In Perl, how can I limit the number of places after the decimal point but have no trailing zeroes?

问题

This question is similar to "dropping trailing ‘.0’ from floats", but for Perl and with a maximum number of digits after the decimal.

I'm looking for a way to convert numbers to string format, dropping any redundant '0', including not just right after the decimal. And still with a maximum number of digital, e.g. 3

The input data is floats. Desired output:

0         -> 0
0.1       -> 0.1
0.11      -> 0.11
0.111     -> 0.111
0.1111111 -> 0.111

回答1:

You can also use Math::Round to do this:

$ perl -MMath::Round=nearest -e 'print nearest(.001, 0.1), "\n"'
0.1
$ perl -MMath::Round=nearest -e 'print nearest(.001, 0.11111), "\n"'
0.111

回答2:

Use the following directly:

my $s = sprintf('%.3f', $f);
$s =~ s/\.?0*$//;

print $s

...or define a subroutine to do it more generically:

sub fstr {
  my ($value,$precision) = @_;
  $precision ||= 3;
  my $s = sprintf("%.${precision}f", $value);
  $s =~ s/\.?0*$//;
  $s
}

print fstr(0) . "\n";
print fstr(1) . "\n";
print fstr(1.1) . "\n";
print fstr(1.12) . "\n";
print fstr(1.123) . "\n";
print fstr(1.12345) . "\n";
print fstr(1.12345, 2) . "\n";
print fstr(1.12345, 10) . "\n";

Prints:

回答3:

You can use "sprintf" combined with "eval".

my $num = eval sprintf('%.3f', $raw_num);

For example:

#!/usr/bin/perl 

my @num_array = (
    0, 1, 1.0, 0.1, 0.10, 0.11, 0.111, 0.1110, 0.1111111
);


for my $raw_num (@num_array) {
    my $num = eval sprintf('%.3f', $raw_num);
    print $num . "\n";
}

outputs:

回答4:

This will give you the output your looking for,

sub dropTraillingZeros{
$_ = shift;
s/(\d*\.\d{3})(.*)/$1/;
s/(\d*\.\d)(00)/$1/;
s/(\d*\.\d{2})(0)/$1/;
print "$_\n";
}
dropTraillingZeros(0);
dropTraillingZeros(0.1);
dropTraillingZeros(0.11);
dropTraillingZeros(0.111);
dropTraillingZeros(0.11111111);

来源：https://stackoverflow.com/questions/571329/in-perl-how-can-i-limit-the-number-of-places-after-the-decimal-point-but-have-n

标签

perl

printf

decimal-point