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Script for finding standard deviation

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Top Forums Shell Programming and Scripting Script for finding standard deviation
# 1  
Old 09-11-2008
Script for finding standard deviation
I have a CSV file that looks like
Code:
 
0,0,0,0,1,0,1,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0
10,11,7,0,4,12,2,3,7,0,11,3,12,4,0,5,5,4,5,0,8,6,12,0,9,3,3,0,2,7,8
19,11,7,0,4,14,16,10,8,2,13,7,15,6,0,76,6,4,10,0,18,10,17,1,11,3,3,0,9,9,8
22,11,13,1,5,14,16,10,9,10,13,7,16,6,0,59,6,4,10,0,18,13,17,1,11,3,3,0,12,9,10
22,11,13,1,5,14,16,10,9,10,13,7,16,6,22,90,6,4,10,0,18,13,17,1,11,3,4,0,12,9,10
41,18,27,9,27,41,59,20,27,54,63,34,28,43,40,131,7,8,19,0,62,16,30,23,25,3,4,9,24,12,19
42,18,27,9,27,41,59,20,27,55,68,36,28,46,41,132,7,8,19,13,64,16,31,25,25,3,4,9,24,12,19
125,124,78,62,97,87,145,70,87,119,150,124,99,95,41,175,85,58,57,88,142,83,92,102,107,80,45,64,64,94, 89
125,126,78,62,99,87,145,70,87,119,161,124,99,95,41,175,85,58,58,88,142,84,112,103,108,80,68,64,65,98 ,89
189,254,164,153,192,153,230,132,188,163,210,210,167,198,93,235,146,110,97,130,211,107,181,140,151,11 9,105,105,178,126,165
189,324,168,192,194,159,233,132,192,169,244,210,167,201,103,235,147,152,180,181,213,107,192,190,212, 119,119,126,195,126,166
189,324,168,255,194,225,233,141,192,230,244,260,167,201,172,283,181,206,217,216,261,107,192,235,212, 119,169,197,264,189,229
366,438,315,319,382,287,398,320,416,382,407,397,342,448,276,392,297,368,237,347,336,332,384,405,412, 284,329,350,396,326,356

I need to find the stadard deviation for each individual row. Here is the code I have so far. I can't get the square root to work and also I can't get any floating point numbers.
Code:
 
for i in `cat file.csv ` 
do
     x1=0
     x2=0
     sigma=0
     IFS=, 
     for f in $i 
          do  
          let x1=$x1+$f
          let x2=$f*$f+$x2
     done 
     let x1=$x1/30
     let x2=$x2/30
     let sigma=sqrt($x2-$x1*$x1)
     echo "Mean = " $x1
     echo "Standard Deviation = " $sigma
done

# 2  
Old 09-11-2008
The shell does only integer arithmetic operations. You need to use awk or perl or some other envrionment that supports FP operations.
# 3  
Old 09-11-2008
Ok. Can anyone help me rewrite my above script into awk or perl. awk would be preferred? Thanks.
# 4  
Old 09-11-2008
using your algorithm.... in awk which supports FP.
Code:
awk -F','  '{ sum=0; sumsq=0;
                for(i=1; i<=NF;i++) {sum+=$i; sumsq+=$i*$i}
                printf("mean=%f  stddev= %f\n", sum/NF, sqrt(sumsq - (sum*sum)) )
              } ' file.csv

# 5  
Old 09-11-2008
Tools 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!Smilie
Last edited by joeyg; 09-11-2008 at 12:44 PM.. Reason: added comment at end
# 6  
Old 09-11-2008
Quote:
Originally Posted by jim mcnamara
using your algorithm.... in awk which supports FP.
Code:
awk -F','  '{ sum=0; sumsq=0;
                for(i=1; i<=NF;i++) {sum+=$i; sumsq+=$i*$i}
                printf("mean=%f  stddev= %f\n", sum/NF, sqrt(sumsq - (sum*sum)) )
              } ' file.csv

This awk code throws the following error.
awk: The sqrt parameter to a math library function is not in the domain.

This means that the portion to find the average works fine but because sqrt throws an error the std deviation does not work. I think this is because sumsq - (sum*sum) is a negative number.
Last edited by RJ17; 09-11-2008 at 01:16 PM..
# 7  
Old 09-11-2008
You are correct. I copied your algorithm - it needs checks.

Code:
awk -F','  '{ sum=0; sumsq=0;
                for(i=1; i<=NF;i++) {sum+=$i; sumsq+=$i*$i}
                printf("mean=%f  stddev= %f\n", sum/NF, 
                sqrt(  ( (sumsq - (sum*sum))< 0) 
                           ? sumsq - (sum*sum)*-1 : sumsq -(sum*sum) )
              } ' file.csv

This should prevent domain errors.... the fact that there are a lot of zero values means the sum of squares can be very small number. You could also use a function like this placed at the top of the awk code block
function abs(n) { return (n <0)? n*=-1 : n}


Code:
awk -F','  '{ function abs(n) { return (n <0)? n*=-1 : n}

                sum=0; sumsq=0;
                for(i=1; i<=NF;i++) {sum+=$i; sumsq+=$i*$i}
                printf("mean=%f  stddev= %f\n", sum/NF, sqrt(abs(sumsq - (sum*sum))) )
              } ' file.csv
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