Dick Moores wrote:
> Here's a function I wrote some time ago, and just discovered that in
> one important category of cases, long numbers with a decimal point,
> it doesn't do what I intended.
>
> =====================================================
> def numberRounding(n, significantDigits=4):
> """
> Rounds a number (float or integer, negative or positive) to any number
> of
> significant digits. If an integer, there is no limitation on it's size.
> """
> import decimal
> def d(x):
> return decimal.Decimal(str(x))
> decimal.getcontext().prec = significantDigits
> return d(n)/1
> ======================================================
>
> Now, print
> numberRounding(232.3452345230987987098709879087098709870987098745234,
> 30) prints
> 232.345234523
The problem is that
232.3452345230987987098709879087098709870987098745234 is a float which
cannot represent this number exactly. Just typing it at the interpreter
prompt shows the problem:
>>> 232.3452345230987987098709879087098709870987098745234
232.34523452309881
>>> str(_)
'232.345234523'
So the precision you want is lost immediately when the constant is created.
>
> whereas if the first argument is enclosed in quotes, it does what I
> indended. Thus:
> print
> numberRounding('232.3452345230987987098709879087098709870987098745234',
> 30) prints
> 232.345234523098798709870987909 .
>
> So my question is, how can I revise numberRounding() so that it is
> not necessary to employ the quotes.
You can't. A float simply can't represent the number you want and the
function has no way to access the textual representation of the number.
> Or alternatively, is there a way
> to non-manually put quotes around an argument that is a long decimal?
No
Kent
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