Question

我正在考虑找到两个列表中差异的好方法

问题在于：

两个列表包含一些字符串，其中前3个数字/字符（*分隔）表示唯一键（后跟文本String =＆＃34; key1 * key2 * key3 * text＆＃34;）。

这是字符串示例：

AA1*1D*4*The quick brown fox*****CC*3456321234543~

其中＆＃34; * AA1 * 1D * 4 *＆＃34;是一个独特的关键

List1：＆＃34; index1 * index2 * index3＆＃34;，＆＃34; index2 * index2 * index3＆＃34;，＆＃34; index3 * index2 * index3＆＃34;

List2：＆＃34; index2 * index2 * index3＆＃34;，＆＃34; index1 * index2 * index3＆＃34;，＆＃34; index3 * index2 * index3＆＃34;，＆＃34; index4 *索引2 * INDEX3＆＃34;

我需要匹配两个列表中的索引并进行比较。

如果1个列表中的所有3个索引都匹配另一个列表中的3个索引，我需要跟踪新列表中的两个字符串条目
如果一个列表中有一组索引不会出现在另一个列表中，我需要跟踪一侧并在另一侧保留一个空条目。（上例中的＃4）

返回列表

这是我到目前为止所做的，但我在这里有点挣扎：

        List<String> Base = baseListCopy.Except(resultListCopy, StringComparer.InvariantCultureIgnoreCase).ToList(); //Keep unique values(keep differences in lists)
        List<String> Result = resultListCopy.Except(baseListCopy, StringComparer.InvariantCultureIgnoreCase).ToList(); //Keep unique values (keep differences in lists)

        List<String[]> blocksComparison = new List<String[]>(); //we container for non-matching blocks; so we could output them later

        //if both reports have same amount of blocks
        if ((Result.Count > 0 || Base.Count > 0) && (Result.Count == Base.Count))
        {
            foreach (String S in Result)
            {
                String[] sArr = S.Split('*');
                foreach (String B in Base)
                {
                    String[] bArr = B.Split('*');

                    if (sArr[0].Equals(bArr[0]) && sArr[1].Equals(bArr[1]) && sArr[2].Equals(bArr[2]) && sArr[3].Equals(bArr[3]))
                    {
                        String[] NA = new String[2]; //keep results
                        NA[0] = B; //[0] for base
                        NA[1] = S; //[1] for result
                        blocksComparison.Add(NA);
                        break;
                    }
                }
            }
        }

你能为这个过程建议一个好的算法吗？

谢谢

Answer 1

您可以使用HashSet。

为List1创建一个HashSet。记住index1 * index2 * index3与index3 * index2 * index1不同。

现在迭代第二个列表。

Create Hashset for List1.

foreach(string in list2)
{
    if(hashset contains string)
       //Add it to the new list.
}

Answer 2

List one = new List();
List two = new List();
List three = new List();
HashMap<String,Integer> intersect = new HashMap<String,Integer>();

for(one: String index)
{
    intersect.put(index.next,intersect.get(index.next) + 1);
}

for(two: String index)
{
    if(intersect.containsKey(index.next))
    {
        three.add(index.next);
    }
}

Answer 3

如果我正确理解您的问题，您希望能够通过“密钥”前缀比较元素，而不是整个字符串内容。如果是这样，实现自定义相等比较器将允许您轻松利用LINQ集算法。

这个程序......

class EqCmp : IEqualityComparer<string> {

    public bool Equals(string x, string y) {
        return GetKey(x).SequenceEqual(GetKey(y));
    }

    public int GetHashCode(string obj) {
        // Using Sum could cause OverflowException.
        return GetKey(obj).Aggregate(0, (sum, subkey) => sum + subkey.GetHashCode());
    }

    static IEnumerable<string> GetKey(string line) {
        // If we just split to 3 strings, the last one could exceed the key, so we split to 4.
        // This is not the most efficient way, but is simple.
        return line.Split(new[] { '*' }, 4).Take(3);
    }

}

class Program {

    static void Main(string[] args) {

        var l1 = new List<string> {
            "index1*index1*index1*some text",
            "index1*index1*index2*some text ** test test test",
            "index1*index2*index1*some text",
            "index1*index2*index2*some text",
            "index2*index1*index1*some text"
        };

        var l2 = new List<string> {
            "index1*index1*index2*some text ** test test test",
            "index2*index1*index1*some text",
            "index2*index1*index2*some text"
        };

        var eq = new EqCmp();

        Console.WriteLine("Elements that are both in l1 and l2:");
        foreach (var line in l1.Intersect(l2, eq))
            Console.WriteLine(line);

        Console.WriteLine("\nElements that are in l1 but not in l2:");
        foreach (var line in l1.Except(l2, eq))
            Console.WriteLine(line);

        // Etc...

    }

}

...打印以下结果：

Elements that are both in l1 and l2:
index1*index1*index2*some text ** test test test
index2*index1*index1*some text

Elements that are in l1 but not in l2:
index1*index1*index1*some text
index1*index2*index1*some text
index1*index2*index2*some text

找到两个列表中的差异

3 个答案: