I have a string in Perl that contains a small number:
$num = "0.00000001";
When I make a numeric operation on it, it becomes number in exponential format:
$num = $num * 100;
print "$num\n";
The result is: 1e-06
The question is,
how to get this number be printed in floating-point format, i.e. 0.000001.
I know I can do it for specific number with sprintf("%.6f", $num),
but I'd like to have a generic solution,
so I won't have to determine each time how many digits to show after the decimal point
(like 6 in the above sprintf example)
When you apply a numeric operation to $num, it becomes a floating-point number. 1e-06 and 0.000001 are textual representations of that number; the stored value doesn't distinguish between them.
If you simply print or stringify the number, it uses a default format which, as you've seen, results in "1e-06". Using sprintf with a format of "%f" will give you a reasonable result; sprintf("%f", $num) yields "0.000001".
But the "%f" format can lose information. For example:
$num = "0.00000001";
printf("%f\n", $num);
prints:
0.000000
You say you want to print without having to determine each time how many digits to show after the decimal point. Something has to make that determination, and there's no universally correct way to do so. The obvious thing to do is print just the significant digits, omitting trailing zeros, but that presents some problems. How many digits do you print for 1.0/3.0, whose decimal representation has an infinite sequence of 3s? And 0.00000001 can't be represented exactly in binary floating-point:
$num = "0.00000001";
printf("%f\n", $num);
printf("%.78f\n", $num);
prints:
0.000000
0.000000010000000000000000209225608301284726753266340892878361046314239501953125
Use rodents of unusual size:
$ perl -Mbigrat -E'$num = 0.00000001; $num *= 100; say $num'
0.000001
Related
I would like to normalize the variable from ie. 00000000.1, to 0.1 using Perl
my $number = 000000.1;
$number =\~ s/^0+(\.\d+)/0$1/;
Is there any other solution to normalize floats lower than 1 by removing upfront zeros than using regex?
When I try to put those kind of numbers into an example function below
test(00000000.1, 0000000.025);
sub test {
my ($a, $b) = #_;
print $a, "\n";
print $b, "\n";
print $a + $b, "\n";
}
I get
01
021
22
which is not what is expected.
A number with leading zeros is interpreted as octal, e.g. 000000.1 is 01. I presume you have a string as input, e.g. my $number = "000000.1". With this your regex is:
my $number = "000000.1";
$number =~ s/^0+(?=0\.\d+)//;
print $number;
Output:
0.1
Explanation of regex:
^0+ -- 1+ 0 digits
(?=0\.\d+) -- positive lookahead for 0. followed by digits
Learn more about regex: https://twiki.org/cgi-bin/view/Codev/TWikiPresentation2018x10x14Regex
Simplest way, force it to be treated as a number and it will drop the leading zeros since they are meaningless for decimal numbers
my $str = '000.1';
...
my $num = 0 + $str;
An example,† to run from the command-line:
perl -wE'$n = shift; $n = 0 + $n; say $n' 000.1
Prints 0.1
Another, more "proper" way is to format that string ('000.1' and such) using sprintf. Then you do need to make a choice about precision, but that is often a good idea anyway
my $num = sprintf "%f", $str; # default precision
Or, if you know how many decimal places you want to keep
my $num = sprintf "%.3f", $str;
† The example in the question is really invalid. An unquoted string of digits which starts with a zero (077, rather than '077') would be treated as an octal number except that the decimal point (in 000.1) renders that moot as octals can't be fractional; so, Perl being Perl, it is tortured into a number somehow, but possibly yielding unintended values.
I am not sure how one could get an actual input like that. If 000.1 is read from a file or from the command-line or from STDIN ... it will be a string, an equivalent of assigning '000.1'
See Scalar value constructors in perldata, and for far more detail, perlnumber.
As others have noted, in Perl, leading zeros produce octal numbers; 10 is just a decimal number ten but 010 is equal to decimal eight. So yeah, the numbers should be in quotes for the problem to make any sense.
But the other answers don’t explain why the printed results look funny. Contrary to Peter Thoeny’s comment and zdim’s answer, there is nothing ‘invalid’ about the numbers. True, octals can’t be floating point, but Perl does not strip the . to turn 0000000.025 into 025. What happens is this:
Perl reads the run of zeros and recognises it as an octal number.
Perl reads the dot and parses it as the concatenation operator.
Perl reads 025 and again recognises it as an octal number.
Perl coerces the operands to strings, i.e. the decimal value of the numbers in string form; 0000000 is, of course, '0' and 025 is '21'.
Perl concatenates the two strings and returns the result, i.e. '021'.
And without error.
(As an exercise, you can check something like 010.025 which, for the same reason, turns into '821'.)
This is why $a and $b are each printed with a leading zero. Also note that, to evaluate $a + $b, Perl coerces the strings to numbers, but since leading zeros in strings do not produce octals, '01' + '021' is the same as '1' + '21', returning 22.
I have one floating point number in Perl and after multiplying with negative integer, all trailing zeroes are getting removed automatically. However I still need those extra zeroes.
Example:
my $float = 1.40000;
my $multiply = -1 * $float;
print "Negative number: $multiply"; //-1.4
Is there any way to get -1.40000?
You need to format your output using printf.
printf( "Negative number: %.4f\n", $multiply );
Learn more about printf with perldoc -f printf or here.
I am using sprintf in Perl to convert decimal numbers to their hex equivalents, as in:
my $positive = sprintf("X'%02X'", 1); # X'01'
my $negative = sprintf("X'%02X'", -1); # X'FFFFFFFFFFFFFFFF'
Why does the negative number always get sign-extended to the 8-byte representation of the number, even if the precision specified in sprintf is smaller?
Is there a way to avoid this so I don't have to substring the value every time? The same result occurs even if I remove the 0, as in: sprintf("X'%2X'", -1).
Well, %x specifically takes an unsigned integer, so when you give it the bit pattern for a signed integer, it treats all of those bits as an unsigned one. The length of the output probably depends on how big your compiler made numbers (although I'm guessing most modern things are the same now). Perl stores most of its numbers as doubles, so that's the size of the number. It's one of few places where Perl exposes the underlying architecture.
This means that you need to handle the sign yourself and use the absolute value of the number:
sprintf( "%s%2x", ($n < 0 ? '-' : ''), abs($n) );
As for the sprintf field specifiers, you are not specifying a precision there. That's minimum width. The %f and %g specifiers may have a maximum number of decimal places, but the entire string can still overflow.
But, it turns out that your problem is different. You want the values from -128 to 127 it their signed integer form. So, -1 gets FF. You need to add a little bit to get that.
my $s = sprintf( "%2x", $n );
$s = substr( $s, -2 ) if $n < 0;
I am trying to remove trailing zeroes from decimal numbers.
For eg: If the input number is 0.0002340000, I would like the output to be 0.000234
I am using sprintf("%g",$number), but that works for the most part, except sometimes it converts the number into an exponential value with E-. How can I have it only display as a full decimal number?
Numbers don't have trailing zeroes. Trailing zeroes can only occur once you represent the number in decimal, a string. So the first step is to convert the number to a string if it's not already.
my $s = sprintf("%.10f", $n);
(The solution is suppose to work with the OP's inputs, and his inputs appear to have 10 decimal places. If you want more digits to appear, use the number of decimal places you want to appear instead of 10. I thought this was obvious. If you want to be ridiculous like #asjo, use 324 decimal places for the doubles if you want to make sure not to lose any precision you didn't already lose.)
Then you can delete the trailing zeroes.
$s =~ s/0+\z// if $s =~ /\./;
$s =~ s/\.\z//;
or
$s =~ s/\..*?\K0+\z//;
$s =~ s/\.\z//;
or
$s =~ s/\.(?:|.*[^0]\K)0*\z//;
To avoid scientific notation for numbers use the format conversion %f instead of %g.
A lazy way could be simply: $number=~s/0+$// (substitute trailing zeroes by nothing).
The solution is easier than you might think.
Instead of using %g use %f and it will result in the behavior you are looking for. %f will always output your floating decimal in "fixed decimal notation".
What does the documentation say about %g vs %f?
As you may notice in the below table %g will result in either the same as %f or %e (when appropriate).
Ff you'd want to force the use of fixed decimal notation use the appropriate format identifier, which in this case is %f.
sprintf - perldoc.perl.org
%% a percent sign
%c a character with the given number
%s a string
%d a signed integer, in decimal
%u an unsigned integer, in decimal
%o an unsigned integer, in octal
%x an unsigned integer, in hexadecimal
%e a floating-point number, in scientific notation
%f a floating-point number, in fixed decimal notation
%g a floating-point number, in %e or %f notation
What about TIMTOWTDI; aren't we writing perl?
Yes, as always there are more than one ways of doing it.
If you'd just like to trim the trailing decimal-point zeros from a string you could use a regular expression such as the below.
$number = "123000.321000";
$number =~ s/(\.\d+?)0+$/$1/;
$number # is now "12300.321"
Remember that floating point values in perl doesn't have trailing decimals, unless you are dealing with a string. With that said; a string is not a number, even though it can explicitly and implicitly be converted to one.
The simplest way is probably to multiply by 1.
Original:
my $num = sprintf("%.10f", 0.000234000001234);
print($num);
#output
0.0002340000
With multiplying:
my $num = sprintf("%.10f", 0.000234000001234) * 1;
print($num);
#output
0.000234
The whole point of the %g format is to use a fixed point notation when it is reasonable and to use exponential notation when fixed point is not reasonable. So, you need to know the range of values you'll be dealing with.
Clearly, you could write a regular expression to post-process the string from sprintf(), removing the trailing zeroes:
my $str = sprintf("%g", $number);
$str =~ s/0+$//;
If you always want a fixed point number, use '%f', possibly with number of decimal places that you want; you might still need to remove trailing zeroes. If you always want exponential notation, use '%e'.
An easy way:
You could cheat a little. Add 1 to avoid the number breaking into scientific notation. Then manipulate the number as a string (thereby making perl convert it into a string).
$n++;
$n =~ s/^(\d+)(?=\.)/$1 - 1/e;
print $n;
A "proper" way:
For a more "proper" solution, counting the number of decimal places to use with %f would be optimal. It turned out getting the correct number of decimal points is trickier than one would think. Here's an attempt:
use strict;
use warnings;
use v5.10;
my $n = 0.000000234;
say "Original: $n";
my $l = getlen($n);
printf "New : %.${l}f\n", $n;
sub getlen {
my $num = shift;
my $dec = 0;
return 0 unless $num; # no 0 values allowed
return 0 if $num >= 1; # values above 1 don't need this computation
while ($num < 1) {
$num *= 10;
$dec++;
}
$num =~ s/\d+\.?//; # must have \.? to accommodate e.g. 0.01
$dec += length $num;
return $dec;
}
Output:
Original: 2.34e-007
New : 0.000000234
The value can have trailing zeroes only if it is a string.
You can add 0 to it. It will be coverted to a numerical value, not showing any trailing zeroes.
the following perl code converts a float number to the wrong integer number
use strict;
my $zahl =297607.22000;
$zahl=$zahl * 100;
print "$zahl\n";
my $text=sprintf ("%017d",$zahl);
print $text;
The output of this is :
29760722
00000000029760721
The thing is, you can change the given number to other numbers and it works.
Any idea what is wrong here or does Perl simply do it wrong?
Thanks for your help!
This is related to a FAQ (Why am I getting long decimals). $zahl is not rounded properly, it is rounded down to the next lower integer.
22/100 is a periodic number in binary just like 1/3 is a periodic number in decimal. It would take infinite storage to store it exactly in a floating point number.
$ perl -e'$_="297607.22000"; $_*=100; printf "%.20f\n", $_'
29760721.99999999627470970154
int and sprintf %d truncate decimals, so you end up with 29760721. print and sprintf %f round, so you can get the desired result.
$ perl -e'$_="297607.22000"; $_*=100; printf "%017.0f\n", $_'
00000000029760722
When you are doing your floating point multiplication by 100 the result will be something like 29760721.9999999963. Then when you do the %d conversion to an integer this is truncated to 29760721.
Try sprintf('%.10f', $zahl) and you should be able to see this.
You have to be really careful with floating point numbers and treating them as fixed point. Due to various conversions that may take place in the builtins, there may be times where one integer conversion is not exactly the same as another. It appears that this happens many times with x.22 numbers:
use strict;
my $n = 0;
for (0 .. 10_000_000) {
my $float = 100 * "$_.22";
my $str = "$float";
my $int = int $float;
if ($str ne $int) {
$n++;
#say "$float, $str, $int";
}
}
say "n = $n";
which prints
n = 76269
on my system.
A careful look at the Perl source would be required to see where the exact conversion difference is.
I would recommend that if you are going to be working with fixed point numbers, to convert them all to integers (using a common conversion function, preferably looking at the source numbers as strings), and then work with them all under the use integer; pragma which will disable floating point numbers.