On Fri, Jun 1, 2012 at 1:01 PM, krastanov.ste...@gmail.com
<krastanov.ste...@gmail.com> wrote:
> Not much better, however:
>
> array = np.ceil(1000*array)/1000
>
> On 1 June 2012 08:45, Bharath M R <catchmrbhar...@gmail.com> wrote:
>> Hi,
>> I was trying to implement interval arithmetic using numpy for plotting. I
>> could not possibly find a way
>> to **round up** a floating point value. ie sin(<1, 1>) = <0.841, 0.842>
>> rounded to 3 decimals. I would like
Defining the problem well is important. Do you really want to a) round
up, b) just need the interval, width=1/1000, that contains a given
point, or c) want the interval of width 1/1000 that has the point as
close to the center as possible?
The last routine I sent calculated b, this one calculates c:
def bounds(x, n):
m = 10**-n
a = x - m/2
return a, a+m
>>> bounds(sin(1).n(),3)
(0.840970984807897, 0.841970984807897)
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