Normalize values to a smaller range
By : Dasaraden Mauree
Date : March 29 2020, 07:55 AM
Hope this helps I'm actually surprised that you are not getting collisions before you normalize the range of hash values. It looks like you are using an unnormalized range of [0,2^32). Looking at the Birthday Problem chart here the probability of collision with 4*10^7 elements should be higher than 75%. In any case, normalizing the hash outputs to a range equal to the size of the set of elements is practically guaranteeing a non-trivial number of collisions. Unless you're willing to use a counter for your hash values I don't see how you'll be able to avoid that. EDIT: Saw your edit. Even with a range 100 times the number of elements (which is about 4*10*9) you are still likely to get a lot of collisions. As indicated above, the probability of one or more collisions is well over 75%.
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Normalize values within a range
By : Tom Chan
Date : March 29 2020, 07:55 AM
Hope this helps Is there a formula that will normalize values within a range so their average will equal the middle of the range? , You can subtract the average and add 3: code :
select 3 + (t.x - avg(t.x) over ())
from t;
select (max(t.x) over () + min(t.x) over ()) / 2 + (t.x - avg(t.x) over ())
from t;
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Normalize values to a range between -1 and 1
By : user2169695
Date : March 29 2020, 07:55 AM
this one helps. I want to process joystick values in such a way that the "normalized" values lay between -1 and 1 (including numbers with decimal places, for example 0.0129). , This should do the trick: code :
float[] NormalizeStickData(float[] stickData)
{
return new[]
{
Normalize(stickData[0], 35, 228, 128, -1, 1, 0),
Normalize(stickData[1], 27, 220, 128, -1, 1, 0)
};
}
float Normalize(float value, float oldMin, float oldMax, float oldMid, float newMin, float newMax, float newMid)
{
if (value < oldMid)
return Interpolate(value, oldMin, oldMid, newMin, newMid);
else if (value > oldMid)
return Interpolate(value, oldMid, oldMax, newMid, newMax);
else
return newMid;
}
float Interpolate(float value, float oldMin, float oldMax, float newMin, float newMax)
{
return (float)(newMin + (newMax - newMin)*(value - oldMin)/(oldMax - oldMin));
}
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Normalize values in the range of [0 -1]
By : patel saurabh
Date : March 29 2020, 07:55 AM
wish help you to fix your issue If I understood you correctly, you should use a Sigmoid function, which refers to the special case of the Logistic function. Sigmoid and other logistic functions are often used in Neural networks to shrink (compress or normalize) input range of data (for example, to [-1,1] or [0,1] range).
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How to normalize an image to a range of values
By : BifiBifi
Date : March 29 2020, 07:55 AM
This might help you What you want is cv::normalize. Also, its Java equivalent can be found here.
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