How do I find if a string(A) exists in another string(B) using c#?

Question

Before marking this as duplicate, please read the details here.

Example 1:

String A: The seven habits of highly effective people.

String B: "This is a sample text. There is only one product in it. It is a book. The book is The seven habits of highly effective people."

Example 2:

String A: The seven habits of highly effective people.

String B: "This is a sample text. There is only one product in it. It is a book. The book is The seven habits of highly effective peopl."

Now solving the above examples with a code like
B.Contains(A)
will give the correct results. However the same code will return "false" as output in Example 2.

How do I resolve this problem?

There is an "e" missing in example 2 and I am aware about it and that's the problem. How do I compare one string with another where string A is nearly identical with a "part of string B"?

@user1039119 - Same code returns "false" as output in Example 2,as the complete string is not there. what you want to achieve ? — Bibhu
– Bibhu, Commented Sep 11, 2013 at 8:24
At the end of string B in ex.2, you have peopl. not people. — jwaliszko
– jwaliszko, Commented Sep 11, 2013 at 8:25
The strings in example 2 are obviously different - if you want to get matches for "nearly indentical" strings it gets difficult very fast, simply because defining "nearly identical" is fun. — Christian Sauer
– Christian Sauer, Commented Sep 11, 2013 at 8:25
What you're looking for is measuring how similar two strings are, then setting some threshold for how similar "similar enough" is. This question looks like a good place to start — BambooleanLogic
– BambooleanLogic, Commented Sep 11, 2013 at 8:29

Simon Whitehead · Accepted Answer · 2013-09-11 08:34:53Z

2

As stated in my comment.. the Levenshtein Distance algorithm (and similar ones) compute differences between strings and return a numerical result (wiki: http://en.m.wikipedia.org/wiki/Levenshtein_distance).

However, I would definitely apply benchmarking and caching strategies for these algorithms. They are decent with small input.. but when I have implemented it I have had to make sure I cache results / lookups. Your large example will not perform "fast".. depending on what "fast" is for your use case.

answered Sep 11, 2013 at 8:34

Simon Whitehead

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2 Comments

milan m Over a year ago

Hi Simon, thanks for your answer. I have gone through the details about the algorithm you mentioned. However, the problem I am facing is not about comparing two nearly identical strings. The Problem is comparing one string with "part of another string" which is nearly identical. As shown in the example I mentioned, only a part of String B matches string A. hence the distance algorithm may not give accurate results.

I4V Over a year ago

@user1039119 you can calculate it, for example, bool contains = Math.Abs(Math.Abs(strB.Length - strA.Length) - levenshteinDistance) < 3;

rekaszeru · Accepted Answer · 2013-09-11 12:02:29Z

1

You can use string.compare, Find below few examples which may help you.

string a = "a"; 
string b = "b"; 
int c;

c = string.Compare(a, b);
Console.WriteLine(c);

c = string.CompareOrdinal(b, a);
Console.WriteLine(c);

c = a.CompareTo(b);
Console.WriteLine(c);

c = b.CompareTo(a);
Console.WriteLine(c);

edited Sep 11, 2013 at 12:02

rekaszeru

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answered Sep 11, 2013 at 8:39

ZubinAmit

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Comments

Chopchop · Accepted Answer · 2013-09-11 08:50:19Z

0

What you are looking for looks like a search engine with score rate.

I used the Levenshtein Distance methode to search/compare string that looks like the same but who are not.

there is an example at the following link :

http://www.dotnetperls.com/levenshtein

answered Sep 11, 2013 at 8:50

Chopchop

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Comments

milan m · Accepted Answer · 2013-09-19 06:04:01Z

0

I am answering my own question.

I was looking for a solution to compare one string with another where string A is nearly identical with a "part of string B".

This is how I resolved the issue.

I applied the "Longest Common Substring" algorithm and founded the longest common substring between the two strings.
Then I used "Levenshtein Distance algorithm" to compare my String A with the "Longest Common Substring" found from step 1.
If the result available from the algorithm mentioned in step 2 is above certain threshold, then it implies that the string A exists in String B.
Problem Solved.

I have worked on the problem for one day and I have found decent results for the problem.

edited Sep 19, 2013 at 6:04

answered Sep 12, 2013 at 5:10

milan m

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