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perlfaq4.pod
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=head1 NAME
perlfaq4 - Data Manipulation
=head1 DESCRIPTION
This section of the FAQ answers questions related to manipulating
numbers, dates, strings, arrays, hashes, and miscellaneous data issues.
=head1 Data: Numbers
=head2 Why am I getting long decimals (eg, 19.9499999999999) instead of the numbers I should be getting (eg, 19.95)?
For the long explanation, see David Goldberg's "What Every Computer
Scientist Should Know About Floating-Point Arithmetic"
(http://docs.sun.com/source/806-3568/ncg_goldberg.html).
Internally, your computer represents floating-point numbers in binary.
Digital (as in powers of two) computers cannot store all numbers
exactly. Some real numbers lose precision in the process. This is a
problem with how computers store numbers and affects all computer
languages, not just Perl.
L<perlnumber> shows the gory details of number representations and
conversions.
To limit the number of decimal places in your numbers, you can use the
C<printf> or C<sprintf> function. See the L<"Floating Point
Arithmetic"|perlop> for more details.
printf "%.2f", 10/3;
my $number = sprintf "%.2f", 10/3;
=head2 Why is int() broken?
Your C<int()> is most probably working just fine. It's the numbers that
aren't quite what you think.
First, see the answer to "Why am I getting long decimals
(eg, 19.9499999999999) instead of the numbers I should be getting
(eg, 19.95)?".
For example, this
print int(0.6/0.2-2), "\n";
will in most computers print 0, not 1, because even such simple
numbers as 0.6 and 0.2 cannot be presented exactly by floating-point
numbers. What you think in the above as 'three' is really more like
2.9999999999999995559.
=head2 Why isn't my octal data interpreted correctly?
(contributed by brian d foy)
You're probably trying to convert a string to a number, which Perl only
converts as a decimal number. When Perl converts a string to a number, it
ignores leading spaces and zeroes, then assumes the rest of the digits
are in base 10:
my $string = '0644';
print $string + 0; # prints 644
print $string + 44; # prints 688, certainly not octal!
This problem usually involves one of the Perl built-ins that has the
same name a Unix command that uses octal numbers as arguments on the
command line. In this example, C<chmod> on the command line knows that
its first argument is octal because that's what it does:
%prompt> chmod 644 file
If you want to use the same literal digits (644) in Perl, you have to tell
Perl to treat them as octal numbers either by prefixing the digits with
a C<0> or using C<oct>:
chmod( 0644, $file); # right, has leading zero
chmod( oct(644), $file ); # also correct
The problem comes in when you take your numbers from something that Perl
thinks is a string, such as a command line argument in C<@ARGV>:
chmod( $ARGV[0], $file); # wrong, even if "0644"
chmod( oct($ARGV[0]), $file ); # correct, treat string as octal
You can always check the value you're using by printing it in octal
notation to ensure it matches what you think it should be. Print it
in octal and decimal format:
printf "0%o %d", $number, $number;
=head2 Does Perl have a round() function? What about ceil() and floor()? Trig functions?
Remember that C<int()> merely truncates toward 0. For rounding to a
certain number of digits, C<sprintf()> or C<printf()> is usually the
easiest route.
printf("%.3f", 3.1415926535); # prints 3.142
The C<POSIX> module (part of the standard Perl distribution)
implements C<ceil()>, C<floor()>, and a number of other mathematical
and trigonometric functions.
use POSIX;
$ceil = ceil(3.5); # 4
$floor = floor(3.5); # 3
In 5.000 to 5.003 perls, trigonometry was done in the C<Math::Complex>
module. With 5.004, the C<Math::Trig> module (part of the standard Perl
distribution) implements the trigonometric functions. Internally it
uses the C<Math::Complex> module and some functions can break out from
the real axis into the complex plane, for example the inverse sine of
2.
Rounding in financial applications can have serious implications, and
the rounding method used should be specified precisely. In these
cases, it probably pays not to trust whichever system rounding is
being used by Perl, but to instead implement the rounding function you
need yourself.
To see why, notice how you'll still have an issue on half-way-point
alternation:
for ($i = 0; $i < 1.01; $i += 0.05) { printf "%.1f ",$i}
0.0 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.4 0.4 0.5 0.5 0.6 0.7 0.7
0.8 0.8 0.9 0.9 1.0 1.0
Don't blame Perl. It's the same as in C. IEEE says we have to do
this. Perl numbers whose absolute values are integers under 2**31 (on
32 bit machines) will work pretty much like mathematical integers.
Other numbers are not guaranteed.
=head2 How do I convert between numeric representations/bases/radixes?
As always with Perl there is more than one way to do it. Below are a
few examples of approaches to making common conversions between number
representations. This is intended to be representational rather than
exhaustive.
Some of the examples later in L<perlfaq4> use the C<Bit::Vector>
module from CPAN. The reason you might choose C<Bit::Vector> over the
perl built in functions is that it works with numbers of ANY size,
that it is optimized for speed on some operations, and for at least
some programmers the notation might be familiar.
=over 4
=item How do I convert hexadecimal into decimal
Using perl's built in conversion of C<0x> notation:
$dec = 0xDEADBEEF;
Using the C<hex> function:
$dec = hex("DEADBEEF");
Using C<pack>:
$dec = unpack("N", pack("H8", substr("0" x 8 . "DEADBEEF", -8)));
Using the CPAN module C<Bit::Vector>:
use Bit::Vector;
$vec = Bit::Vector->new_Hex(32, "DEADBEEF");
$dec = $vec->to_Dec();
=item How do I convert from decimal to hexadecimal
Using C<sprintf>:
$hex = sprintf("%X", 3735928559); # upper case A-F
$hex = sprintf("%x", 3735928559); # lower case a-f
Using C<unpack>:
$hex = unpack("H*", pack("N", 3735928559));
Using C<Bit::Vector>:
use Bit::Vector;
$vec = Bit::Vector->new_Dec(32, -559038737);
$hex = $vec->to_Hex();
And C<Bit::Vector> supports odd bit counts:
use Bit::Vector;
$vec = Bit::Vector->new_Dec(33, 3735928559);
$vec->Resize(32); # suppress leading 0 if unwanted
$hex = $vec->to_Hex();
=item How do I convert from octal to decimal
Using Perl's built in conversion of numbers with leading zeros:
$dec = 033653337357; # note the leading 0!
Using the C<oct> function:
$dec = oct("33653337357");
Using C<Bit::Vector>:
use Bit::Vector;
$vec = Bit::Vector->new(32);
$vec->Chunk_List_Store(3, split(//, reverse "33653337357"));
$dec = $vec->to_Dec();
=item How do I convert from decimal to octal
Using C<sprintf>:
$oct = sprintf("%o", 3735928559);
Using C<Bit::Vector>:
use Bit::Vector;
$vec = Bit::Vector->new_Dec(32, -559038737);
$oct = reverse join('', $vec->Chunk_List_Read(3));
=item How do I convert from binary to decimal
Perl 5.6 lets you write binary numbers directly with
the C<0b> notation:
$number = 0b10110110;
Using C<oct>:
my $input = "10110110";
$decimal = oct( "0b$input" );
Using C<pack> and C<ord>:
$decimal = ord(pack('B8', '10110110'));
Using C<pack> and C<unpack> for larger strings:
$int = unpack("N", pack("B32",
substr("0" x 32 . "11110101011011011111011101111", -32)));
$dec = sprintf("%d", $int);
# substr() is used to left pad a 32 character string with zeros.
Using C<Bit::Vector>:
$vec = Bit::Vector->new_Bin(32, "11011110101011011011111011101111");
$dec = $vec->to_Dec();
=item How do I convert from decimal to binary
Using C<sprintf> (perl 5.6+):
$bin = sprintf("%b", 3735928559);
Using C<unpack>:
$bin = unpack("B*", pack("N", 3735928559));
Using C<Bit::Vector>:
use Bit::Vector;
$vec = Bit::Vector->new_Dec(32, -559038737);
$bin = $vec->to_Bin();
The remaining transformations (e.g. hex -> oct, bin -> hex, etc.)
are left as an exercise to the inclined reader.
=back
=head2 Why doesn't & work the way I want it to?
The behavior of binary arithmetic operators depends on whether they're
used on numbers or strings. The operators treat a string as a series
of bits and work with that (the string C<"3"> is the bit pattern
C<00110011>). The operators work with the binary form of a number
(the number C<3> is treated as the bit pattern C<00000011>).
So, saying C<11 & 3> performs the "and" operation on numbers (yielding
C<3>). Saying C<"11" & "3"> performs the "and" operation on strings
(yielding C<"1">).
Most problems with C<&> and C<|> arise because the programmer thinks
they have a number but really it's a string. The rest arise because
the programmer says:
if ("\020\020" & "\101\101") {
# ...
}
but a string consisting of two null bytes (the result of C<"\020\020"
& "\101\101">) is not a false value in Perl. You need:
if ( ("\020\020" & "\101\101") !~ /[^\000]/) {
# ...
}
=head2 How do I multiply matrices?
Use the C<Math::Matrix> or C<Math::MatrixReal> modules (available from CPAN)
or the C<PDL> extension (also available from CPAN).
=head2 How do I perform an operation on a series of integers?
To call a function on each element in an array, and collect the
results, use:
@results = map { my_func($_) } @array;
For example:
@triple = map { 3 * $_ } @single;
To call a function on each element of an array, but ignore the
results:
foreach $iterator (@array) {
some_func($iterator);
}
To call a function on each integer in a (small) range, you B<can> use:
@results = map { some_func($_) } (5 .. 25);
but you should be aware that the C<..> operator creates an array of
all integers in the range. This can take a lot of memory for large
ranges. Instead use:
@results = ();
for ($i=5; $i <= 500_005; $i++) {
push(@results, some_func($i));
}
This situation has been fixed in Perl5.005. Use of C<..> in a C<for>
loop will iterate over the range, without creating the entire range.
for my $i (5 .. 500_005) {
push(@results, some_func($i));
}
will not create a list of 500,000 integers.
=head2 How can I output Roman numerals?
Get the L<http://www.cpan.org/modules/by-module/Roman> module.
=head2 Why aren't my random numbers random?
If you're using a version of Perl before 5.004, you must call C<srand>
once at the start of your program to seed the random number generator.
BEGIN { srand() if $] < 5.004 }
5.004 and later automatically call C<srand> at the beginning. Don't
call C<srand> more than once--you make your numbers less random,
rather than more.
Computers are good at being predictable and bad at being random
(despite appearances caused by bugs in your programs :-). see the
F<random> article in the "Far More Than You Ever Wanted To Know"
collection in L<http://www.cpan.org/misc/olddoc/FMTEYEWTK.tgz>, courtesy
of Tom Phoenix, talks more about this. John von Neumann said, "Anyone
who attempts to generate random numbers by deterministic means is, of
course, living in a state of sin."
If you want numbers that are more random than C<rand> with C<srand>
provides, you should also check out the C<Math::TrulyRandom> module from
CPAN. It uses the imperfections in your system's timer to generate
random numbers, but this takes quite a while. If you want a better
pseudorandom generator than comes with your operating system, look at
"Numerical Recipes in C" at L<http://www.nr.com/>.
=head2 How do I get a random number between X and Y?
To get a random number between two values, you can use the C<rand()>
built-in to get a random number between 0 and 1. From there, you shift
that into the range that you want.
C<rand($x)> returns a number such that C<< 0 <= rand($x) < $x >>. Thus
what you want to have perl figure out is a random number in the range
from 0 to the difference between your I<X> and I<Y>.
That is, to get a number between 10 and 15, inclusive, you want a
random number between 0 and 5 that you can then add to 10.
my $number = 10 + int rand( 15-10+1 ); # ( 10,11,12,13,14, or 15 )
Hence you derive the following simple function to abstract
that. It selects a random integer between the two given
integers (inclusive), For example: C<random_int_between(50,120)>.
sub random_int_between {
my($min, $max) = @_;
# Assumes that the two arguments are integers themselves!
return $min if $min == $max;
($min, $max) = ($max, $min) if $min > $max;
return $min + int rand(1 + $max - $min);
}
=head1 Data: Dates
=head2 How do I find the day or week of the year?
The C<localtime> function returns the day of the year. Without an
argument C<localtime> uses the current time.
$day_of_year = (localtime)[7];
The C<POSIX> module can also format a date as the day of the year or
week of the year.
use POSIX qw/strftime/;
my $day_of_year = strftime "%j", localtime;
my $week_of_year = strftime "%W", localtime;
To get the day of year for any date, use C<POSIX>'s C<mktime> to get
a time in epoch seconds for the argument to C<localtime>.
use POSIX qw/mktime strftime/;
my $week_of_year = strftime "%W",
localtime( mktime( 0, 0, 0, 18, 11, 87 ) );
The C<Date::Calc> module provides two functions to calculate these.
use Date::Calc;
my $day_of_year = Day_of_Year( 1987, 12, 18 );
my $week_of_year = Week_of_Year( 1987, 12, 18 );
=head2 How do I find the current century or millennium?
Use the following simple functions:
sub get_century {
return int((((localtime(shift || time))[5] + 1999))/100);
}
sub get_millennium {
return 1+int((((localtime(shift || time))[5] + 1899))/1000);
}
On some systems, the C<POSIX> module's C<strftime()> function has been
extended in a non-standard way to use a C<%C> format, which they
sometimes claim is the "century". It isn't, because on most such
systems, this is only the first two digits of the four-digit year, and
thus cannot be used to reliably determine the current century or
millennium.
=head2 How can I compare two dates and find the difference?
(contributed by brian d foy)
You could just store all your dates as a number and then subtract.
Life isn't always that simple though. If you want to work with
formatted dates, the C<Date::Manip>, C<Date::Calc>, or C<DateTime>
modules can help you.
=head2 How can I take a string and turn it into epoch seconds?
If it's a regular enough string that it always has the same format,
you can split it up and pass the parts to C<timelocal> in the standard
C<Time::Local> module. Otherwise, you should look into the C<Date::Calc>
and C<Date::Manip> modules from CPAN.
=head2 How can I find the Julian Day?
(contributed by brian d foy and Dave Cross)
You can use the C<Time::JulianDay> module available on CPAN. Ensure
that you really want to find a Julian day, though, as many people have
different ideas about Julian days. See
http://www.hermetic.ch/cal_stud/jdn.htm for instance.
You can also try the C<DateTime> module, which can convert a date/time
to a Julian Day.
$ perl -MDateTime -le'print DateTime->today->jd'
2453401.5
Or the modified Julian Day
$ perl -MDateTime -le'print DateTime->today->mjd'
53401
Or even the day of the year (which is what some people think of as a
Julian day)
$ perl -MDateTime -le'print DateTime->today->doy'
31
=head2 How do I find yesterday's date?
X<date> X<yesterday> X<DateTime> X<Date::Calc> X<Time::Local>
X<daylight saving time> X<day> X<Today_and_Now> X<localtime>
X<timelocal>
(contributed by brian d foy)
Use one of the Date modules. The C<DateTime> module makes it simple, and
give you the same time of day, only the day before.
use DateTime;
my $yesterday = DateTime->now->subtract( days => 1 );
print "Yesterday was $yesterday\n";
You can also use the C<Date::Calc> module using its C<Today_and_Now>
function.
use Date::Calc qw( Today_and_Now Add_Delta_DHMS );
my @date_time = Add_Delta_DHMS( Today_and_Now(), -1, 0, 0, 0 );
print "@date_time\n";
Most people try to use the time rather than the calendar to figure out
dates, but that assumes that days are twenty-four hours each. For
most people, there are two days a year when they aren't: the switch to
and from summer time throws this off. Let the modules do the work.
If you absolutely must do it yourself (or can't use one of the
modules), here's a solution using C<Time::Local>, which comes with
Perl:
# contributed by Gunnar Hjalmarsson
use Time::Local;
my $today = timelocal 0, 0, 12, ( localtime )[3..5];
my ($d, $m, $y) = ( localtime $today-86400 )[3..5];
printf "Yesterday: %d-%02d-%02d\n", $y+1900, $m+1, $d;
In this case, you measure the day starting at noon, and subtract 24
hours. Even if the length of the calendar day is 23 or 25 hours,
you'll still end up on the previous calendar day, although not at
noon. Since you don't care about the time, the one hour difference
doesn't matter and you end up with the previous date.
=head2 Does Perl have a Year 2000 or 2038 problem? Is Perl Y2K compliant?
(contributed by brian d foy)
Perl itself never had a Y2K problem, although that never stopped people
from creating Y2K problems on their own. See the documentation for
C<localtime> for its proper use.
Starting with Perl 5.11, C<localtime> and C<gmtime> can handle dates past
03:14:08 January 19, 2038, when a 32-bit based time would overflow. You
still might get a warning on a 32-bit C<perl>:
% perl5.11.2 -E 'say scalar localtime( 0x9FFF_FFFFFFFF )'
Integer overflow in hexadecimal number at -e line 1.
Wed Nov 1 19:42:39 5576711
On a 64-bit C<perl>, you can get even larger dates for those really long
running projects:
% perl5.11.2 -E 'say scalar gmtime( 0x9FFF_FFFFFFFF )'
Thu Nov 2 00:42:39 5576711
You're still out of luck if you need to keep track of decaying protons
though.
=head1 Data: Strings
=head2 How do I validate input?
(contributed by brian d foy)
There are many ways to ensure that values are what you expect or
want to accept. Besides the specific examples that we cover in the
perlfaq, you can also look at the modules with "Assert" and "Validate"
in their names, along with other modules such as C<Regexp::Common>.
Some modules have validation for particular types of input, such
as C<Business::ISBN>, C<Business::CreditCard>, C<Email::Valid>,
and C<Data::Validate::IP>.
=head2 How do I unescape a string?
It depends just what you mean by "escape". URL escapes are dealt
with in L<perlfaq9>. Shell escapes with the backslash (C<\>)
character are removed with
s/\\(.)/$1/g;
This won't expand C<"\n"> or C<"\t"> or any other special escapes.
=head2 How do I remove consecutive pairs of characters?
(contributed by brian d foy)
You can use the substitution operator to find pairs of characters (or
runs of characters) and replace them with a single instance. In this
substitution, we find a character in C<(.)>. The memory parentheses
store the matched character in the back-reference C<\g1> and we use
that to require that the same thing immediately follow it. We replace
that part of the string with the character in C<$1>.
s/(.)\g1/$1/g;
We can also use the transliteration operator, C<tr///>. In this
example, the search list side of our C<tr///> contains nothing, but
the C<c> option complements that so it contains everything. The
replacement list also contains nothing, so the transliteration is
almost a no-op since it won't do any replacements (or more exactly,
replace the character with itself). However, the C<s> option squashes
duplicated and consecutive characters in the string so a character
does not show up next to itself
my $str = 'Haarlem'; # in the Netherlands
$str =~ tr///cs; # Now Harlem, like in New York
=head2 How do I expand function calls in a string?
(contributed by brian d foy)
This is documented in L<perlref>, and although it's not the easiest
thing to read, it does work. In each of these examples, we call the
function inside the braces used to dereference a reference. If we
have more than one return value, we can construct and dereference an
anonymous array. In this case, we call the function in list context.
print "The time values are @{ [localtime] }.\n";
If we want to call the function in scalar context, we have to do a bit
more work. We can really have any code we like inside the braces, so
we simply have to end with the scalar reference, although how you do
that is up to you, and you can use code inside the braces. Note that
the use of parens creates a list context, so we need C<scalar> to
force the scalar context on the function:
print "The time is ${\(scalar localtime)}.\n"
print "The time is ${ my $x = localtime; \$x }.\n";
If your function already returns a reference, you don't need to create
the reference yourself.
sub timestamp { my $t = localtime; \$t }
print "The time is ${ timestamp() }.\n";
The C<Interpolation> module can also do a lot of magic for you. You can
specify a variable name, in this case C<E>, to set up a tied hash that
does the interpolation for you. It has several other methods to do this
as well.
use Interpolation E => 'eval';
print "The time values are $E{localtime()}.\n";
In most cases, it is probably easier to simply use string concatenation,
which also forces scalar context.
print "The time is " . localtime() . ".\n";
=head2 How do I find matching/nesting anything?
This isn't something that can be done in one regular expression, no
matter how complicated. To find something between two single
characters, a pattern like C</x([^x]*)x/> will get the intervening
bits in $1. For multiple ones, then something more like
C</alpha(.*?)omega/> would be needed. But none of these deals with
nested patterns. For balanced expressions using C<(>, C<{>, C<[> or
C<< < >> as delimiters, use the CPAN module Regexp::Common, or see
L<perlre/(??{ code })>. For other cases, you'll have to write a
parser.
If you are serious about writing a parser, there are a number of
modules or oddities that will make your life a lot easier. There are
the CPAN modules C<Parse::RecDescent>, C<Parse::Yapp>, and
C<Text::Balanced>; and the C<byacc> program. Starting from perl 5.8
the C<Text::Balanced> is part of the standard distribution.
One simple destructive, inside-out approach that you might try is to
pull out the smallest nesting parts one at a time:
while (s/BEGIN((?:(?!BEGIN)(?!END).)*)END//gs) {
# do something with $1
}
A more complicated and sneaky approach is to make Perl's regular
expression engine do it for you. This is courtesy Dean Inada, and
rather has the nature of an Obfuscated Perl Contest entry, but it
really does work:
# $_ contains the string to parse
# BEGIN and END are the opening and closing markers for the
# nested text.
@( = ('(','');
@) = (')','');
($re=$_)=~s/((BEGIN)|(END)|.)/$)[!$3]\Q$1\E$([!$2]/gs;
@$ = (eval{/$re/},$@!~/unmatched/i);
print join("\n",@$[0..$#$]) if( $$[-1] );
=head2 How do I reverse a string?
Use C<reverse()> in scalar context, as documented in
L<perlfunc/reverse>.
$reversed = reverse $string;
=head2 How do I expand tabs in a string?
You can do it yourself:
1 while $string =~ s/\t+/' ' x (length($&) * 8 - length($`) % 8)/e;
Or you can just use the C<Text::Tabs> module (part of the standard Perl
distribution).
use Text::Tabs;
@expanded_lines = expand(@lines_with_tabs);
=head2 How do I reformat a paragraph?
Use C<Text::Wrap> (part of the standard Perl distribution):
use Text::Wrap;
print wrap("\t", ' ', @paragraphs);
The paragraphs you give to C<Text::Wrap> should not contain embedded
newlines. C<Text::Wrap> doesn't justify the lines (flush-right).
Or use the CPAN module C<Text::Autoformat>. Formatting files can be
easily done by making a shell alias, like so:
alias fmt="perl -i -MText::Autoformat -n0777 \
-e 'print autoformat $_, {all=>1}' $*"
See the documentation for C<Text::Autoformat> to appreciate its many
capabilities.
=head2 How can I access or change N characters of a string?
You can access the first characters of a string with substr().
To get the first character, for example, start at position 0
and grab the string of length 1.
$string = "Just another Perl Hacker";
$first_char = substr( $string, 0, 1 ); # 'J'
To change part of a string, you can use the optional fourth
argument which is the replacement string.
substr( $string, 13, 4, "Perl 5.8.0" );
You can also use substr() as an lvalue.
substr( $string, 13, 4 ) = "Perl 5.8.0";
=head2 How do I change the Nth occurrence of something?
You have to keep track of N yourself. For example, let's say you want
to change the fifth occurrence of C<"whoever"> or C<"whomever"> into
C<"whosoever"> or C<"whomsoever">, case insensitively. These
all assume that $_ contains the string to be altered.
$count = 0;
s{((whom?)ever)}{
++$count == 5 # is it the 5th?
? "${2}soever" # yes, swap
: $1 # renege and leave it there
}ige;
In the more general case, you can use the C</g> modifier in a C<while>
loop, keeping count of matches.
$WANT = 3;
$count = 0;
$_ = "One fish two fish red fish blue fish";
while (/(\w+)\s+fish\b/gi) {
if (++$count == $WANT) {
print "The third fish is a $1 one.\n";
}
}
That prints out: C<"The third fish is a red one."> You can also use a
repetition count and repeated pattern like this:
/(?:\w+\s+fish\s+){2}(\w+)\s+fish/i;
=head2 How can I count the number of occurrences of a substring within a string?
There are a number of ways, with varying efficiency. If you want a
count of a certain single character (X) within a string, you can use the
C<tr///> function like so:
$string = "ThisXlineXhasXsomeXx'sXinXit";
$count = ($string =~ tr/X//);
print "There are $count X characters in the string";
This is fine if you are just looking for a single character. However,
if you are trying to count multiple character substrings within a
larger string, C<tr///> won't work. What you can do is wrap a while()
loop around a global pattern match. For example, let's count negative
integers:
$string = "-9 55 48 -2 23 -76 4 14 -44";
while ($string =~ /-\d+/g) { $count++ }
print "There are $count negative numbers in the string";
Another version uses a global match in list context, then assigns the
result to a scalar, producing a count of the number of matches.
$count = () = $string =~ /-\d+/g;
=head2 How do I capitalize all the words on one line?
X<Text::Autoformat> X<capitalize> X<case, title> X<case, sentence>
(contributed by brian d foy)
Damian Conway's L<Text::Autoformat> handles all of the thinking
for you.
use Text::Autoformat;
my $x = "Dr. Strangelove or: How I Learned to Stop ".
"Worrying and Love the Bomb";
print $x, "\n";
for my $style (qw( sentence title highlight )) {
print autoformat($x, { case => $style }), "\n";
}
How do you want to capitalize those words?
FRED AND BARNEY'S LODGE # all uppercase
Fred And Barney's Lodge # title case
Fred and Barney's Lodge # highlight case
It's not as easy a problem as it looks. How many words do you think
are in there? Wait for it... wait for it.... If you answered 5
you're right. Perl words are groups of C<\w+>, but that's not what
you want to capitalize. How is Perl supposed to know not to capitalize
that C<s> after the apostrophe? You could try a regular expression:
$string =~ s/ (
(^\w) #at the beginning of the line
| # or
(\s\w) #preceded by whitespace
)
/\U$1/xg;
$string =~ s/([\w']+)/\u\L$1/g;
Now, what if you don't want to capitalize that "and"? Just use
L<Text::Autoformat> and get on with the next problem. :)
=head2 How can I split a [character] delimited string except when inside [character]?
Several modules can handle this sort of parsing--C<Text::Balanced>,
C<Text::CSV>, C<Text::CSV_XS>, and C<Text::ParseWords>, among others.
Take the example case of trying to split a string that is
comma-separated into its different fields. You can't use C<split(/,/)>
because you shouldn't split if the comma is inside quotes. For
example, take a data line like this:
SAR001,"","Cimetrix, Inc","Bob Smith","CAM",N,8,1,0,7,"Error, Core Dumped"
Due to the restriction of the quotes, this is a fairly complex
problem. Thankfully, we have Jeffrey Friedl, author of
I<Mastering Regular Expressions>, to handle these for us. He
suggests (assuming your string is contained in C<$text>):
@new = ();
push(@new, $+) while $text =~ m{
"([^\"\\]*(?:\\.[^\"\\]*)*)",? # groups the phrase inside the quotes
| ([^,]+),?
| ,
}gx;
push(@new, undef) if substr($text,-1,1) eq ',';
If you want to represent quotation marks inside a
quotation-mark-delimited field, escape them with backslashes (eg,
C<"like \"this\"">.
Alternatively, the C<Text::ParseWords> module (part of the standard
Perl distribution) lets you say:
use Text::ParseWords;
@new = quotewords(",", 0, $text);
=head2 How do I strip blank space from the beginning/end of a string?
(contributed by brian d foy)
A substitution can do this for you. For a single line, you want to
replace all the leading or trailing whitespace with nothing. You
can do that with a pair of substitutions:
s/^\s+//;
s/\s+$//;
You can also write that as a single substitution, although it turns
out the combined statement is slower than the separate ones. That
might not matter to you, though:
s/^\s+|\s+$//g;
In this regular expression, the alternation matches either at the
beginning or the end of the string since the anchors have a lower
precedence than the alternation. With the C</g> flag, the substitution
makes all possible matches, so it gets both. Remember, the trailing
newline matches the C<\s+>, and the C<$> anchor can match to the
absolute end of the string, so the newline disappears too. Just add
the newline to the output, which has the added benefit of preserving
"blank" (consisting entirely of whitespace) lines which the C<^\s+>
would remove all by itself:
while( <> ) {
s/^\s+|\s+$//g;
print "$_\n";
}
For a multi-line string, you can apply the regular expression to each
logical line in the string by adding the C</m> flag (for
"multi-line"). With the C</m> flag, the C<$> matches I<before> an
embedded newline, so it doesn't remove it. This pattern still removes
the newline at the end of the string:
$string =~ s/^\s+|\s+$//gm;
Remember that lines consisting entirely of whitespace will disappear,
since the first part of the alternation can match the entire string
and replace it with nothing. If you need to keep embedded blank lines,
you have to do a little more work. Instead of matching any whitespace
(since that includes a newline), just match the other whitespace:
$string =~ s/^[\t\f ]+|[\t\f ]+$//mg;
=head2 How do I pad a string with blanks or pad a number with zeroes?
In the following examples, C<$pad_len> is the length to which you wish
to pad the string, C<$text> or C<$num> contains the string to be padded,
and C<$pad_char> contains the padding character. You can use a single
character string constant instead of the C<$pad_char> variable if you
know what it is in advance. And in the same way you can use an integer in
place of C<$pad_len> if you know the pad length in advance.
The simplest method uses the C<sprintf> function. It can pad on the left
or right with blanks and on the left with zeroes and it will not
truncate the result. The C<pack> function can only pad strings on the
right with blanks and it will truncate the result to a maximum length of
C<$pad_len>.
# Left padding a string with blanks (no truncation):
$padded = sprintf("%${pad_len}s", $text);
$padded = sprintf("%*s", $pad_len, $text); # same thing
# Right padding a string with blanks (no truncation):
$padded = sprintf("%-${pad_len}s", $text);
$padded = sprintf("%-*s", $pad_len, $text); # same thing
# Left padding a number with 0 (no truncation):
$padded = sprintf("%0${pad_len}d", $num);
$padded = sprintf("%0*d", $pad_len, $num); # same thing
# Right padding a string with blanks using pack (will truncate):
$padded = pack("A$pad_len",$text);
If you need to pad with a character other than blank or zero you can use
one of the following methods. They all generate a pad string with the
C<x> operator and combine that with C<$text>. These methods do
not truncate C<$text>.
Left and right padding with any character, creating a new string:
$padded = $pad_char x ( $pad_len - length( $text ) ) . $text;
$padded = $text . $pad_char x ( $pad_len - length( $text ) );
Left and right padding with any character, modifying C<$text> directly:
substr( $text, 0, 0 ) = $pad_char x ( $pad_len - length( $text ) );
$text .= $pad_char x ( $pad_len - length( $text ) );
=head2 How do I extract selected columns from a string?
(contributed by brian d foy)
If you know the columns that contain the data, you can
use C<substr> to extract a single column.
my $column = substr( $line, $start_column, $length );
You can use C<split> if the columns are separated by whitespace or
some other delimiter, as long as whitespace or the delimiter cannot
appear as part of the data.
my $line = ' fred barney betty ';
my @columns = split /\s+/, $line;
# ( '', 'fred', 'barney', 'betty' );
my $line = 'fred||barney||betty';
my @columns = split /\|/, $line;
# ( 'fred', '', 'barney', '', 'betty' );