UNIX for Dummies Questions & Answers

I have a file with a single column of numbers. Starting from the 11th line, I'd like to find the line(s) on which the number meets two criteria: 1) it is less than the 10 numbers that precede it, 2) it is less than the number that follows it. So, for the following, the script should return line 18. I can't think of way of doing this without a grotesquely long 'if' statement.

(start of file)
69.9776
69.8288
69.5982
69.1929
68.7545
68.4256
67.9714
67.3566
66.76
65.9346
65.1717
64.2631
63.3557
62.0736
60.8232
59.9634
59.1894
58.3003
58.4929
59.6617
59.987
...

Thanks!

try this:

Code:

awk '
     { arr[NR]=$0
       if(NR<11) 
          {next}
       val=NR-1
       retval=(arr[NR]>arr[val])? 1: 0
       for(i=val -10; i<val && retval; i++)
       {
          if(arr[i]<=arr[val]) 
              {retval=0 }
       }
       if(retval)
         { print arr[val] }
       } ' inputfile > outputfile

Thanks a lot!