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I am beginner in working with bash script. I have a data file:

i    z    r                    
0   -30   3.5                 
0   -29.5 3.4                 
0   -29   3.6                  
...                             
0    29.5 4.2                  
0    30   4.6                 
1   -30   2.8                  
1   -29.5 3.4                    
....
1    30   5.2
2   -30   2.5
2   -29.5 2.6

And I'm trying to rearrange data into a file like this:

i     z     r
0    -30    3.5
1    -30    2.8
2    -30    2.5
...
0    -29.5  3.4
1    -29.5  3.4
2    -29.5  2.6
...

I used the script below but it cant not run like what i desire.

awk '{
  for(z=/-30/; z<=30; z+=0.5){
    for(i=0; i<=17489;i++){
      if($1==$i  && $2==$z)
        print $1 $2 $3
    }
  }
}' data.dat >> avg_SD.dat

please help me! Thank you so much.

share|improve this question
1  
I've edited as best I can to make this make sense but you need to explain the transformation between the two files. What is it that you're trying to do? –  Oli Jun 23 at 9:55
    
Thank for your edition. It looks better. In the above form: in per variale $i (i from 0 to 17489, step 1), i have calculate z and r value, respectively (z from -30 to 30, step 0.5). after finish z loop, the i was increased and repeat z loop. know, I want to modified in this way: in per variable $z (z from -30 to 30, step 0.5), there will have a r value per i frame. That is what i trying to do. –  user297072 Jun 23 at 10:15
    
How did you calculate the r value? And why sometimes "z += 0.5" and sometimes not and why the sign of z somtimes is changed? Your example is not really clear i can't see any logic how you calculate the new z and the new r values. –  TuKsn Jun 23 at 10:35
    
I use another script to calculate r and file out of this script is data.dat (above form). However calculating the average and standard divition of r value are difficult. So I have to modify it into below form. –  user297072 Jun 23 at 10:37
    
Ok if I understand you right, you are trying to loop through each line of your data.dat file, increase i+1 (in each block) and send z and r to another script (or function) to calculate them, then write the result of the changes in a new file? –  TuKsn Jun 23 at 10:48

1 Answer 1

From the example and your script it seems, that your original file is sorted by the first column ( i ) but you want it to be primarily sorted by the second column ( z ) instead and than by i.

This can be achieved much easier with sort

tail -n +2 data.dat | sort -n -k 2 -k 1 >> avg_SD.dat

It is also probably much faster than looping through two loops and doing at least one million comparisions for every line in the input file.

Explanation:

  • tail -n +2 data.dat prints data.dat starting with the second line, thus removing the header.

  • | redirects the output of the command on the left to the input of the command on the right.

  • sort -n -k 2 -k 1 sorts the input numerically (-n), first by the second field (-k 2) than by the first field (-k 1)

  • >> avg_SD.dat prints the output to avg_SD.dat keeping previous content. If you want to overwrite the previous content, use a single > instead of >>.

If there is no header in the original file, you do not need tail and can pass data.dat directly to sort as an argument:

sort -n -k 2 -k 1 data.dat >> avg_SD.dat
share|improve this answer
    
thank Adaephon so much. It works. If I want my file start form z = -30 and finish when z = 30, how can I do? other hand, can I use sort command to calculate average and Standard division for r. for ex: i want to calculate r for every frame number 0. –  user297072 Jun 23 at 11:59
    
To select only lines with -30 <= z <= 30 you can use awk '(-30 <= $2 && $2 <= 30) {print $0}'. And no, sort is only good at sorting. awk could be used to do these calculations, but it seems to me that a more general programming language might serve your needs better, for example python or perl. –  Adaephon Jun 25 at 20:12

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