perl regex - convert multiple decimal numbers on a line to nearest integer











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I am practicing a Perl regex question where I have lines in the following format that are read in from stdin:



I bought 2.3kg of oranges for $13.50.



It takes 4.5 hours to get from the city to the sea.



She moved in 2010, and left in 2014.



And I have to convert all the decimal numbers in each line to their nearest integer. So the expected output would look like:



I bought 2kg of oranges for $14.00.



It takes 5 hours to get from the city to the sea.



She moved in 2010, and left in 2014.



I have the following code:



#!/usr/bin/perl -w

use strict;

use warnings;

use Math::Round;



my @lines = <STDIN>; 

chomp @lines;



foreach my $line (@lines) {

    $line =~ s/(d+.?d*)/hello/g;

}


The part which I am not sure how to do is actually replace the decimal numbers in each line with their integer version. My approach is to read all the lines from stdin into an array, and then for each match on each line substitute the decimal number with its nearest whole number. I know you can round floating point numbers in Perl using the round() function.



However, using the round() function as the replacement text within the 's/(d+.?d*)//g' would not work. Using the substitution operator seemed to be the simplest approach, but I don't think I will be able to use it.
I am not exactly sure what other methods I could use to solve this problem. Any insights would be really appreciated.






regex perl







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  • 1




    You say "integer" but give $14.00 example? If you really mean to round prices to 0 cents you need to modify @ikegami solution, perhaps by adding an optional dollar match ($?) and then in the replacement have two cases, s/../$1 ? to-zero-cents : to-int/ge.
    – zdim
    Nov 10 at 8:13






  • 1




    A clumsy way to do the "zero-cents" above is to nest sprintf's, one for rounding the other for precision: s/($?)(d+.d+)/$1 ? $1.sprintf("%.2f",sprintf "%.0f", $2) : sprintf("%.0f",$2)/ge. There are other ways.
    – zdim
    Nov 10 at 8:14

















up vote
1
down vote

favorite
1












I am practicing a Perl regex question where I have lines in the following format that are read in from stdin:



I bought 2.3kg of oranges for $13.50.



It takes 4.5 hours to get from the city to the sea.



She moved in 2010, and left in 2014.



And I have to convert all the decimal numbers in each line to their nearest integer. So the expected output would look like:



I bought 2kg of oranges for $14.00.



It takes 5 hours to get from the city to the sea.



She moved in 2010, and left in 2014.



I have the following code:



#!/usr/bin/perl -w

use strict;

use warnings;

use Math::Round;



my @lines = <STDIN>; 

chomp @lines;



foreach my $line (@lines) {

    $line =~ s/(d+.?d*)/hello/g;

}


The part which I am not sure how to do is actually replace the decimal numbers in each line with their integer version. My approach is to read all the lines from stdin into an array, and then for each match on each line substitute the decimal number with its nearest whole number. I know you can round floating point numbers in Perl using the round() function.



However, using the round() function as the replacement text within the 's/(d+.?d*)//g' would not work. Using the substitution operator seemed to be the simplest approach, but I don't think I will be able to use it.
I am not exactly sure what other methods I could use to solve this problem. Any insights would be really appreciated.






regex perl







share|improve this question


















  • 1




    You say "integer" but give $14.00 example? If you really mean to round prices to 0 cents you need to modify @ikegami solution, perhaps by adding an optional dollar match ($?) and then in the replacement have two cases, s/../$1 ? to-zero-cents : to-int/ge.
    – zdim
    Nov 10 at 8:13






  • 1




    A clumsy way to do the "zero-cents" above is to nest sprintf's, one for rounding the other for precision: s/($?)(d+.d+)/$1 ? $1.sprintf("%.2f",sprintf "%.0f", $2) : sprintf("%.0f",$2)/ge. There are other ways.
    – zdim
    Nov 10 at 8:14















up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I am practicing a Perl regex question where I have lines in the following format that are read in from stdin:



I bought 2.3kg of oranges for $13.50.



It takes 4.5 hours to get from the city to the sea.



She moved in 2010, and left in 2014.



And I have to convert all the decimal numbers in each line to their nearest integer. So the expected output would look like:



I bought 2kg of oranges for $14.00.



It takes 5 hours to get from the city to the sea.



She moved in 2010, and left in 2014.



I have the following code:



#!/usr/bin/perl -w

use strict;

use warnings;

use Math::Round;



my @lines = <STDIN>; 

chomp @lines;



foreach my $line (@lines) {

    $line =~ s/(d+.?d*)/hello/g;

}


The part which I am not sure how to do is actually replace the decimal numbers in each line with their integer version. My approach is to read all the lines from stdin into an array, and then for each match on each line substitute the decimal number with its nearest whole number. I know you can round floating point numbers in Perl using the round() function.



However, using the round() function as the replacement text within the 's/(d+.?d*)//g' would not work. Using the substitution operator seemed to be the simplest approach, but I don't think I will be able to use it.
I am not exactly sure what other methods I could use to solve this problem. Any insights would be really appreciated.






regex perl







share|improve this question













I am practicing a Perl regex question where I have lines in the following format that are read in from stdin:



I bought 2.3kg of oranges for $13.50.



It takes 4.5 hours to get from the city to the sea.



She moved in 2010, and left in 2014.



And I have to convert all the decimal numbers in each line to their nearest integer. So the expected output would look like:



I bought 2kg of oranges for $14.00.



It takes 5 hours to get from the city to the sea.



She moved in 2010, and left in 2014.



I have the following code:



#!/usr/bin/perl -w

use strict;

use warnings;

use Math::Round;



my @lines = <STDIN>; 

chomp @lines;



foreach my $line (@lines) {

    $line =~ s/(d+.?d*)/hello/g;

}


The part which I am not sure how to do is actually replace the decimal numbers in each line with their integer version. My approach is to read all the lines from stdin into an array, and then for each match on each line substitute the decimal number with its nearest whole number. I know you can round floating point numbers in Perl using the round() function.



However, using the round() function as the replacement text within the 's/(d+.?d*)//g' would not work. Using the substitution operator seemed to be the simplest approach, but I don't think I will be able to use it.
I am not exactly sure what other methods I could use to solve this problem. Any insights would be really appreciated.






regex perl




regex perl






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asked Nov 10 at 5:20









ceno980

805




805








  • 1




    You say "integer" but give $14.00 example? If you really mean to round prices to 0 cents you need to modify @ikegami solution, perhaps by adding an optional dollar match ($?) and then in the replacement have two cases, s/../$1 ? to-zero-cents : to-int/ge.
    – zdim
    Nov 10 at 8:13






  • 1




    A clumsy way to do the "zero-cents" above is to nest sprintf's, one for rounding the other for precision: s/($?)(d+.d+)/$1 ? $1.sprintf("%.2f",sprintf "%.0f", $2) : sprintf("%.0f",$2)/ge. There are other ways.
    – zdim
    Nov 10 at 8:14
















  • 1




    You say "integer" but give $14.00 example? If you really mean to round prices to 0 cents you need to modify @ikegami solution, perhaps by adding an optional dollar match ($?) and then in the replacement have two cases, s/../$1 ? to-zero-cents : to-int/ge.
    – zdim
    Nov 10 at 8:13






  • 1




    A clumsy way to do the "zero-cents" above is to nest sprintf's, one for rounding the other for precision: s/($?)(d+.d+)/$1 ? $1.sprintf("%.2f",sprintf "%.0f", $2) : sprintf("%.0f",$2)/ge. There are other ways.
    – zdim
    Nov 10 at 8:14










1




1




You say "integer" but give $14.00 example? If you really mean to round prices to 0 cents you need to modify @ikegami solution, perhaps by adding an optional dollar match ($?) and then in the replacement have two cases, s/../$1 ? to-zero-cents : to-int/ge.
– zdim
Nov 10 at 8:13




You say "integer" but give $14.00 example? If you really mean to round prices to 0 cents you need to modify @ikegami solution, perhaps by adding an optional dollar match ($?) and then in the replacement have two cases, s/../$1 ? to-zero-cents : to-int/ge.
– zdim
Nov 10 at 8:13




1




1




A clumsy way to do the "zero-cents" above is to nest sprintf's, one for rounding the other for precision: s/($?)(d+.d+)/$1 ? $1.sprintf("%.2f",sprintf "%.0f", $2) : sprintf("%.0f",$2)/ge. There are other ways.
– zdim
Nov 10 at 8:14






A clumsy way to do the "zero-cents" above is to nest sprintf's, one for rounding the other for precision: s/($?)(d+.d+)/$1 ? $1.sprintf("%.2f",sprintf "%.0f", $2) : sprintf("%.0f",$2)/ge. There are other ways.
– zdim
Nov 10 at 8:14














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Use s///e to put code as the replacement expression. The code is expected to return the value to use as the replacement.



 s/(d+.d+)/ sprintf("%.0f", $1) /eg


(Feel free to use Math::Round's round if you wish. I just used that with which I am familiar.)






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    active

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    up vote
    2
    down vote



    accepted










    Use s///e to put code as the replacement expression. The code is expected to return the value to use as the replacement.



     s/(d+.d+)/ sprintf("%.0f", $1) /eg


    (Feel free to use Math::Round's round if you wish. I just used that with which I am familiar.)






    share|improve this answer

























      up vote
      2
      down vote



      accepted










      Use s///e to put code as the replacement expression. The code is expected to return the value to use as the replacement.



       s/(d+.d+)/ sprintf("%.0f", $1) /eg


      (Feel free to use Math::Round's round if you wish. I just used that with which I am familiar.)






      share|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Use s///e to put code as the replacement expression. The code is expected to return the value to use as the replacement.



         s/(d+.d+)/ sprintf("%.0f", $1) /eg


        (Feel free to use Math::Round's round if you wish. I just used that with which I am familiar.)






        share|improve this answer












        Use s///e to put code as the replacement expression. The code is expected to return the value to use as the replacement.



         s/(d+.d+)/ sprintf("%.0f", $1) /eg


        (Feel free to use Math::Round's round if you wish. I just used that with which I am familiar.)







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 10 at 5:39









        ikegami

        260k11174394




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