# How to detect a double's significant digits

Discussion in 'Python' started by mrstephengross, May 5, 2005.

1. ### mrstephengrossGuest

Hi all... How can I find out the number of significant digits (to the
right of the decimal place, that is) in a double? At least, I *think*
that's what I'm asking for. For instance:

0.103 --> 3
0.0103 --> 4
0.00103 --> 5
0.000103 --> 6
0.0000103 --> 7

--Steve ()

mrstephengross, May 5, 2005

2. ### James StroudGuest

Significant digits are an accounting concept. As such, it is up to the
accountant to keep track of these as only she knows the precision of her
measurements.

Koan for the day:

What are the significant digits of 0.1?

Hint:

>>> 0.1

James

On Thursday 05 May 2005 10:37 am, so sayeth mrstephengross:
> Hi all... How can I find out the number of significant digits (to the
> right of the decimal place, that is) in a double? At least, I *think*
> that's what I'm asking for. For instance:
>
> 0.103 --> 3
> 0.0103 --> 4
> 0.00103 --> 5
> 0.000103 --> 6
> 0.0000103 --> 7
>
> --Steve ()

--
James Stroud, Ph.D.
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095

http://www.jamesstroud.com/

James Stroud, May 5, 2005

3. ### mrstephengrossGuest

So how can I get the kind of information I want then?

For example:

0.103 --> 3
0.0103 --> 4
0.00103 --> 5
0.000103 --> 6
0.0000103 --> 7

Any ideas?
--Steve

mrstephengross, May 5, 2005
4. ### philGuest

fl = 1.0002
x = str(fl)
pos = x.find('.')
print len( x[pos+1:] )
>>> 4

mrstephengross wrote:

> Hi all... How can I find out the number of significant digits (to the
> right of the decimal place, that is) in a double? At least, I *think*
> that's what I'm asking for. For instance:
>
> 0.103 --> 3
> 0.0103 --> 4
> 0.00103 --> 5
> 0.000103 --> 6
> 0.0000103 --> 7
>
> --Steve ()
>
>

phil, May 5, 2005
5. ### mrstephengrossGuest

Ok, that won't work. First of all, str() is not a function. If I want
to convert the float into a string, the conversion function will have
to use some kind of numeric precision, which will screw things up.
Consider this:

float f = 1.004;
ostringstream s;
s << f;
cout << s.str();

The above code may produce "1.004", or "1.0040", or "1.00400",
depending on the stream's precision setting. I need a way to detect the
number of digits to the right of decimal point *prior* to doing any
kind of string conversion.

--Steve

mrstephengross, May 5, 2005
6. ### Steven BethardGuest

mrstephengross wrote:
> So how can I get the kind of information I want then?
>
> For example:
>
> 0.103 --> 3
> 0.0103 --> 4
> 0.00103 --> 5
> 0.000103 --> 6
> 0.0000103 --> 7

Beware that this is probably only relevant if you have your numbers as
strings, not as floats:

py> 0.103
0.10299999999999999

But, assuming you have your numbers as strings, I would suggest looking
at str.split() and len(). I'd give you an example, but this sounds
kinda like a homework assignment.

STeVe

Steven Bethard, May 5, 2005
7. ### Steven BethardGuest

mrstephengross wrote:
> First of all, str() is not a function.

Yes it is.

> float f = 1.004;
> ostringstream s;
> s << f;
> cout << s.str();

This doesn't look like Python to me. Are you sure you're on the right
newsgroup?

STeVe

Steven Bethard, May 5, 2005
8. ### mrstephengrossGuest

>But, assuming you have your numbers as strings, I would suggest
looking
at str.split() and len().

Well, the numbers are in fact stored as numbers, so string processing
won't work.

>I'd give you an example, but this sounds kinda like a homework

assignment.

The task may sound like it comes from class, but I can assure you that
I am indeed a professional developer. I'm doing some rather intricate
text processing / rendering stuff these days, and C++ is unfortunately
none too handy for that sort of thing. Unfortunately, I have to use it

Thanks,
--Steve

mrstephengross, May 5, 2005
9. ### Steven BethardGuest

mrstephengross wrote:
> Well, the numbers are in fact stored as numbers, so string processing
> won't work.

What kind of numbers? Python floats?

STeVe

Steven Bethard, May 5, 2005
10. ### Steven BethardGuest

mrstephengross wrote:
>>But, assuming you have your numbers as strings, I would suggest

>
> looking
> at str.split() and len().
>
> Well, the numbers are in fact stored as numbers, so string processing
> won't work.

py> def digits(f):
.... return len(str(f).split('.')[1].rstrip('0'))
....
py> for f in [0.103, 0.1030, 0.0103, 0.010300]:
.... print f, digits(f)
....
0.103 3
0.103 3
0.0103 4
0.0103 4

I believe the rstrip is unnecessary because I think str() will never
produce additional following zeros, but you can guarantee it by calling
rstrip if you want.

Note that, for example, 0.103 and 0.1030 are identical as far as Python
is concerned, so I hope you're not hoping to show a difference between
these two...

STeVe

Steven Bethard, May 5, 2005
11. ### Charles KrugGuest

On 5 May 2005 10:37:00 -0700, mrstephengross <> wrote:
> Hi all... How can I find out the number of significant digits (to the
> right of the decimal place, that is) in a double? At least, I *think*
> that's what I'm asking for. For instance:
>
> 0.103 --> 3
> 0.0103 --> 4
> 0.00103 --> 5
> 0.000103 --> 6
> 0.0000103 --> 7
>
> --Steve ()
>

I would say that each of these examples has three signficant figures.
Each of them can be expressed as:

1.03e+n

For any integer n.

The fact that you've only shown the cases where n \in {-1, -2, -3, -4,
-5 . . } doesn't change the generality of the answer.

Charles Krug, May 5, 2005
12. ### Grant EdwardsGuest

On 2005-05-05, mrstephengross <> wrote:
>>But, assuming you have your numbers as strings, I would suggest

> looking
> at str.split() and len().
>
> Well, the numbers are in fact stored as numbers,

Then your question is in fact meaningless. The related
question that can be answered is "where is the least
significant '1' bit in the IEEE representation". If that's

> so string processing won't work.

>>I'd give you an example, but this sounds kinda like a homework
>> assignment.

>
> The task may sound like it comes from class, but I can assure
> you that I am indeed a professional developer. I'm doing some
> rather intricate text processing / rendering stuff these days,
> and C++ is unfortunately none too handy for that sort of
> thing. Unfortunately, I have to use it for the task.

--
Grant Edwards grante Yow! Hmmm... A hash-singer
at and a cross-eyed guy were
visi.com SLEEPING on a deserted
island, when...

Grant Edwards, May 5, 2005
13. ### mrstephengrossGuest

>This doesn't look like Python to me. Are you sure you're on the right
newsgroup?

Er, ok, I'm an idiot. This was all supposed to be on comp.lang.c++, but
obviously I posted on the wrong one. Sorry for all the hassle. In
python, this stuff is a heck of a lot easier.

--Steve

mrstephengross, May 5, 2005
14. ### philGuest

Bollocks, works here.

That looks like Java!!! Aaaihh!

mrstephengross wrote:

> Ok, that won't work. First of all, str() is not a function. If I want
> to convert the float into a string, the conversion function will have
> to use some kind of numeric precision, which will screw things up.
> Consider this:
>
> float f = 1.004;
> ostringstream s;
> s << f;
> cout << s.str();
>
> The above code may produce "1.004", or "1.0040", or "1.00400",
> depending on the stream's precision setting. I need a way to detect the
> number of digits to the right of decimal point *prior* to doing any
> kind of string conversion.
>
> --Steve
>
>

phil, May 5, 2005
15. ### Peter OttenGuest

mrstephengross wrote:

> This was all supposed to be on comp.lang.c++, but

http://mail.python.org/pipermail/python-dev/2004-March/043703.html

http://www.netlib.org/fp/

"""
file g_fmt.c
by David Gay
for ANSI C or C++ source for function g_fmt(char *, double):
, with help from dtoa, g_fmt(buf, x) sets buf to the shortest
, decimal string that correctly rounds to x and returns buf.
"""

Peter

Peter Otten, May 5, 2005
16. ### Fredrik LundhGuest

"mrstephengross" wrote:

> >But, assuming you have your numbers as strings, I would suggest

> looking
> at str.split() and len().
>
> Well, the numbers are in fact stored as numbers, so string processing
> won't work.

if they're not strings, your question is meaningless. as others have
pointed out, the exact internal representation for 0.103 is more like
0.10299999999999999433786257441170164383947849273681640625
which has a lot more than 3 digits...

> >I'd give you an example, but this sounds kinda like a homework

> assignment.
>
> The task may sound like it comes from class, but I can assure you that
> I am indeed a professional developer.

well professional or not, you clearly need to refresh your floating point
skills. I suggest reading the following documents before proceeding:

http://docs.python.org/tut/node16.html
http://www.lahey.com/float.htm

</F>

Fredrik Lundh, May 5, 2005
17. ### Jeremy BowersGuest

On Thu, 05 May 2005 18:42:17 +0000, Charles Krug wrote:

> On 5 May 2005 10:37:00 -0700, mrstephengross <>
> wrote:
>> Hi all... How can I find out the number of significant digits (to the
>> right of the decimal place, that is) in a double? At least, I *think*
>> that's what I'm asking for. For instance:
>>
>> 0.103 --> 3
>> 0.0103 --> 4
>> 0.00103 --> 5
>> 0.000103 --> 6
>> 0.0000103 --> 7
>>
>> --Steve ()
>>
>>

> I would say that each of these examples has three signficant figures. Each
> of them can be expressed as:
>
> 1.03e+n
>
> For any integer n.

You beat me to it.

Step one for mrstephengross is to *rigorously* define what he means by
"significant digits", then go from there. Since I think he mentioned
something about predicting how much space it will take to print out, my
suggestion is to run through whatever printing routines there are and get
a string out, the measure the string, as anything else will likely be
wrong. If that's not possible with the formatting library, you've already
lost; you'll have to completely correctly re-implement the formatting
library, and not only is that a major PITA, you almost never get it
bug-for-bug right...

Jeremy Bowers, May 5, 2005
18. ### Erik Max FrancisGuest

Jeremy Bowers wrote:

> Step one for mrstephengross is to *rigorously* define what he means by
> "significant digits", then go from there. Since I think he mentioned
> something about predicting how much space it will take to print out, my
> suggestion is to run through whatever printing routines there are and get
> a string out, the measure the string, as anything else will likely be
> wrong. If that's not possible with the formatting library, you've already
> lost; you'll have to completely correctly re-implement the formatting
> library, and not only is that a major PITA, you almost never get it
> bug-for-bug right...

Especially since all of his examples have the same number of significant
digits (3), as the term is usually meant. Zeroes to the right are
significant, not zeroes to the left.

--
Erik Max Francis && && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
What is it that shapes a species?
-- Louis Wu

Erik Max Francis, May 5, 2005
19. ### Grant EdwardsGuest

On 2005-05-05, Jeremy Bowers <> wrote:

> Since I think he mentioned something about predicting how much
> space it will take to print out, my suggestion is to run
> through whatever printing routines there are and get a string
> out,

A step which will require him to tell the printing routine how
many digits he wants printed.

--
Grant Edwards grante Yow! FUN is never having
at to say you're SUSHI!!
visi.com

Grant Edwards, May 6, 2005
20. ### Grant EdwardsGuest

On 2005-05-05, Erik Max Francis <> wrote:

> Especially since all of his examples have the same number of
> significant digits (3), as the term is usually meant. Zeroes
> to the right are significant, not zeroes to the left.

And only the person who performed the measurement knows how
many of the zeros to the right are significant.

--
Grant Edwards grante Yow! They don't hire