Finding Differences in Two Lists
I am thinking of a good way to find differences in two lists
here is the problem:
The two lists have multiple strings where the first 3 numbers / characters (* delimited) represent a unique key (followed by the text String = "key1 * key2 * key3 * text").
Here's an example line:
AA1*1D*4*The quick brown fox*****CC*3456321234543~
where "* AA1 * 1D * 4 *" is a unique key
List1: "index1 * index2 * index3", "index2 * index2 * index3", "index3 * index2 * index3"
List2: "index2 * index2 * index3", "index1 * index2 * index3", "index3 * index2 * index3", "index4 * index2 * index3"
I need to map the indices in both lists and compare them.
-
If all 3 indices from 1 list match 3 indices from another list, I need to keep track of both strings in the new list
-
If there is no set of indices in one list that does not appear in the other, then I need to keep track of one side and keep an empty entry on the other side. (# 4 in the above example)
return list
This is what I've done so far, but I'm kind of struggling here:
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;
}
}
}
}
Could you suggest a good algorithm for this process?
thank
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If I understand your question correctly, you would like to compare items by their "key" prefix, not by the entire content of the string. If so, implementing a custom equality mapping will allow you to easily use LINQ collection algorithms.
This program...
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...
}
}
... prints the following output:
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
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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);
}
}
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