[RESOLVED] Loop through array to get min and max values

[DylanM]

Hi. I need to loop through different parts of an array to get Max and Min values. Any help greatly appreciated.

[Joacim Andersson]

Post moved to its own thread. Do not hijack someone else's thread to ask your own question and never be rude to someone just because they pointed this out to you. Your rude post has been deleted, keep the language civil in the future.

[Shaggy Hiker]

For the entire array, you could use .Max or .Min. Since you only want to search through a portion of the array you'd have to go about it a bit differently. Something like this should do:

Dim maxV as Integer = Integer.MinValue
Dim minV as Integer = Integer.MaxValue

For x = lowIndex to highIndex
 If yourArray(x) > maxV Then
  maxV = yourArray(x)
 End If
 If yourArray(x) < minV Then
  minV = yourArray(x)
 End If
Next

There are lots of assumptions in that code, such as assuming that the array contains integers. If that assumption is wrong, then the types would have to be changed to the correct type. This also retains the maximum and minimum values between lowIndex and highIndex, but it doesn't retain which index had the low and high values.

[dday9]

I would personally do this:

Dim intArray() As Integer = {56, 298, -14, 20}

Array.Sort(intArray)

Console.WriteLine(String.Format("Min: {0} Max: {1}", intArray(0), intArray(intArray.Length - 1)))

[Shaggy Hiker]

That works if the search is for the min/max of the entire array, and you don't need to retain the original order of items in the array. Lord only knows what the OP wants.

[kaliman79912]

Expanding on the solution proposed by Shaggy

Dim maxV As Integer = Integer.MinValue
Dim minV As Integer = Integer.MaxValue

For x As Integer = lowIndex To highIndex
    maxV = Math.Max(maxV, yourArray(x))
    minV = Math.Min(minV, yourArray(x))
Next

[dday9]

That works if the search is for the min/max of the entire array, and you don't need to retain the original order of items in the array. Lord only knows what the OP wants.

That's true. It's all very vague at this point.

[kaliman79912]

I would personally do this:

Dim intArray() As Integer = {56, 298, -14, 20}

Array.Sort(intArray)

Console.WriteLine(String.Format("Min: {0} Max: {1}", intArray(0), intArray(intArray.Length - 1)))

Yep, that would work. I don't know if the performace is great if you need to sort the array. And would that be acceptable? I mean, to modifiy the order of the elements? Maybe, if the entire array is to be checked, as Shaggy said, you cold just

maxV = yourArray.Max
minV = yourArray.Min

[Shaggy Hiker]

Expanding on the solution proposed by Shaggy

Dim maxV As Integer = Integer.MinValue
Dim minV As Integer = Integer.MaxValue

For x As Integer = lowIndex To highIndex
    maxV = Math.Max(maxV, yourArray(x))
    minV = Math.Min(minV, yourArray(x))
Next

When I saw that approach, I felt that I had to compare my code to that code. This code is certainly cleaner than what I had written, so I was curious as to whether or not it performed better or worse. Therefore, I put both sets of code into my test app and compared the performance.

What I wrote ends up being about 17% faster, which is a pretty trivial amount. It was consistent, though. Once again it shows that VB gives you many ways to solve a problem, some are verbose and others are concise. However, verbose always seems to top concise. I don't see why that should be.

[kaliman79912]

When I saw that approach, I felt that I had to compare my code to that code. This code is certainly cleaner than what I had written, so I was curious as to whether or not it performed better or worse. Therefore, I put both sets of code into my test app and compared the performance.

What I wrote ends up being about 17% faster, which is a pretty trivial amount. It was consistent, though. Once again it shows that VB gives you many ways to solve a problem, some are verbose and others are concise. However, verbose always seems to top concise. I don't see why that should be.

I have thought about that issue a lot. And I came to the conclusion that when the application is not performance critical, then using easier to read code is best; but when seconds or even milliseconds are important, then you go for the most efficient code. It is a mater of choice of course.

[techgnome]

And just because we're throwing things around... here's yet another way:

Dim intArr As Integer() = {1, 2, 3, 6, 32, 16, 8, 9, 45, 335, 7, 878, 4, 566}

Dim startIdx As Integer = 5
Dim endIdx As Integer = 10

Dim intMax As Integer = intArr.Take(endIdx).Reverse.Take(startIdx).Reverse.Max
Dim intMin As Integer = intArr.Take(endIdx).Reverse.Take(startIdx).Reverse.Min

MessageBox.Show(String.Format("Max: {0} -- Min: {1}", intMax, intMin))

-tg

[Shaggy Hiker]

Woof. You couldn't throw that far enough for my taste. It's certainly concise, but readable....not so much.

[kaliman79912]

And just because we're throwing things around... here's yet another way:

Dim intArr As Integer() = {1, 2, 3, 6, 32, 16, 8, 9, 45, 335, 7, 878, 4, 566}

Dim startIdx As Integer = 5
Dim endIdx As Integer = 10

Dim intMax As Integer = intArr.Take(endIdx).Reverse.Take(startIdx).Reverse.Max
Dim intMin As Integer = intArr.Take(endIdx).Reverse.Take(startIdx).Reverse.Min

MessageBox.Show(String.Format("Max: {0} -- Min: {1}", intMax, intMin))

-tg

Very interesting, can you explain how this one works?

[passel]

The truth is, it doesn't work.
Using the indicies 5 and 10 make it seem to work, but that is only for that case.
This is the corrected version with different values that would show the original failing.

Dim intArr As Integer() = {1, 2, 3, 6, 32, 16, 8, 9, 45, 335, 7, 878, 4, 566}

Dim startIdx As Integer = 2
Dim endIdx As Integer = 8

Dim intMax As Integer = intArr.Take(endIdx).Reverse.Take(endIdx - startIdx).Reverse.Max
Dim intMin As Integer = intArr.Take(endIdx).Reverse.Take(endIdx - startIdx).Reverse.Min

MessageBox.Show(String.Format("Max: {0} -- Min: {1}", intMax, intMin))

If you put your cursor on Take, and hit F1, and after reading what Take does, then do the same on Reverse, and you should be able to figure out what it is doing. That might be more entertaining than just explaining it.
Also, if you wanted to include the endIdx value in the test, then you would have to increment the endIdx value by 1 before using the code as written (as take would return 8 values, not including the 8th index).

[techgnome]

Bleh... yeah, I see why it doesn't work ... I just happened to hit the right numbers with 5/10 where it did work... even still... not that hard to fix - as you noted in the code... :P

-tg

[Joacim Andersson]

Reversing the array 4 times can be very slow if the array is large. If you want to go with this it's better to do it this way:

    Dim intArr As Integer() = {1, 2, 3, 6, 32, 16, 8, 9, 45, 335, 7, 878, 4, 566}

    Dim startIdx As Integer = 2
    Dim endIdx As Integer = 8

    Dim subsetArr As IEnumerable(Of Integer) = intArr.Take(endIdx + 1).Reverse.Take(endIdx - startIdx + 1).Reverse
    Dim intMin As Integer = subsetArr.Min
    Dim intMax As Integer = subsetArr.Max
    MessageBox.Show(String.Format("Max: {0} -- Min: {1}", intMax, intMin))

Also note that I changed the code so that it actually works. You should Take(endIdx + 1) and Take(endIdx - startIdx + 1) since 0 is the first index in the array.

[DylanM]

Joacim Andersson, I'll do as Im please. **** off

[dbasnett]

Joacim Andersson, I'll do as Im please. **** off

WOW! Let me guess. This is the only question you intend on asking here.

[techgnome]

Joacim Andersson, I'll do as Im please. **** off

Wow... I think JA's reply was more directed at me... so for you to go off on a mod like that... just wow...


-tg

[vbfbryce]

No, he's still hung up on post #2, but still...

WOW

[Joacim Andersson]

Joacim Andersson, I'll do as Im please. **** off

I'm sorry but in this forum you can't do as you please. We do have rules that you agreed to when you signed up. Since I already warned you about using foul language you may now go and do what you want somewhere else. Bye.

[kaliman79912]

I forsee the future... the list of members of the forum will decrease by one really soon

[vbfbryce]

Can you give me tonight's Powerball numbers, too?!

[techgnome]

I forsee the future... the list of members of the forum will decrease by one really soon

Already done...

Can you give me tonight's Powerball numbers, too?!

1,2,3,4,5,6,7,8,9,10,11,12,13,14,.... oh... you meant tonight WINNING Powerball numbers...

-tg

[vbfbryce]

Yeah, the WINNING numbers!

I'll give you 1% of my winnings if you do!

[dday9]

I love how this thread became a CC thread after the OP's second post. By the way I'll talk that 1% but do they have to be winning numbers?

[vbfbryce]

Have to match at least 5 of the numbers, yes!

[Shaggy Hiker]

Let's just let this one die. It should have been locked anyways. I gave a serious answer because, while I felt the OP was a bit nutty, I thought they were fumbling towards reasonability. I'm sorry to see that was not the case and that the OP ended up just being pushy and uncivil.

Let's just let it fade away.

[Joacim Andersson]

Well I actually thought that by just marking this thread as resolved (since the OP actually got his question answered before he insulted me) would be enough to end the discussion but since it apparently hasn't (and since this is not a CC thread) I guess I have to close it. (However this is not the CC forum and I shouldn't have to stop the discussion by closing the thread, you guys should know better!)