How do you "add 1" to last digit of a fraction such that

     0.0012 gives  0.0013

     0.99999 gives  1

Update: rephrased question

sub incdec {

  $number = shift;

  # do something if not /^\d*\.\d+$/

  return $number+1 if $number =~ /^\d+$/;

  $number =~ /\.(\d+)$/;

  return $number + ("1E-".length($1);
}

for $n (1, .004, 1.99, 1.79, 1.009999) {

  print "$n++ = ".incdec($n)."\n";
}
[download]

gives:

1++ = 2
0.004++ = 0.005
1.99++ = 2
1.79++ = 1.8
1.009999++ = 1.01
[download]

Update: use E notation

--
Steve Marvell

Thanks.

  (0.1 ** length($1)) could also be written:
 ("1E-" . length($1))

update: This is what I came up with:

$frac = shift;

$frac =~ s/(\d+).$/$1/;   # remove last digit & get fraction size
$frac +=  "1E-" . length($1);   # add 1E-(size) => 0.0001

print $frac;
[download]

Works for "0.123999", "10.12399" and ".12399"

The result is weird when the number does not have a fractional part, but that is not my case.

Don't reinvoke on the same variable...

(1++) ++ = 3
(0.004++) ++ = 0.006
[download]

But...

(1.99++) ++ = 3
(1.79++) ++ = 1.9
(1.009999++) ++ = 1.02
[download]

--
Tommy
Too stupid to live.
Too stubborn to die.

If not, why not just use sprintf

like perl -e 'printf("%0.5f",1.233394494955);
So what did I miss?

--

Brother Frankus.

¤

I want to round a number, keeping as many digits as possible, if the number ends in /999.$/ .

So I'd need something like a "%0.(max)f" format (maybe it exists?)

• an arbitrarily long string of zeroes and chop them off starting with the first one and
• an arbitrarily long strong of nines and chop them off starting with the first one, then add one to the end of the remaining number to make it round up.

You didn't say, was it necessary to retain the tiny amount that you adjusted by? My suggestion doesn't keep that information -- I assumed it was irrelevant.

--t. alex

"Mud, mud, glorious mud. Nothing quite like it for cooling the blood!"
--Michael Flanders and Donald Swann

was it necessary to retain the tiny amount that you adjusted by?

No, that is just some quick hack.

add one to the end

Is there an easy way to do that? (besides using substr and x)

Otherwise, I'm guessing this is just for aesthetics of presentation? I guess I'm just too steeped in scientific ambivalence when it comes to the beauty of numbers. I'd just pick either the number of sig figs you want to use, or the number of decimal places, and use sprintf.

This is not criticism -- I am curious as to what you want to do with your numbers.

Matt

-- Mike

--
just,my${.02}

Found a way to round it safely:

my $frac = shift;
if ($frac =~ /\..{5,}000.$/) {
    $frac =~ s/.$//;
}
elsif ($frac =~ /(\..{5,}999)(.)$/) {
    $frac = $frac + ( (10-$2) . "E-" . length($1));
}
print "rounded to $frac";
[download]

note: $frac can be an integer; $frac must be positive