{list1[x1].Name}\t{list2[x2].Name}");
++x1; ++x2;
}
else
{
r.Add($"{list1[x1].Name}\tNULL");
++x1;
}
}
}
<强>解释
这个想法是管理名单中的职位 - 即我们是否应该提升职位。一旦查找了所有位置，循环就会退出。
答案 4 :(得分：0)
你问过LINQ函数，没有，但你可以扩展它，所以它可以用于你想要这个技巧的任何两个序列。
您所要做的就是编写IEnumerable的扩展函数，类似于所有其他LINQ函数。
请参阅Extension Methods Demystified
public static class MyEnumerableExtensions
{
public IEnumerable<System.Tuple<T, T>> EqualityZip<T>(this IEnumerable<T> sourceA,
IEnumerable<T> sourceB)
{
// TODO: check for parameters null
var enumeratorA = sourceA.GetEnumerator();
var enumeratorB = sourceB.GetEnumerator();
// enumerate as long as we have elements in A and in B:
bool aAvailable = enumeratorA.MoveNext();
bool bAvailable = enumeratorB.MoveNext();
while (aAvailable && bAvailable)
{ // we have an A element and a B element
T a = enumeratorA.Current;
T b = enumeratorB.Current;
// compare the two elements:
if (a == b)
{ // equal: return tuple (a, b)
yield return Tuple.Create(a, b)
}
else
{ // not equal, return (a, null)
yield return Tuple.Create(a, (T)null)
}
// move to the next element
aAvailable = enumeratorA.MoveNext();
bAvailable = enumeratorB.MoveNext();
}
// now either we are out of A or out of B
while (aAvailable)
{ // we still have A but no B, return (A, null)
T A = enumeratorA.Current;
yield return Tuple.Create(A, (T)null);
aAvailable = enumeratorA.MoveNext();
}
while (bAvailable)
{ // we don't have A, but there are still B, return (null, B)
T B = enumeratorB.Current;
yield return Tuple.Create((T)null, B);
bAvailable = enumeratorB.MoveNext();
}
// if there are still A elements without B element: return (a, null)
while (enumaratorA.Nex
}
}
用法：
var sequenceA = ...
var sequenceB = ...
var result = sequenceA.EqualityZip(sequenceB);
TODO：使函数更好，可以比较两个不同的类，KeySelectors选择A和B的比较键以及IEqualityCompare：
public static IEnumerable<Tuple<TA, TB> EqualityZip<TA, TB, TKey>(
this IEnumerable<TA> sourceA, // the first sequence
this IEnumerable<TB> sourceB, // the second sequence
Func<TA, TKey> keySelectorA, // the property of sourceA to take
Func<TB, TKey> keySelectorB, // the property of sourceB to take
IEqualityComparer<TKey> comparer)
{
// TODO: ArgumentNullException if arguments null
if (comparer==null) comparer = EqualityCompare<TKey>.Default;
var enumeratorA = sourceA.GetEnumerator();
var enumeratorB = sourceB.GetEnumerator();
// enumerate as long as we have elements in A and in B:
bool aAvailable = enumeratorA.MoveNext();
bool bAvailable = enumeratorB.MoveNext();
while (aAvailable && bAvailable)
{ // we have an A element and a B element
TKey keyA = keySelectorA(enumeratorA.Current);
TKey keyB = keySelectorB(enumeratorB.Current);
if (comparer.Equals(keyA, keyB)
{
yield return Tuple.Create(Ta, Tb)
}
else
等