awk print column for a given range

Question

My data looks like that:

0.000000 0.071429 0.071429 0.857143 
0.071429 0.428571 0.071429 0.428571 
0.357143 0.214286 0.357143 0.071429 
0.000000 0.714286 0.000000 0.285714 
0.000000 0.571429 0.000000 0.428571 
0.428571 0.357143 0.071429 0.142857 
0.000000 0.071429 0.071429 0.857143 
0.071429 0.000000 0.928571 0.000000 
0.000000 0.071429 0.000000 0.928571 
0.000000 0.285714 0.000000 0.714286 
0.142857 0.000000 0.785714 0.071429 

I want it to look like that:

AC name_of_the_file.txt
00 0.000000 0.071429 0.071429 0.857143 
01 0.071429 0.428571 0.071429 0.428571 
02 0.357143 0.214286 0.357143 0.071429 
03 0.000000 0.714286 0.000000 0.285714 
04 0.000000 0.571429 0.000000 0.428571 
05 0.428571 0.357143 0.071429 0.142857 
06 0.000000 0.071429 0.071429 0.857143 
07 0.071429 0.000000 0.928571 0.000000 
08 0.000000 0.071429 0.000000 0.928571 
09 0.000000 0.285714 0.000000 0.714286 
10 0.142857 0.000000 0.785714 0.071429  
XX
//

How can I awk $1 for a range (from 00 till the file ends)?

Answer 1

One way:

awk 'FNR == 1 { print FILENAME } { printf "%02d %s\n", FNR - 1, $0 }' infile

Output:

infile
00 00 0.000000 0.071429 0.071429 0.857143 
01 01 0.071429 0.428571 0.071429 0.428571 
02 02 0.357143 0.214286 0.357143 0.071429 
03 03 0.000000 0.714286 0.000000 0.285714 
04 04 0.000000 0.571429 0.000000 0.428571 
05 05 0.428571 0.357143 0.071429 0.142857 
06 06 0.000000 0.071429 0.071429 0.857143 
07 07 0.071429 0.000000 0.928571 0.000000 
08 08 0.000000 0.071429 0.000000 0.928571 
09 09 0.000000 0.285714 0.000000 0.714286 
10 10 0.142857 0.000000 0.785714 0.071429