On 2006-02-28, [EMAIL PROTECTED] <[EMAIL PROTECTED]> 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 ?
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