Rounding up to the nearest exact logarithmic decade

[Derek Basch]

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

Thanks Everyone!
Derek Basch

 

[Fredrik Lundh]

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
... return 10**math.ceil(math.log10(x))
... 10000.0

?

</F>

 

[jao]

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:

How about this:

def roundup(x):
if x < 1:
return 1
else:
return '1' + ('0' * len(str(int(x))))

Jack Orenstein

 

[Felipe Almeida Lessa]

Em Ter, 2006-02-28 às 17:47 -0500, (e-mail address removed) escreveu:

How about this:

def roundup(x):
if x < 1:
return 1
else:
return '1' + ('0' * len(str(int(x))))

No dice. First, it returns an int for some cases and a string for the
others. Second, casting from str to int and vice-versa and concatenating
strings won't perform any good. I wouldn't like this hack on my code.

My 2¢,
Felipe.

--
"Quem excele em empregar a força militar subjulga os exércitos dos
outros povos sem travar batalha, toma cidades fortificadas dos outros
povos sem as atacar e destrói os estados dos outros povos sem lutas
prolongadas. Deve lutar sob o Céu com o propósito primordial da
'preservação'. Desse modo suas armas não se embotarão, e os ganhos
poderão ser preservados. Essa é a estratégia para planejar ofensivas."

-- Sun Tzu, em "A arte da guerra"

 

[Derek Basch]

Thanks effbot. I knew their had to be something buried in the math
module that could help. ceil() it is!

</dTb>

 

[Grant Edwards]

How about this:

def roundup(x):
if x < 1:
return 1
else:
return '1' + ('0' * len(str(int(x))))

Interesting approach. You're converting to string at the same
time. I would guess the OP wanted a float.

 

[johnzenger]

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 ?

 

[Grant Edwards]

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?

 

[Fredrik Lundh]

BTW, are the python-dev guys aware that 10 ** -1 = 0.10000000000000001 ?

http://docs.python.org/tut/node16.html

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