Derek Basch wrote: > Given a value (x) that is within the range (1e-1, 1e7) how do I round > (x) up to the closest exact logarithmic decade? For instance: > > 10**3 = 1000 > x = 4978 > 10**4 = 10000 > x = 10000
how about
>>> import math
>>> def roundup(x):
... return 10**math.ceil(math.log10(x))
...
>>> roundup(4978)
10000.0
?
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