Rounding up to the nearest exact logarithmic decade

[Grant Edwards]

On 2006-02-28, johnzenger at gmail.com <johnzenger at gmail.com> wrote:
> I like Fredrik's solution.  If for some reason you are afraid of
> logarithms, you could also do:
>
>>>> x = 4978
>>>> decades = [10 ** n for n in xrange(-1,8)]
>>>> import itertools
>>>> itertools.ifilter(lambda decade: x < decade, decades).next()
> 10000
>
> BTW, are the python-dev guys aware that 10 ** -1 = 0.10000000000000001 ?

You're joking, right?

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