如何删除具有 2 个常见元素的数组/结构列表的元素
本文关键字:元素 常见 数组 列表 结构 何删除 删除 | 更新日期: 2023-09-27 18:30:42
有一个结构列表或者一个数组 列出每个包含 3 个元素的列表,例如
12 8 7
5 1 0
7 3 2
10 6 5
6 2 1
8 4 3
6 1 5
7 2 6
8 3 7
9 4 8
11 7 6
13 9 8
11 6 10
12 7 11
13 8 12
14 9 13
我想摆脱列表中有 2 个常见子项的项目,在我想删除的示例中
5 1 0
6 2 1
6 1 5
7 3 2
7 2 6
8 4 3
8 3 7 has 2 same items as row 7,3,2
9 4 8 has 2 same items as row 8,4,3
10 6 5
11 7 6
11 6 10 has 2 same items as row 11,7,6
12 7 11 has 2 same items as row 11,7,10
12 8 7
13 8 12
13 9 8
14 9 13 has 2 same items as row 13,9,8
因此,使用结构方法,我正在考虑按元素 A 对列表进行排序,然后循环和比较元素,如果当前元素有 2 个值等于列表中的其他元素,我不将其添加到结果列表中,但是我卡住了,不知道是否有更好的方法
struct S
{
public int A;
public int B;
public int C;
}
public void test()
{
List<S> DataItems = new List<S>();
DataItems.Add(new S { A = 1, B = 2, C=3} );
DataItems.Add(new S { A = 12, B = 8, C = 7 });
DataItems.Add(new S { A = 5, B = 1, C = 0 });
DataItems.Add(new S { A = 7, B = 3, C = 2 });
DataItems.Add(new S { A = 10, B = 6, C = 5 });
DataItems.Add(new S { A = 6, B = 2, C = 1 });
DataItems.Add(new S { A = 8, B = 4, C = 3 });
DataItems.Add(new S { A = 6, B = 1, C = 5 });
DataItems.Add(new S { A = 7, B = 2, C = 6 });
DataItems.Add(new S { A = 8, B = 3, C = 7 });
DataItems.Add(new S { A = 9, B = 4, C = 8 });
DataItems.Add(new S { A = 11, B = 7, C = 6 });
DataItems.Add(new S { A = 13, B = 9, C = 8 });
DataItems.Add(new S { A = 11, B = 6, C = 10 });
DataItems.Add(new S { A = 12, B = 7, C = 11 });
DataItems.Add(new S { A = 13, B = 8, C = 12 });
DataItems.Add(new S { A = 14, B = 9, C = 13 });
var sortedList = DataItems.OrderBy(x => x.A);
List<S> resultList = new List<S>();
for (int i = 0; i < sortedList.Count (); i++)
{
for (int j = i+1; j < sortedList.Count(); j++)
{
if (sortedList.ElementAt(i).A == sortedList.ElementAt(j).A || sortedList.ElementAt(i).A == sortedList.ElementAt(j).B || sortedList.ElementAt(i).A == sortedList.ElementAt(j).C)
{
//ONE HIT, WAIT OTHER
}
}
}
}
有没有更有效的方法来获取列表,而无需具有 2 个相同项目的项目,以便我会得到,而不是硬编码解决方案?
5 1 0
6 2 1
6 1 5
7 3 2
7 2 6
8 4 3
10 6 5
11 7 6
12 8 7
13 8 12
13 9 8
给定一个项目...
{ A = 1, B = 2, C = 3 }
您有 3 种可能的组合可以在另一个项目中重复,例如
AB, AC & BC which is {1, 2}, {1, 3} & {2, 3}
所以我要做的是遍历你的列表,将这些组合添加到带有分隔符字符的字典中(首先是最小的数字,所以如果 B
"1-2", "1-3", "2-3"
现在,当您添加每个项目时,请检查键是否已存在,如果存在,则可以忽略该项(不要将其添加到结果列表中)。
在性能方面,这将是一次遍历整个列表,并使用字典检查具有 2 个常用数字的项目。
解决它的一种方法是在struct S
中引入中间方法:
public struct S {
public int A;
public int B;
public int C;
public bool IsSimilarTo(S s) {
int similarity = HasElement(A, s) ? 1 : 0;
similarity += HasElement(B, s) ? 1 : 0;
return similarity >= 2 ? true : HasElement(C, s);
}
public bool HasElement(int val, S s) {
return val == s.A || val == s.B || val == s.C;
}
public int HasSimilarInList(List<S> list, int index) {
if (index == 0)
return -1;
for (int i = 0; i < index; ++i)//compare with the previous items
if (IsSimilarTo(list[i]))
return i;
return -1;
}
}
然后你可以这样解决它,而无需订购:
public void test() {
List<S> DataItems = new List<S>();
DataItems.Add(new S { A = 1, B = 2, C = 3 });
DataItems.Add(new S { A = 12, B = 8, C = 7 });
DataItems.Add(new S { A = 5, B = 1, C = 0 });
DataItems.Add(new S { A = 7, B = 3, C = 2 });
DataItems.Add(new S { A = 10, B = 6, C = 5 });
DataItems.Add(new S { A = 6, B = 2, C = 1 });
DataItems.Add(new S { A = 8, B = 4, C = 3 });
DataItems.Add(new S { A = 6, B = 1, C = 5 });
DataItems.Add(new S { A = 7, B = 2, C = 6 });
DataItems.Add(new S { A = 8, B = 3, C = 7 });
DataItems.Add(new S { A = 9, B = 4, C = 8 });
DataItems.Add(new S { A = 11, B = 7, C = 6 });
DataItems.Add(new S { A = 13, B = 9, C = 8 });
DataItems.Add(new S { A = 11, B = 6, C = 10 });
DataItems.Add(new S { A = 12, B = 7, C = 11 });
DataItems.Add(new S { A = 13, B = 8, C = 12 });
DataItems.Add(new S { A = 14, B = 9, C = 13 });
int index = 1; //0-th element does not need to be checked
while (index < DataItems.Count) {
int isSimilarTo = DataItems[index].HasSimilarInList(DataItems, index);
if (isSimilarTo == -1) {
++index;
continue;
}
DataItems.RemoveAt(index);
}
}