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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
602
The f format lets you specify a particular number of decimal places to round its argument to. Perl looks at the following digit, rounds up if it is 5 or greater, and rounds down otherwise. Three functions that may be useful if you want to round a floating-point value to an integral value are int , ceil , and floor .
Using Math.round() method is another method to limit the decimal places in Java. If we want to round a number to 1 decimal place, then we multiply and divide the input number by 10.0 in the round() method. Similarly, for 2 decimal places, we can use 100.0, for 3 decimal places, we can use 1000.0, and so on.
To limit the number of digits up to 2 places after the decimal, the toFixed() method is used. The toFixed() method rounds up the floating-point number up to 2 places after the decimal.
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:
0
1
1.1
1.12
1.123
1.123
1.12
1.12345
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