How do I find an item in a list(of T)

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Hello, I'm trying to understand how to check duplicates names in a list(of T). I have a name that's not in the list yet, and before I put it in the list, I need to check for duplicates before adding the new information to the list.

Here's some code that Griff helped me out with yesterday.

Dim contacts As New List(Of Contact)()
Dim newContact As New Contact()

newContact.FirstName = "joe"
newContact.LastName = "smith"
newContact.PhoneNumber = "234345435"
newContact.Birthday = Date.Now.AddYears(-20)
contacts.Add(newContact)

What I want to know is, does the last name in the existing class match the the last name that's in memory that I haven't instantiated yet?

I appreciate any help at all,
thanks

What I have tried:

I've tried the above code and I've been searching the net for days. Most of the answers are about using a predicate and I don't know how to use that either. So I would like to know how to do it without it first if possible.

Posted 13-Apr-16 1:52am

jumper777

Updated 13-Apr-16 2:22am

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F-ES Sitecore 13-Apr-16 7:56am

I won't post this as a solution as I don't know the vb.net, but it's going to be like

if not contacts.Any(c => c.LastName.Equals(nameToCheck)) then
' lastname doesn't exist in the collection
end if

If you check out the overloads of Equals you can make it case insensitive if that's what you need.

Sergey Alexandrovich Kryukov 13-Apr-16 9:07am

Why not avoiding duplicates in first place, when you add the items?
If you need fast search in pretty big data sets, lists are not the best. Then you should consider associative collections based on hash; dictionaries, sets, sorted lists, sorted dictionaries...
—SA

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CHill60 · Accepted Answer · 2016-04-13T02:22:00

Solution 1

Try the following:

VB

If Not contacts.Any(Function(obj) obj.LastName = "smith") Then
    Dim newContact As New Contact()
    newContact.FirstName = "joe"
    newContact.LastName = "smith"
    newContact.PhoneNumber = "234345435"
    newContact.Birthday = Date.Now.AddYears(-20)
    contacts.Add(newContact)
End If

The bit in brackets after Any is a Lambda expression[^]
It's essentially saying - here is a Function that is going to take a parameter obj and with obj we want to look at the LastName property and compare it to the text "smith"(you would probably want to convert to lowercase or capitals before doing the comparison).

The compiler knows that LastName can be a property of obj because we are using contacts.Any and it knows that contacts is a List of Contact. We are using .Any because we don't actually need a list of matches, we just need to know if there is a match.
And note the Not because the .Any will return True if it does find a match

Posted 13-Apr-16 2:22am

CHill60

Comments

Sergey Alexandrovich Kryukov 13-Apr-16 9:10am

5ed.
Two points: 1) if the task of removing duplicates appears, it usually indicates a big flaw in a wider-scope design; duplicates should be prevented, not detected and removed; 2) for bigger data sets and search, lists are not suitable; the associative collections based on hash and buckets should be used.
—SA

CHill60 13-Apr-16 9:11am

Thank you - and well said on the other points raised.

jumper777 13-Apr-16 12:04pm

Thank you Sergey. I am trying to avoid duplicates. And yes it's before the record is saved to the list. I try to validate when the save button is clicked. This is when I'm trying to decide whether to save the data or not.

Thanks for telling me about the "Any" part. I'm pretty new at lists. And I've heard about Lambda, but haven't studied it yet. At them moment, this program is not a real world program. I'm trying to teach myself concepts right now so when an actual program comes up, I will be prepared. I'm going to try your solution right now.
Thank you very much.

jumper777 13-Apr-16 12:13pm

Sergey, you are my hero man. I just marked your answer as the accepted solution. I did have to change one "tiny" part. The "NOT" operator that you talked about is actually reversed. If you use Not, it will add the the new data to the list, but if you remove it, it does not. But it only took me a one time through the debugger to find that out.

Can't thank you enough.

CHill60 13-Apr-16 16:17pm

That's good use of the debugger. Nice one.
However I'm not SA ;-)

jumper777 13-Apr-16 17:09pm

Sergey, I have to apologize about the "Not" statement. I think I was wrong and I hope you didn't think I was rude, because I really do thank you for your help.