我有一个二维字符串数组：string[,] arr = new string[3,3]{ { "a", "b", "c" }, { "a", "b", "c" }, { "d", "e", "f" }, }我想得到的结果是：a,b,c = 2d,e,f = 1我试过：List<List<string>> lLString = new List<List<string>>();string[,] stringArray2D = new string[3, 3] { { "a", "b", "c" }, { "a", "b", "c" }, { "d", "e", "f" }, };for (int i = 0; i < stringArray2D.GetLength(0); i++) { List<string> temp = new List<string>(); for (int j = 0; j < stringArray2D.GetLength(1); j++) { temp.Add(stringArray2D[i,j]); } lLString.Add(temp);}试图通过这样做来解决它：lLString.GroupBy (ls => ls);但似乎不对
3 回答
蛊毒传说
TA贡献1895条经验 获得超3个赞
您可以直接使用锯齿状数组，例如：
string[][] arrayJagged = new[]
{
new[] { "a", "b", "c" },
new[] { "a", "b", "c" },
new[] { "d", "e", "f" },
new[] { "g", "h", "i" },
};
但是你想在过去的废墟中挖掘并使用多维数组，只是为了让一切变得更加复杂。
string[,] stringArray2D = new string[,]
{
{ "a", "b", "c" },
{ "a", "b", "c" },
{ "d", "e", "f" },
{ "g", "h", "i" },
};
然后我们必须将多维数组转换为锯齿状数组：
string[][] arrayJagged = new string[stringArray2D.GetLength(0)][];
int length2 = stringArray2D.GetLength(1);
for (int i = 0; i < arrayJagged.Length; i++)
{
arrayJagged[i] = new string[length2];
for (int j = 0; j < length2; j++)
{
arrayJagged[i][j] = stringArray2D[i, j];
}
}
请注意，您实际上并不需要 a，List<List<string>>因为最终锯齿状数组的维度是预先确定的。
然后你可以.GroupBy()使用锯齿状数组，并.Count()在每组上做一个：
var grouped = arrayJagged.GroupBy(x => x, ArrayEqualityComparer<string>.Default)
.Select(x => new
{
x.Key,
Count = x.Count()
})
.ToArray();
请注意，.NET 没有用于数组的默认相等比较器，因此您需要定义一个，以显示.GroupBy()如何检查元素的相等性：
// Simple T[] IEqualityComparer<>
public sealed class ArrayEqualityComparer<T> : IEqualityComparer<T[]>, IEqualityComparer
{
// One instance is more than enough
public static readonly ArrayEqualityComparer<T> Default = new ArrayEqualityComparer<T>();
// We lazily define it if necessary
private readonly IEqualityComparer<T> equalityComparer;
public ArrayEqualityComparer()
{
equalityComparer = EqualityComparer<T>.Default;
}
public ArrayEqualityComparer(IEqualityComparer<T> equalityComparer)
{
this.equalityComparer = equalityComparer;
}
/* IEqualityComparer<T[]> */
public bool Equals(T[] x, T[] y)
{
if (x == null)
{
if (y == null)
{
return true;
}
return false;
}
if (y == null)
{
return false;
}
return EqualsNotNull(x, y);
}
public int GetHashCode(T[] obj)
{
unchecked
{
int hash = 17;
if (obj != null)
{
// This one will help distinguish between null and empty:
// hash(null) == 17, hash(empty) == 17 * 23
hash = (hash * 23) + obj.Length;
for (int i = 0; i < obj.Length; i++)
{
hash = (hash * 23) + obj[i].GetHashCode();
}
}
return hash;
}
}
/* IEqualityComparer */
bool IEqualityComparer.Equals(object x, object y)
{
if (x == y)
{
return true;
}
if (x == null || y == null)
{
return false;
}
var x2 = x as T[];
if (x2 == null)
{
throw new ArgumentException("x");
}
var y2 = y as T[];
if (y2 == null)
{
throw new ArgumentException("y");
}
return EqualsNotNull(x2, y2);
}
int IEqualityComparer.GetHashCode(object obj)
{
T[] obj2;
if (obj != null)
{
obj2 = obj as T[];
if (obj2 == null)
{
throw new ArgumentException("obj");
}
}
else
{
obj2 = null;
}
return GetHashCode(obj2);
}
/* Implementation */
private bool EqualsNotNull(T[] x, T[] y)
{
if (x.Length != y.Length)
{
return false;
}
if (x.Length != 0)
{
for (int i = 0; i < x.Length; i++)
{
if (!equalityComparer.Equals(x[i], y[i]))
{
return false;
}
}
}
return true;
}
}
千巷猫影
TA贡献1829条经验 获得超7个赞
创建一个覆盖等于的类
public static void GroupList()
{
List<ListComp> lLString = new List<ListComp>();
string[,] stringArray2D = new string[3, 3] { { "a", "b", "c" },
{ "a", "b", "c" },
{ "d", "e", "f" },
};
for (int i = 0; i<stringArray2D.GetLength(0); i++) {
ListComp temp = new ListComp();
for (int j = 0; j<stringArray2D.GetLength(1); j++) {
temp.Add(stringArray2D[i, j]);
}
lLString.Add(temp);
}
var gr = lLString.GroupBy(i => i);
foreach (var g in gr)
{
Debug.WriteLine($"{g.Key} = {g.Count()}");
}
Debug.WriteLine("");
}
public class ListComp : List<string>
{
public override string ToString()
{
return string.Join(",", this);
}
public override bool Equals(object obj)
{
ListComp listComp = obj as ListComp;
if (listComp == null)
return false;
else
return Equals(listComp);
}
public bool Equals(ListComp listComp)
{
if (listComp == null)
return false;
return this.SequenceEqual(listComp);
}
public override int GetHashCode()
{
int hash = 1;
foreach(string s in this)
{
hash *= s.GetHashCode();
}
return hash;
}
public ListComp (List<string> listComp)
{
this.AddRange(listComp);
}
public ListComp()
{
}
}
婷婷同学_
TA贡献1844条经验 获得超8个赞
使用类似于您使用的循环的基本 for 循环创建字符串列表。然后使用GroupBy.
string[,] arr = new string[3, 3]
{
{ "a", "b", "c" },
{ "a", "b", "c" },
{ "d", "e", "f" },
};
int dim1 = arr.GetLength(0);
int dim2 = arr.GetLength(1);
List<string> temp = new List<string>();
string st;
for (int i = 0; i < dim1; i++)
{
st = "";
for (int j = 0; j < dim2; j++)
st += arr[i, j] + ",";
temp.Add(st);
}
var gr = temp.GroupBy(i => i);
foreach (var g in gr)
Console.WriteLine($"{g.Key} = {g.Count()}");
输出：
a,b,c, = 2
d,e,f, = 1
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