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Shell Programming and Scripting
Script for finding standard deviation
Post 302235196 by joeyg on Thursday 11th of September 2008 10:38:25 AM
09-11-2008
Perhaps way out in left field, but
Depending on the accuracy required, you might consider
(a) For each of your values, multiplying by 100 or 1000 prior to beginning any math. Then know that you have to remove the extra digits and they are after the decimal point. For example 3/2 = 1 in integer, but 300/2 = 150 or adjusted 1.50
(b) An approximation for square root can be done in two parts. First off, add up all odd numbers until you are greater than the starting number. For example, sqrt of 10 would give you 1+3+5+7 and those four pieces are greater than the 10 you started with, so sqrt=3 (one less) as integer. Perhaps easier to see in the following [to get the integer part]
1 sqrt = 1+3 (more), so one digit is 1
2 sqrt = 1+3 (more), so 1
3 sqrt = 1+3 (more), so 1
4 sqrt = 1+3+5 (more), so 2 (again think one less)
5 sqrt = 1+3+5 (more), so 2
...
9 sqrt = 1+3+5+7 (more), so 3
To get to the decimal part there is another strange methodology involving looking at remaindors. In short the sqrt of 5 starts off with a 2 as seen above. Adding 1+3+5=9 and that is 4 too many (9-5). My last number in the 1+3+5 was a 5 and if I have 4 too many, I only needed a 1 (5-4=1). Take the 1 and the 5 and do 1/5 = .2
Add the first 2 to the .2 and you get 2.2 vs. actual of 2.23
For 8, start with the 2 as the integer. That is 1 too many (9-8). My last number was 5 again (in 1+3+5), so I only needed 4. Take that 4 and 5 to get to 4/5 = .8
Add the first 2 to this .8 and you get 2.8 vs actual 2.82
This is normally within a couple hundredths of the pure answer.
***
And I knew by the time I could write all that up, someone would have a program solution. But what the heck, if you can follow the logic of what I wrote for approximating sqrt, then you might agree it to be a cool function!
(a) For each of your values, multiplying by 100 or 1000 prior to beginning any math. Then know that you have to remove the extra digits and they are after the decimal point. For example 3/2 = 1 in integer, but 300/2 = 150 or adjusted 1.50
(b) An approximation for square root can be done in two parts. First off, add up all odd numbers until you are greater than the starting number. For example, sqrt of 10 would give you 1+3+5+7 and those four pieces are greater than the 10 you started with, so sqrt=3 (one less) as integer. Perhaps easier to see in the following [to get the integer part]
1 sqrt = 1+3 (more), so one digit is 1
2 sqrt = 1+3 (more), so 1
3 sqrt = 1+3 (more), so 1
4 sqrt = 1+3+5 (more), so 2 (again think one less)
5 sqrt = 1+3+5 (more), so 2
...
9 sqrt = 1+3+5+7 (more), so 3
To get to the decimal part there is another strange methodology involving looking at remaindors. In short the sqrt of 5 starts off with a 2 as seen above. Adding 1+3+5=9 and that is 4 too many (9-5). My last number in the 1+3+5 was a 5 and if I have 4 too many, I only needed a 1 (5-4=1). Take the 1 and the 5 and do 1/5 = .2
Add the first 2 to the .2 and you get 2.2 vs. actual of 2.23
For 8, start with the 2 as the integer. That is 1 too many (9-8). My last number was 5 again (in 1+3+5), so I only needed 4. Take that 4 and 5 to get to 4/5 = .8
Add the first 2 to this .8 and you get 2.8 vs actual 2.82
This is normally within a couple hundredths of the pure answer.
***
And I knew by the time I could write all that up, someone would have a program solution. But what the heck, if you can follow the logic of what I wrote for approximating sqrt, then you might agree it to be a cool function!
Last edited by joeyg; 09-11-2008 at 11:44 AM.. Reason: added comment at end
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LEARN ABOUT OPENSOLARIS
mlib_imagestddev
mlib_ImageStdDev(3MLIB) mediaLib Library Functions mlib_ImageStdDev(3MLIB) NAME
mlib_ImageStdDev - image standard deviation SYNOPSIS
cc [ flag... ] file... -lmlib [ library... ] #include <mlib.h> mlib_status mlib_ImageStdDev(mlib_d64 *sdev, const mlib_image *img, const mlib_d64 *mean); DESCRIPTION
The mlib_ImageStdDev() function computes the standard deviation for each channel in the source image. It uses the following equation: 1 w-1 h-1 sdev[i] = { ----- * SUM SUM (img[x][y][i] - mean[i])**2 }**0.5 w*h x=0 y=0 where, in the case of mean == NULL, 1 w-1 h-1 mean[i] = ----- * SUM SUM img[x][y][i] w*h x=0 y=0 PARAMETERS
The function takes the following arguments: sdev Pointer to standard deviation array, whose size is the number of channels in the source image. sdev[i] contains the standard devia- tion of channel i. img Pointer to input image. mean Pointer to pre-computed mean array for each channel. (If NULL, it will be computed.) mean[i] contains the mean of channel i. RETURN VALUES
The function returns MLIB_SUCCESS if successful. Otherwise it returns MLIB_FAILURE. ATTRIBUTES
See attributes(5) for descriptions of the following attributes: +-----------------------------+-----------------------------+ | ATTRIBUTE TYPE | ATTRIBUTE VALUE | +-----------------------------+-----------------------------+ |Interface Stability |Committed | +-----------------------------+-----------------------------+ |MT-Level |MT-Safe | +-----------------------------+-----------------------------+ SEE ALSO
mlib_ImageMean(3MLIB), mlib_ImageMean_Fp(3MLIB), mlib_ImageStdDev_Fp(3MLIB), attributes(5) SunOS 5.11 2 Mar 2007 mlib_ImageStdDev(3MLIB)