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Full Discussion: How to replace all but the first 3 characters with sed?
Top Forums Shell Programming and Scripting How to replace all but the first 3 characters with sed? Post 302935266 by Scrutinizer on Saturday 14th of February 2015 12:17:38 AM
Old 02-14-2015
Quote:
Originally Posted by disedorgue
Ok, I understand better, so here:

Time reference:
Code:
time awk '{x=substr($0,N+1); gsub(".","*",x); print substr($0,1,N) x}' N=3 file1 > /dev/null
real	0m0.038s
user	0m0.037s
sys	0m0.000s

[..]
And a perl version seems faster that awk:
Code:
time perl -ne 'print substr($_,0,3)."*"x(length()-4)."\n"' file1 >/dev/null 
real	0m0.014s
user	0m0.010s
sys	0m0.000s

Regards.
Indeed the perl is faster because of the powerful repetition (x) feature, but it is not entirely correct, since it does not work correctly for string lengths <=2 (spurious newline)

So I used
Code:
$ time perl -nle 'print substr($_,0,3)."*"x(length()-3)' file > /dev/null

real	0m0.016s
user	0m0.012s
sys	0m0.003s

instead.

The only way I could get close, was with mawk :
Code:
$ time mawk -v n=3 '{m=length()-n+1; $0=substr($0,1,3)} m>1{$m=x}1' OFS=\* file > /dev/null

real	0m0.019s
user	0m0.016s
sys	0m0.002s

Which was marginally slower.

Quote:
Originally Posted by DGPickett
Try C:
Code:
#include <stdio.h>
#include <string.h>
 
int main(){
[..]

Interestingly on OSX 10.10.2 (with LLVM 6.0 compiler) I got:
Code:
$ time ./a.out < file > /dev/null

real	0m0.045s
user	0m0.039s
sys	0m0.002s

On CentOS 6.6 it was fastest though..
Code:
$ time mawk -v n=3 '{m=length()-n+1; $0=substr($0,1,3)} m>1{$m=x}1' OFS=\* file > /dev/null

real	0m0.035s
user	0m0.034s
sys	0m0.001s
$ time perl -nle 'print substr($_,0,3)."*"x(length()-3)' file >/dev/null

real	0m0.026s
user	0m0.022s
sys	0m0.004s
$ time ./a.out < file > /dev/null

real	0m0.021s
user	0m0.019s
sys	0m0.002s

(with gcc 4.4.7)

So which compiler is being used can make quite a difference
Last edited by Scrutinizer; 02-14-2015 at 02:32 AM..
 

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