I have several places where I want to cut my string in several parts.
For example:
$string= "AACCAAGTAA";
#cut_places= {0,4, 8 };
My $string should look like this: AACC AAGT AA;
How can I do that?
To populate an array, use round parentheses, not curly brackets (they're used for hash references).
One possible way is to use substr where the first argument is the position, so you can use the array elements. You just need to compute the length by subtracting the position from the following one; and to be able to compute the last length, you need the length of the whole string, too:
#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
my $string = 'AACCAAGTAA';
my #cut_places = (0, 4, 8);
push #cut_places, length $string;
my #parts = map {
substr $string, $cut_places[$_], $cut_places[$_+1] - $cut_places[$_]
} 0 .. $#cut_places - 1;
say for #parts;
If the original array contained lengths instead of positions, the code would be much easier.
#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
my $string = 'AACCAAGTAA';
my #lengths = (4, 4, 2); # 4, 4, 4 would work, too
my #parts = unpack join("", map "A$_", #lengths), $string;
say for #parts;
See unpack for details.
Here's a solution that starts by calculating the forward differences in the list of positions. The length of the string is first appended to the end of the list of it doesn't already span the full string
The differences are then used to build an unpack format string, which is used to build the required sequence of substrings.
I have written the functionality as a do block, which would be simple to convert to a subroutine if desired.
use strict;
use warnings 'all';
use feature 'say';
my $string = 'AACCAAGTAA';
my #cut_places = ( 0, 4, 8 );
my #parts = do {
my #places = #cut_places;
my $len = length $string;
push #places, $len unless $places[-1] >= $len;
my #w = map { $places[$_]-$places[$_-1] } 1 .. $#places;
my $patt = join ' ', map { "A$_" } #w;
unpack $patt, $string;
};
say "#parts";
output
AACC AAGT AA
Work out the lengths of needed parts first, then all methods are easier. Here regex is used
use warnings;
use strict;
use feature 'say';
my $string = 'AACCAAGTAA';
my #pos = (0, 4, 8);
my #lens = do {
my $prev = shift #pos;
"$prev", map { my $e = $_ - $prev; $prev = $_; $e } #pos;
};
my $patt = join '', map { '(.{'.$_.'})' } #lens;
my $re = qr/$patt/;
my #parts = grep { /./ } $string =~ /$re(.*)/g;
say for #parts;
The lengths #lens are computed by subtracting the successive positions, 2-1, 3-2 (etc). I use do merely so that the #prev variable, unneeded elsewhere, doesn't "pollute" the rest of the code.
The "$prev" is quoted so that it is evaluated first, before it changes in map.
The matches returned by regex are passed through grep to filter out empty string(s) due to the 0 position (or whenever successive positions are the same).
This works for position arrays of any lengths, as long as positions are consistent with a string.
Using Perl RegEx, How to find largest superstring in a sentence, when superstring is a repeatation of 1 or more substring.
For Ex:
$sentence = "zsabcxyzabcabcabccde_xdrabcabcrte__23abcerabcabccbabacxyz";
$subStr = "abc";
I want to find all the occurrences of abc and largest one in that.
Output:
abc
abcabcabc
abcabc
abc
abcabc
Largest string is abcabcabc
Compile a regex using a quantifier. + says 'one or more'.
So your "substr" becomes ((?:abc)+) the outer brackets to 'capture', the inner brackets non capturing. Otherwise you'll also get the 'partial' hits in the array - although the net result isn't changed much, because the longest hit will still sort to the top.
For example:
#!/usr/bin/env perl
use strict;
use warnings;
use Data::Dumper;
my $sentence = "zsabcxyzabcabcabccde_xdrabcabcrte__23abcerabcabccbabacxyz";
my $substring = "abc";
my $regex = qr/((?:$substring)+)/;
my #matches = $sentence =~ m/$regex/g;
print Dumper \#matches;
#Then sort it:
my ( $longest ) = sort { length ( $b ) <=> length ( $a ) } #matches;
print $longest,"\n";
Try as below one
use warnings;
use strict;
my $sentence = "zsabcxyzabcabcabccde_xdrabcabcrte__23abcerabcabccbabacxyz";
my ($larg) = sort{length($b)<=>length($a)} $sentence =~ m/((?:abc)+)/g;
print $larg,"\n";
If don't want to store it means, make a loop
use warnings;
use strict;
my $sentence = "zsabcxyzabcabcabccde_xdrabcabcrte__23abcerabcabccbabacxyzabcabcabcabc";
my $longstr;
my $len = 0;
while($sentence=~m/((?:abc)+)/g)
{
$longstr = $1 and $len = length($1) if(length($1) > $len)
}
the above one is in single regex with (?{}) but not recommended
my $sentence = "zsabcxyzabcabcabccde_xdrabcabcrte__23abcerabcabccbabacxyzabcabcabcabc";
my $lar = 0;
my $larg;
$sentence=~m/((?:abc)+)(?{ $larg = $1 and $lar=(length $1) if(length $1 > $lar )}) \G/x;
Thanks to #mkHun and #Sobrique...
I have used #matches = $str =~ m/(abc)+/ng; and then sorting.
/n makes it looks much simpler which is available from 5.22, i think.
The best way is next: ((?:abc)+) with sorting after finding matches
Using awk, I can print a number with commas as thousands separators.
(with a export LC_ALL=en_US.UTF-8 beforehand).
awk 'BEGIN{printf("%\047d\n", 24500)}'
24,500
I expected the same format to work with Perl, but it does not:
perl -e 'printf("%\047d\n", 24500)'
%'d
The Perl Cookbook offers this solution:
sub commify {
my $text = reverse $_[0];
$text =~ s/(\d\d\d)(?=\d)(?!\d*\.)/$1,/g;
return scalar reverse $text;
}
However I am assuming that since the printf option works in awk, it should also work in Perl.
The apostrophe format modifier is a non-standard POSIX extension.
The documentation for Perl's printf has this to say about such extensions
Perl does its own "sprintf" formatting: it emulates the C
function sprintf(3), but doesn't use it except for
floating-point numbers, and even then only standard modifiers
are allowed. Non-standard extensions in your local sprintf(3)
are therefore unavailable from Perl.
The Number::Format module will do this for you, and it takes its default settings from the locale, so is as portable as it can be
use strict;
use warnings 'all';
use v5.10.1;
use Number::Format 'format_number';
say format_number(24500);
output
24,500
A more perl-ish solution:
$a = 12345678; # no comment
$b = reverse $a; # $b = '87654321';
#c = unpack("(A3)*", $b); # $c = ('876', '543', '21');
$d = join ',', #c; # $d = '876,543,21';
$e = reverse $d; # $e = '12,345,678';
print $e;
outputs 12,345,678.
I realize this question was from almost 4 years ago, but since it comes up in searches, I'll add an elegant native Perl solution I came up with. I was originally searching for a way to do it with sprintf, but everything I've found indicates that it can't be done. Then since everyone is rolling their own, I thought I'd give it a go, and this is my solution.
$num = 12345678912345; # however many digits you want
while($num =~ s/(\d+)(\d\d\d)/$1\,$2/){};
print $num;
Results in:
12,345,678,912,345
Explanation:
The Regex does a maximal digit search for all leading digits. The minimum number of digits in a row it'll act on is 4 (1 plus 3). Then it adds a comma between the two. Next loop if there are still 4 digits at the end (before the comma), it'll add another comma and so on until the pattern doesn't match.
If you need something safe for use with more than 3 digits after the decimal, use this modification: (Note: This won't work if your number has no decimal)
while($num =~ s/(\d+)(\d\d\d)([.,])/$1\,$2$3/){};
This will ensure that it will only look for digits that ends in a comma (added on a previous loop) or a decimal.
Most of these answers assume that the format is universal. It isn't. CLDR uses Unicode information to figure it out. There's a long thread in How to properly localize numbers?.
CPAN has the CLDR::Number module:
#!perl
use v5.10;
use CLDR::Number;
use open qw(:std :utf8);
my $locale = $ARGV[0] // 'en';
my #numbers = qw(
123
12345
1234.56
-90120
);
my $cldr = CLDR::Number->new( locale => $locale );
my $decf = $cldr->decimal_formatter;
foreach my $n ( #numbers ) {
say $decf->format($n);
}
Here are a few runs:
$ perl comma.pl
123
12,345
1,234.56
-90,120
$ perl comma.pl es
123
12.345
1234,56
-90.120
$ perl comma.pl bn
১২৩
১২,৩৪৫
১,২৩৪.৫৬
-৯০,১২০
It seems heavyweight, but the output is correct and you don't have to allow the user to change the locale you want to use. However, when it's time to change the locale, you are ready to go. I also prefer this to Number::Format because I can use a locale that's different from my local settings for my terminal or session, or even use multiple locales:
#!perl
use v5.10;
use CLDR::Number;
use open qw(:std :utf8);
my #locales = qw( en pt bn );
my #numbers = qw(
123
12345
1234.56
-90120
);
my #formatters = map {
my $cldr = CLDR::Number->new( locale => $_ );
my $decf = $cldr->decimal_formatter;
[ $_, $cldr, $decf ];
} #locales;
printf "%10s %10s %10s\n" . '=' x 32 . "\n", #locales;
foreach my $n ( #numbers ) {
printf "%10s %10s %10s\n",
map { $_->[-1]->format($n) } #formatters;
}
The output has three locales at once:
en pt bn
================================
123 123 ১২৩
12,345 12.345 ১২,৩৪৫
1,234.56 1.234,56 ১,২৩৪.৫৬
-90,120 -90.120 -৯০,১২০
Here's an elegant Perl solution I've been using for over 20 years :)
1 while $text =~ s/(.*\d)(\d\d\d)/$1\.$2/g;
And if you then want two decimal places:
$text = sprintf("%0.2f", $text);
1 liner: Use a little loop whith a regex:
while ($number =~ s/^(\d+)(\d{3})/$1,$2/) {}
Example:
use strict;
use warnings;
my #numbers = (12321, 12.12, 122222.3334, '1234abc', '1.1', '1222333444555,666.77');
for(#numbers) {
my $number = $_;
while ($number =~ s/^(\d+)(\d{3})/$1,$2/) {}
print "$_ -> $number\n";
}
Output:
12321 -> 12,321
12.12 -> 12.12
122222.3334 -> 122,222.3334
1234abc -> 1,234abc
1.1 -> 1.1
1222333444555,666.77 -> 1,222,333,444,555,666.77
Pattern:
(\d+)(\d{3})
-> Take all numbers but the last 3 in group 1
-> Take the remaining 3 numbers in group2 on the beginning of $number
-> Followed is ignored
Substitution
$1,$2
-> Put a seperator sign (,) between group 1 and 2
-> The rest remains unchanged
So if you have 12345.67 the numers the regex uses are 12345. The '.' and all followed is ignored.
1. run (12345.67):
-> matches: 12345
-> group 1: 12,
group 2: 345
-> substitute 12,345
-> result: 12,345.67
2. run (12,345.67):
-> does not match!
-> while breaks.
Parting from #Laura's answer, I tweaked the pure perl, regex-only solution to work for numbers with decimals too:
while ($formatted_number =~ s/^(-?\d+)(\d{3}(?:,\d{3})*(?:\.\d+)*)$/$1,$2/) {};
Of course this assumes a "," as thousands separator and a "." as decimal separator, but it should be trivial to use variables to account for that for your given locale(s).
I used the following but it does not works as of perl v5.26.1
sub format_int
{
my $num = shift;
return reverse(join(",",unpack("(A3)*", reverse int($num))));
}
The form that worked for me was:
sub format_int
{
my $num = shift;
return scalar reverse(join(",",unpack("(A3)*", reverse int($num))));
}
But to use negative numbers the code must be:
sub format_int
{
if ( $val >= 0 ) {
return scalar reverse join ",", unpack( "(A3)*", reverse int($val) );
} else {
return "-" . scalar reverse join ",", unpack( "(A3)*", reverse int(-$val) );
}
}
Did somebody say Perl?
perl -pe '1while s/(\d+)(\d{3})/$1,$2/'
This works for any integer.
# turning above answer into a function
sub format_float
# returns number with commas..... and 2 digit decimal
# so format_float(12345.667) returns "12,345.67"
{
my $num = shift;
return reverse(join(",",unpack("(A3)*", reverse int($num)))) . sprintf(".%02d",int(100*(.005+($num - int($num)))));
}
sub format_int
# returns number with commas.....
# so format_int(12345.667) returns "12,345"
{
my $num = shift;
return reverse(join(",",unpack("(A3)*", reverse int($num))));
}
I wanted to print numbers it in a currency format. If it turned out even, I still wanted a .00 at the end. I used the previous example (ty) and diddled with it a bit more to get this.
sub format_number {
my $num = shift;
my $result;
my $formatted_num = "";
my #temp_array = ();
my $mantissa = "";
if ( $num =~ /\./ ) {
$num = sprintf("%0.02f",$num);
($num,$mantissa) = split(/\./,$num);
$formatted_num = reverse $num;
#temp_array = unpack("(A3)*" , $formatted_num);
$formatted_num = reverse (join ',', #temp_array);
$result = $formatted_num . '.'. $mantissa;
} else {
$formatted_num = reverse $num;
#temp_array = unpack("(A3)*" , $formatted_num);
$formatted_num = reverse (join ',', #temp_array);
$result = $formatted_num . '.00';
}
return $result;
}
# Example call
# ...
printf("some amount = %s\n",format_number $some_amount);
I didn't have the Number library on my default mac OS X perl, and I didn't want to mess with that version or go off installing my own perl on this machine. I guess I would have used the formatter module otherwise.
I still don't actually like the solution all that much, but it does work.
This is good for money, just keep adding lines if you handle hundreds of millions.
sub commify{
my $var = $_[0];
#print "COMMIFY got $var\n"; #DEBUG
$var =~ s/(^\d{1,3})(\d{3})(\.\d\d)$/$1,$2$3/;
$var =~ s/(^\d{1,3})(\d{3})(\d{3})(\.\d\d)$/$1,$2,$3$4/;
$var =~ s/(^\d{1,3})(\d{3})(\d{3})(\d{3})(\.\d\d)$/$1,$2,$3,$4$5/;
$var =~ s/(^\d{1,3})(\d{3})(\d{3})(\d{3})(\d{3})(\.\d\d)$/$1,$2,$3,$4,$5$6/;
#print "COMMIFY made $var\n"; #DEBUG
return $var;
}
A solution that produces a localized output:
# First part - Localization
my ( $thousands_sep, $decimal_point, $negative_sign );
BEGIN {
my ( $l );
use POSIX qw(locale_h);
$l = localeconv();
$thousands_sep = $l->{ 'thousands_sep' };
$decimal_point = $l->{ 'decimal_point' };
$negative_sign = $l->{ 'negative_sign' };
}
# Second part - Number transformation
sub readable_number {
my $val = shift;
#my $thousands_sep = ".";
#my $decimal_point = ",";
#my $negative_sign = "-";
sub _readable_int {
my $val = shift;
# a pinch of PERL magic
return scalar reverse join $thousands_sep, unpack( "(A3)*", reverse $val );
}
my ( $i, $d, $r );
$i = int( $val );
if ( $val >= 0 ) {
$r = _readable_int( $i );
} else {
$r = $negative_sign . _readable_int( -$i );
}
# If there is decimal part append it to the integer result
if ( $val != $i ) {
( undef, $d ) = ( $val =~ /(\d*)\.(\d*)/ );
$r = $r . $decimal_point . $d;
}
return $r;
}
The first part gets the symbols used in the current locale to be used on the second part.
The BEGIN block is used to calculate the sysmbols only once at the beginning.
If for some reason there is need to not use POSIX locale, one can ommit the first part and uncomment the variables on the second part to hardcode the sysmbols to be used ($thousands_sep, $thousands_sep and $thousands_sep)
I have a string in Perl that is 23 digits long. I need to cut it apart into different pieces. First 2 digits in one variable, next 3 in another variable, next 4 into another variable, etc. Basically the 23 digits needs to end up as 6 separate variables (2,3,4,4,3,7) characters, in that order.
Any ideas how I can cut the string up like this?
There are lots of ways to do it, but the shortest is probably unpack:
my $string = '1' x 23;
my #values = unpack 'A2A3A4A4A3A7', $string;
If you need separate variables, you can use a list assignment:
my ($v1, $v2, $v3, $v4, $v5, $v6) = unpack 'A2A3A4A4A3A7', $string;
Expanding on Alex's method, rather than specify each start and end, use the list you gave of lengths.
#!/usr/bin/env perl
use strict;
use warnings;
my $string = "abcdefghijklmnopqrstuvw";
my $pos = 0;
my #split = map {
my $start = $pos;
my $end = $_;
$pos += $end;
substr( $string, $start, $end);
} (2,3,4,4,3,7);
print "$_\n" for #split;
This said you probably should look at unpack which is used for fixed width fields. I have no experience with it though.
You could use a regex, viz:
$string =~ /\d{2}\d{3}\d{4}\d{4}\d{3}\d{7}/
and capture each part by surrounding with brackets ().
You then find each capture in the variables $1, $2 ...
or get them all in the returned list
See perldoc perlre
You want to use perldoc substr.
$substring = substr($string, $start, $length);
I'd also use `map' on a list of [start, length] pairs to make your life easier:
$string = "123456789";
#values = map {substr($string, $_->[0], $_->[1])} ([1, 3], [4, 2] , ...);
Here's a sub that will do it, using the already discussed unpack.
sub string_slices {
my $str = shift;
return unpack( join( 'A', '', #_ ), $str );
}