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12-09-2011 09:31 AM

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Hello,

I did a google and forum search for this and came up empty. I'd like to know if there should be a difference in how the two work ?

I'm asking because I was giving the exact same X array to both and was getting two different results, unless I also specified the interval/#bins to be equal for both of them. (interval for Histogram VI and #bins for General Histogram VI)

This also means I'd like to know what the difference is between intervals and #bins ? Are they the same thing with just different names ?

This is more to get a general understanding on how the two VIs work than getting a specific solution for a problem by the way, so any help is appreciated 🙂

Thanks in advance.

Solved! Go to Solution.

12-09-2011 10:38 AM - edited 12-09-2011 10:38 AM

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@Pladio wrote:

[...]

This also means I'd like to know what the difference is between intervals and #bins ? Are they the same thing with just different names ?

For the purposes of understanding these VI's, they are the interchangeable. In the case of statistics, the interval can be thought of "the number of pieces of data you want to spread across a population". When doing this, you are placing the pieces of data into virtual "bins". Hence, they are basically the same thing.

As far as the difference in the two VI's , there's nothing I can say that can't be said by comparing the help for each VI. The advanced VI just gives you tighter control of your data.

Richard

12-09-2011 10:40 AM

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Thanks. Any idea why the same data would give different histograms when not specifying bins ?

Solution

Accepted by topic author Pladio

12-09-2011 12:06 PM

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@Pladio wrote:

Thanks. Any idea why the same data would give different histograms when not specifying bins ?

Yep.

The normal VI, if you do not wire the "interval" input, will give you 10 bins, becaue that input is defaulted to 10.

The advanced VI, if you do not wire the "#bins" input, will give you a certain number of bins based on how many observations are being made, or, the size of the incomming array. This number is 1 + 3.3 log ** n**, where

Richard