The element in the middle of the array

[jiji_830]

plz any suggestion how to solve this :

If the array has an even number of elements, then use (N/2) as the middle element. For example, if the array has
ten elements then the middle-most would be (10/2) = 5 or the 5th element in the array. If the array has an odd number of elements, then the middle will be (N/2) + 0.5. For example, if the array has
seven elements then the middle-most would be (7/2) + 0.5 = 4 or the 4th element in the array.

[crossrulz]

See if this meets your requirements.

The Ramp Pattern stuff was just to make an easy array to verify the rest of the code worked.

[GerdW]

Hi jiji,

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@jiji_830 wrote:

1. If the array has an even number of elements, then use (N/2) as the middle element. For example, if the array has
   ten elements then the middle-most would be (10/2) = 5 or the 5th element in the array.
2. If the array has an odd number of elements, then the middle will be (N/2) + 0.5. For example, if the array has
   seven elements then the middle-most would be (7/2) + 0.5 = 4 or the 4th element in the array.

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1. When the array has an even number of elements then there is no "middle" element! Example: in an array of 4 elements [1 2 3 4] you will pick either the 2nd or the 3rd element as "middle". Your calculation "N/2" would constantly pick the element before the "middle"…
2. When the array has an uneven number of elements then there will be a clear "middle" element - as noted by your calculation…

[Bob_Schor]

I'm not sure where you are having problems -- with the concept of "middle of the array", with the "decision-making" that is described in the algorithm you posted, or "how to program" such an algorithm.

Personally, I think the biggest issue is "What is the middle element of a discrete ordered set of data"?  If I hold up three finger, I think all reasonable people will agree that the second finger, with one finger on either side, is the "middle" finger.  What if I hold up 4 fingers?  Is it the second?  Is it the third?  What is meant by "middle"?

In statistics, there is the concept of the "mean", or "average".  If I number my fingers 1, 2, 3, ..., and take the mean of the 3 finger numbers, I'll compute (1+2+3)/3 = 6/3 = 2.  If I do this for only 2 fingers, I will compute (1+2)/2 = 1.5, which is exactly half-way between 1 and 2.  But if I restrict myself to integers (because I want to identify the "middle finger"), I am forced to use an integer, so I "adopt a convention" and round up, or round down (depending on what convention I choose).

Let's number the elements (or "fingers") starting with 1 and numbering 1, 2, 3, ...  How do we find the "average" (= "middle") value?  Simply sum the series and divide by the number of entries.  Do you remember from High School Algebra the formula for 1+2+...+n?  It is n*(n+1)/2.  If we divide by n, we get the "middle" as (n+1)/2, which is an Integer if n is odd.

Sometimes the more interesting questions are the ones that don't get asked ...

Bob Schor

[altenbach]

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@jiji_830 wrote:

plz any suggestion how to solve this :

If the array has an even number of elements, then use (N/2) as the middle element. For example, if the array has
ten elements then the middle-most would be (10/2) = 5 or the 5th element in the array. If the array has an odd number of elements, then the middle will be (N/2) + 0.5. For example, if the array has
seven elements then the middle-most would be (7/2) + 0.5 = 4 or the 4th element in the array.

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You are making a sequence of incorrect statements and I am not sure what you consider "solve this". (implement the faulty math in LabVIEW? Come up with better definitions?)

As others have said, a 1D array with an even number of elements has no middle element, but you can of course do a linear interpolation between the two values adjacent to the middle. A 1D array with an odd number of elements has a clear middle. In all cases, the indices of the interesting element can be calculated from the array size.

[FireFist-Redhawk]

I feel like this was a school assignment and the problem statement was copy-pasted. Such that the "middle" element here is defined exactly how the problem statement says, much to all of our dismay.

[altenbach]

"Interpolate array" will correctly interpolate for fractional indices, so this will give the value of the middle element (size=odd), or the average of the two elements adjacent to the middle (size=even).

[jiji_830]

thank u all for ur clarification. i miss understanding the problem now I got an idea. i will fixe it. thank u all so much