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"Smoothing" the data

 
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Top Forums UNIX for Dummies Questions & Answers "Smoothing" the data
# 8  
Old 02-21-2012
I used the code that you suggested:
For the data:

Code:
3002.000
3002.300
3002.600
3002.900
3003.200
3003.500
3003.800
3004.100
3004.400
3004.700
3005.000
3005.300
3005.600
3005.900
3006.200
3006.500
3006.800
3007.100
3007.400
3007.700
3008.000
3008.300

Code:
BEGIN { WIDTH=10 }

{       DATA[(++N)%WIDTH]=$1    }

(N>=WIDTH) {
        V=0
        for(X=(N+1); X<=(N+WIDTH); X++)
                V+=DATA[X%WIDTH];

        print V/WIDTH;
}

I was expecting the output:

3002.000
3005.000
3008.000

But I got:

Code:
3003.35
3003.65
3003.95
3004.25
3004.55
3004.85
3005.15
3005.45
3005.75
3006.05
3006.35
3006.65
3006.95
3007.25
3007.55
3007.85
3008.15
3008.45

Maybe I misunderstood the code?
# 9  
Old 02-21-2012
Hi.
Quote:
Originally Posted by cosmologist
I was expecting the output:
3002.000
3005.000
3008.000
Quote:
The SMOOTH function returns a copy of Array smoothed with a boxcar average of the specified width. The result has the same type and dimensions as Array.

-- excerpt from SMOOTH
Yet you are expecting only 3 values back from an input vector of 22 values. Why is that?

Best wishes ... cheers, drl
# 10  
Old 02-21-2012
Quote:
Originally Posted by cosmologist
Maybe I misunderstood the code?
There were other parts than code, which you didn't read. I detailed clearly what I was trying to do, none of which had anything to do with stripping off decimal points -- you wanted a boxcar average...

Code:
echo "3.14159" | awk '{ printf("%d\n", $1) }'
3
$

That still doesn't clear up what you want to happen to the second column of data. You need to tell us what we're supposed to do to that. It's not going to change itself.
# 11  
Old 02-21-2012
Quote:
Originally Posted by Corona688
There were other parts than code, which you didn't read. I detailed clearly what I was trying to do, none of which had anything to do with stripping off decimal points -- you wanted a boxcar average...
I read that part and it made sense, but when I applied it to my data I didn't get what I was expecting, so now thinking about it I might have a problem with the mathematical concept itself..

I understand that this is a Unix forum not a math forum, so I'll try to figure the math part first.

Thanks
# 12  
Old 02-22-2012
Quote:
Originally Posted by cosmologist
I read that part and it made sense, but when I applied it to my data I didn't get what I was expecting
I added three numbers, and divided by three? Smilie That's liable to add more decimal points...
# 13  
Old 02-22-2012
Ok, I know what it is that confused me now Smilie

So in the example that you mentioned

Quote:
Originally Posted by Corona688
For these points:

Code:
1
5
2
6
3
7
4
8
5
9

for a boxcar average of width 3, would you want:

Code:
(1+5+2)/3, (5+2+6)/3, (2+6+3)/3, (6+3+7)/3, (3+7+4)/3, (7+4+8)/3, (4+5+5)/3, (8+5+9)/3

What I was thinking was:

Code:
(1+5+2)/3, (2+6+3)/3, (3+7+4)/3, (4+5+5)/3

[If I am using the mathematical terms correctly this is convolution with a rectangle function, a special case of the boxcar function]..

Is there a way to modify your code to get that? Smilie

Thanks again for all the help!
# 14  
Old 02-22-2012
If you meant 4+8+5, then sure. I'll let it work on multiple columns, too.

Code:
$ cat data
1       5
5       6
2       7
6       8
3       9
7       10
4       11
8       12
5       13
9       14

$ cat smooth2.awk

BEGIN { WIDTH=3; OFS="\t" }

{
        N++;
        for(M=1; M<=NF; M++)    DATA[M,N%WIDTH]=$M
}

(N>=WIDTH) && !((N-WIDTH)%(WIDTH-1)) {

        for(M=1; M<=NF; M++)
        {
                $M=0

                for(X=(N+1); X<=(N+WIDTH); X++)
                        $M+=DATA[M,X%WIDTH];

                $M /= WIDTH;
        }

        print
}

$ awk -f smooth2.awk data

2.66667 6
3.66667 8
4.66667 10
5.66667 12

$

Note the number of decimal places in the second column is a coincidence, those sums just happened to divide nicely by three.
This User Gave Thanks to Corona688 For This Post:
cosmologist
 
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