sub is_integer {
defined $_[0] && $_[0] =~ /^[+-]?\d+$/;
}
sub is_float {
defined $_[0] && $_[0] =~ /^[+-]?\d+(\.\d+)?$/;
}
For the code mentioned above, if we give input as 999999999999999999999999999999999999999999, it is giving output as not real number.
Why it is behaving like that?
I forgot to mention one more thing:
If I am using this code for $x as the above value:
if($x > 0 || $x <= 0 ) {
print "Real";
}
Output is real.
How is this possible?
$ perl -e 'print 999999999999999999999999999999999999999999'
1e+42
i.e. Perl uses scientific representation for this number and that is why your regexp doesn't match.
Use the looks_like_number function from Scalar::Util (which is a core module).
use Scalar::Util qw( looks_like_number );
say "Number" if looks_like_number 999999999999999999999999999999999999999999;
# above prints "Number"
Just to add one more thing. As others have explained, the number you are working with is out of range for a Perl integer (unless you are on a 140 bit machine). Therefore, the variable will be stored as a floating point number. Regular expressions operate on strings. Therefore, the number is converted to its string representation before the regular expression operates on it.
Others have explained what is going on: out of the box, Perl can't handle numbers that large without using scientific notation.
If you need to work with large numbers, take a look at bignum or its components, such as Math::BigInt. For example:
use strict;
use warnings;
use Math::BigInt;
my $big_str = '900000000000000000000000000000000000000';
my $big_num = Math::BigInt->new($big_str);
$big_num ++;
print "Is integer: $big_num\n" if is_integer($big_num);
sub is_integer {
defined $_[0] && $_[0] =~ /^[+-]?\d+$/;
}
Also, you may want to take a look at bignum in the Perl documentation.
Related
My motive is to convert the string number into floating point number while creating a hash.
I have placed my entire code and error below. Please help me to solve this issue.
Sample code
use strict;
use warnings;
use Data::Dumper;
my $price = 8.5;
my $g={};
$g->{'get'}=sprintf('%.02f',$price);
print Dumper($g);
Current output
$VAR1 = {
'get' => '8.50'
};
Expected output
$VAR1 = {
'get' => 8.50
};
Despite the single quotes around 8.50 in the Dumper output, Perl will still treat it as a numeric value when you go to use it:
use strict;
use warnings;
my $price = 8.5;
my $g={};
$g->{'get'}=sprintf('%.02f',$price);
my $x = 5;
printf "%.02f\n", $x + $g->{get};
Outputs:
13.50
use Scalar::Util 'looks_like_number';
.
.
print Dumper($g) =~ s/'(.*?)'/looks_like_number($1)?$1:"'$1'"/ger;
Which changes the output from Dumper before it's printed. It removes both 's of every single quoted string if it looks like a number according to Scalar::Util.
I suspect you're worrying unnecessarily here. Perl treats strings and numbers as largely interchangeable and will generally do the right thing with data of either type. The number of times when you should care if you have a string or a number is tiny.
In fact, even if you explicitly give Perl a number in code like yours, it will be displayed as a string:
$ perl -MData::Dumper -E'say Dumper { get => 8.5 }'
$VAR1 = {
'get' => '8.5'
};
In a text file, I have many lines looking like (a,b,c) where a, b, and c are double precision real numbers, for instance (8.27605704077856,0.505526531790625,1.15577754382534e-05). Is there a simple way to replace numbers smaller than 10e-4 by 0 in Perl?
Edit: For instance, the text file to be treated looks like:
\plotinstruction[color,style,width]
points{
(8.27,0.5,1.1e-05)
(8.26,1,4.1e-06)
(8.25,1.5,3e-06)
}
and I want to write in a new file:
\plotinstruction[color,style,width]
points{
(8.27,0.5,0)
(8.26,1,0)
(8.25,1.5,0)
}
Perhaps I'm missing something, but perhaps use of map would help?
#!/usr/bin/env perl
use strict;
use warnings;
use Data::Dumper;
my #values = (8.27605704077856, 0.505526531790625, 1.15577754382534e-05);
my #filtered_values = map(($_ > 1e-4) ? $_ : 0, #values);
print Dumper \#filtered_values;
Results:
$VAR1 = [
'8.27605704077856',
'0.505526531790625',
0
];
To parse input, you could use a regular expression to extract a comma-separated string of numbers, using split on that to get a Perl list to run map upon.
You can write:
perl -pwe 's/\d[\d.e+-]+/$& < 0.001 && $& > -0.001 ? "0" : $&/ge' < INPUT > OUTPUT
(-p means to read in the input, one line at a time, into $_, run the program, print out $_, and loop again; -w enables warnings; -e means that the program is specified directly as a command-line argument; s/// is a regex-substitution; /g means that it's a "global" substitution; and /e means that the replacement-text should be treated as a full Perl expression, rather than as a string with mere variable interpolation.)
I know that there is a library that do that
use Scalar::Util qw(looks_like_number);
yet I want to do it using perl regular expression. And I want it to work for double numbers not for only integers.
so I want something better than this
$var =~ /^[+-]?\d+$/
thanks.
Constructing a single regular expression to validate a number is really difficult. There simply are too many criteria to consider. Perlfaq4 contains a section "How do I determine whether a scalar is a number/whole/integer/float?
The code from that documentation shows the following tests:
if (/\D/) {print "has nondigits\n" }
if (/^\d+$/) {print "is a whole number\n" }
if (/^-?\d+$/) {print "is an integer\n" }
if (/^[+-]?\d+$/) {print "is a +/- integer\n" }
if (/^-?\d+\.?\d*$/) {print "is a real number\n" }
if (/^-?(?:\d+(?:\.\d*)?|\.\d+)$/) {print "is a decimal number\n"}
if (/^([+-]?)(?=\d|\.\d)\d*(\.\d*)?([Ee]([+-]?\d+))?$/) {
print "is a C float\n"
}
The first test disqualifies an unsigned integer.
The second test qualifies a whole number.
The third test qualifies an integer.
The fourth test qualifies a positive/negatively signed integer.
The fifth test qualifies a real number.
The sixth test qualifies a decimal number.
The seventh test qualifies a number in c-style scientific notation.
So if you were using those tests (excluding the first one) you would have to verify that one or more of the tests passes. Then you've got a number.
Another method, since you don't want to use the module Scalar::Util, you can learn from the code IN Scalar::Util. The looks_like_number() function is set up like this:
sub looks_like_number {
local $_ = shift;
# checks from perlfaq4
return $] < 5.009002 unless defined;
return 1 if (/^[+-]?\d+$/); # is a +/- integer
return 1 if (/^([+-]?)(?=\d|\.\d)\d*(\.\d*)?([Ee]([+-]?\d+))?$/); # a C float
return 1 if ($] >= 5.008 and /^(Inf(inity)?|NaN)$/i)
or ($] >= 5.006001 and /^Inf$/i);
0;
}
You should be able to use the portions of that function that are applicable to your situation.
I would like to point out, however, that Scalar::Util is a core Perl module; it ships with Perl, just like strict does. The best practice of all is probably to just use it.
You should use Regexp::Common, most patterns are more complicated than you realize.
use Regexp::Common;
my $real = 3.14159;
print "Real" if $real =~ /$RE{num}{real}/;
However, the pattern is not anchored by default, so a stricter version is:
my $real_pat = $RE{num}{real};
my $real = 3.14159;
print "Real" if $real =~ /^$real_pat$/;
Well first you should make sure that the number does not contain any commas so you do this:
$var =~ s/,//g; # remove all the commas
Then create another variable to do the rest of the compare.
$var2=$var;
Then remove the . from the new variable yet only once occurrence.
$var2 =~ s/.//; # replace . with nothing to compare yet only once.
now var2 should look like an integer with no "."
so do this:
if($var2 !~ /^[+-]?\d+$/){
print "not valid";
}else{
#use var1
}
you can fix this code and write it as a function if you need to use it more than once.
Cheers!
Here is my code:
#!/usr/bin/env perl
sub start
{
...
}
sub stop
{
...
}
if (defined $ARGV[0])
{
if ($ARGV[0]=='start') {start}
elsif ($ARGV[0]=='stop') {stop}
else {die "Unrecognized command: $ARGV[0]"}
}
Whatever I do though, it always executes &start. Am I doing something wrong?
I'm using Linux Mint 10 and Perl 5.10.1.
You're using numeric comparison to check string equality, which converts the arguments to numbers first before comparing them.
In this case, 'start' is not a number, so it gets converted to 0; the value in $ARGV[0] (which is expected to be a word here) is also converted, resulting in another 0 and a final condition of if (0 == 0), which is always true.
You want to use the eq string-wise comparison operator instead:
if ($ARGV[0] eq 'start') { start }
See the docs for more information on the different comparison operators.
Note that (as has been pointed out in the comments) including use warnings; at the top of your script would have caused perl to warn you about this from the start. Always including use warnings; and use strict; is generally best practice, as it helps catch these kinds of errors early.
The bug is that your program neglected to start with the standard boilerplate for Perl code.
Is there a simple way in Perl that will allow me to determine if a given variable is numeric? Something along the lines of:
if (is_number($x))
{ ... }
would be ideal. A technique that won't throw warnings when the -w switch is being used is certainly preferred.
Use Scalar::Util::looks_like_number() which uses the internal Perl C API's looks_like_number() function, which is probably the most efficient way to do this.
Note that the strings "inf" and "infinity" are treated as numbers.
Example:
#!/usr/bin/perl
use warnings;
use strict;
use Scalar::Util qw(looks_like_number);
my #exprs = qw(1 5.25 0.001 1.3e8 foo bar 1dd inf infinity);
foreach my $expr (#exprs) {
print "$expr is", looks_like_number($expr) ? '' : ' not', " a number\n";
}
Gives this output:
1 is a number
5.25 is a number
0.001 is a number
1.3e8 is a number
foo is not a number
bar is not a number
1dd is not a number
inf is a number
infinity is a number
See also:
perldoc Scalar::Util
perldoc perlapi for looks_like_number
The original question was how to tell if a variable was numeric, not if it "has a numeric value".
There are a few operators that have separate modes of operation for numeric and string operands, where "numeric" means anything that was originally a number or was ever used in a numeric context (e.g. in $x = "123"; 0+$x, before the addition, $x is a string, afterwards it is considered numeric).
One way to tell is this:
if ( length( do { no warnings "numeric"; $x & "" } ) ) {
print "$x is numeric\n";
}
If the bitwise feature is enabled, that makes & only a numeric operator and adds a separate string &. operator, you must disable it:
if ( length( do { no if $] >= 5.022, "feature", "bitwise"; no warnings "numeric"; $x & "" } ) ) {
print "$x is numeric\n";
}
(bitwise is available in perl 5.022 and above, and enabled by default if you use 5.028; or above.)
Check out the CPAN module Regexp::Common. I think it does exactly what you need and handles all the edge cases (e.g. real numbers, scientific notation, etc). e.g.
use Regexp::Common;
if ($var =~ /$RE{num}{real}/) { print q{a number}; }
Usually number validation is done with regular expressions. This code will determine if something is numeric as well as check for undefined variables as to not throw warnings:
sub is_integer {
defined $_[0] && $_[0] =~ /^[+-]?\d+$/;
}
sub is_float {
defined $_[0] && $_[0] =~ /^[+-]?\d+(\.\d+)?$/;
}
Here's some reading material you should look at.
A simple (and maybe simplistic) answer to the question is the content of $x numeric is the following:
if ($x eq $x+0) { .... }
It does a textual comparison of the original $x with the $x converted to a numeric value.
Not perfect, but you can use a regex:
sub isnumber
{
shift =~ /^-?\d+\.?\d*$/;
}
A slightly more robust regex can be found in Regexp::Common.
It sounds like you want to know if Perl thinks a variable is numeric. Here's a function that traps that warning:
sub is_number{
my $n = shift;
my $ret = 1;
$SIG{"__WARN__"} = sub {$ret = 0};
eval { my $x = $n + 1 };
return $ret
}
Another option is to turn off the warning locally:
{
no warnings "numeric"; # Ignore "isn't numeric" warning
... # Use a variable that might not be numeric
}
Note that non-numeric variables will be silently converted to 0, which is probably what you wanted anyway.
rexep not perfect... this is:
use Try::Tiny;
sub is_numeric {
my ($x) = #_;
my $numeric = 1;
try {
use warnings FATAL => qw/numeric/;
0 + $x;
}
catch {
$numeric = 0;
};
return $numeric;
}
Try this:
If (($x !~ /\D/) && ($x ne "")) { ... }
I found this interesting though
if ( $value + 0 eq $value) {
# A number
push #args, $value;
} else {
# A string
push #args, "'$value'";
}
Personally I think that the way to go is to rely on Perl's internal context to make the solution bullet-proof. A good regexp could match all the valid numeric values and none of the non-numeric ones (or vice versa), but as there is a way of employing the same logic the interpreter is using it should be safer to rely on that directly.
As I tend to run my scripts with -w, I had to combine the idea of comparing the result of "value plus zero" to the original value with the no warnings based approach of #ysth:
do {
no warnings "numeric";
if ($x + 0 ne $x) { return "not numeric"; } else { return "numeric"; }
}
You can use Regular Expressions to determine if $foo is a number (or not).
Take a look here:
How do I determine whether a scalar is a number
There is a highly upvoted accepted answer around using a library function, but it includes the caveat that "inf" and "infinity" are accepted as numbers. I see some regex stuff for answers too, but they seem to have issues. I tried my hand at writing some regex that would work better (I'm sorry it's long)...
/^0$|^[+-]?[1-9][0-9]*$|^[+-]?[1-9][0-9]*(\.[0-9]+)?([eE]-?[1-9][0-9]*)?$|^[+-]?[0-9]?\.[0-9]+$|^[+-]?[1-9][0-9]*\.[0-9]+$/
That's really 5 patterns separated by "or"...
Zero: ^0$
It's a kind of special case. It's the only integer that can start with 0.
Integers: ^[+-]?[1-9][0-9]*$
That makes sure the first digit is 1 to 9 and allows 0 to 9 for any of the following digits.
Scientific Numbers: ^[+-]?[1-9][0-9]*(\.[0-9]+)?([eE]-?[1-9][0-9]*)?$
Uses the same idea that the base number can't start with zero since in proper scientific notation you start with the highest significant bit (meaning the first number won't be zero). However, my pattern allows for multiple digits left of the decimal point. That's incorrect, but I've already spent too much time on this... you could replace the [1-9][0-9]* with just [0-9] to force a single digit before the decimal point and allow for zeroes.
Short Float Numbers: ^[+-]?[0-9]?\.[0-9]+$
This is like a zero integer. It's special in that it can start with 0 if there is only one digit left of the decimal point. It does overlap the next pattern though...
Long Float Numbers: ^[+-]?[1-9][0-9]*\.[0-9]+$
This handles most float numbers and allows more than one digit left of the decimal point while still enforcing that the higher number of digits can't start with 0.
The simple function...
sub is_number {
my $testVal = shift;
return $testVal =~ /^0$|^[+-]?[1-9][0-9]*$|^[+-]?[1-9][0-9]*(\.[0-9]+)?([eE]-?[1-9][0-9]*)?$|^[+-]?[0-9]?\.[0-9]+$|^[+-]?[1-9][0-9]*\.[0-9]+$/;
}
if ( defined $x && $x !~ m/\D/ ) {}
or
$x = 0 if ! $x;
if ( $x !~ m/\D/) {}
This is a slight variation on Veekay's answer but let me explain my reasoning for the change.
Performing a regex on an undefined value will cause error spew and will cause the code to exit in many if not most environments. Testing if the value is defined or setting a default case like i did in the alternative example before running the expression will, at a minimum, save your error log.