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Display the last part of a number list

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# 1  
Old 06-26-2016
Display the last part of a number list
Hi ,
i have a file wich have 50+ of numbers like :

Code:
  
0.014544106
0.005464263
0.014526045
0.005484374
0.014539412
0.005467600
0.014558349
0.005452185

i would like to display the list from the 6th bit to the end for example

0.005452185 (should become) 2185.

I've tried with tail -c 4 filename but it display just the last one in the file.

Thanks!

Regards
# 2  
Old 06-26-2016
Would this work?

Code:
grep -o '....$' Board27.numbers

# 3  
Old 06-26-2016
If your system's grep utility does not support the -o option, you could also do this entirely in the shell (with any shell that performs basic POSIX-required parameter expansions) using:
Code:
while read -r line
do	printf '%s\n' "${line#${line%????}}"
done < file

which with your sample input file produces the output:
Code:
4106
4263
6045
4374
9412
7600
8349
2185

This User Gave Thanks to Don Cragun For This Post:
Board27
# 4  
Old 06-27-2016
How about the "Substring Parameter Expansion" (possibly not available in all POSIX compliant shells):

man bash:
Quote:
${parameterSmilieffset}...
If offset evaluates to a number less than zero, the value is used as an offset in characters from the end of the value of parameter.

Code:
printf '%s\n' "${line: -4}"

# 5  
Old 06-27-2016
Thanks a lot!
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