> This is all fun, but what about the context? Your original function
> took an integer, not a string, and thus wasn't charged with measuring
> string length, possibly multiple times. Even so, each of these tests is
> taking around a microsecond. So are you expecting to do anything with
> t
On 10/05/2012 07:23 AM, Albert-Jan Roskam wrote:
> - Original Message -
>
>> From: Asokan Pichai
>> To: tutor@python.org
>> Cc:
>> Sent: Friday, October 5, 2012 11:06 AM
>> Subject: Re: [Tutor] rounding up to the nearest multiple of 8
>>
>>
- Original Message -
> From: Asokan Pichai
> To: tutor@python.org
> Cc:
> Sent: Friday, October 5, 2012 11:06 AM
> Subject: Re: [Tutor] rounding up to the nearest multiple of 8
>
> If you are doing so many times, it may be worth precomputing the padding for a
>
On Fri, Oct 5, 2012 at 4:24 AM, Albert-Jan Roskam wrote:
>
> import timeit
> ver1 = timeit.timeit("""
> import math
> value = "1234"
> value = "%-*s" % (int(math.ceil(len(value)/8.0)*8), value)
> """)
> ver2 = timeit.timeit("""
> value = "1234"
> value = value.ljust( len(value) + (-len(value) % 8)
If you are doing so many times, it may be worth precomputing the padding for a
given length and adding it by a look up on the length.
For example:
SPACE = ' '
MAX = 1000
TAB = 8
paddding = [ SPACE * (n % TAB) for n in range(MAX) ]
.
s = padding[len(s)] + s
.
-
On Fri, Oct 5, 2012 at 2:54 AM, Steven D'Aprano wrote:
>
> py> from __future__ import division
> py> from math import ceil
> py> "%*s" % (int(ceil(len(mystring)/8)*8), mystring)
> '123412341234'
>
>
> Or left-justified:
>
> py> "%-*s" % (int(ceil(len(mystring)/8)*8), mystring)
> '123412341234
- Original Message -
> From: Steven D'Aprano
> To: tutor@python.org
> Cc:
> Sent: Friday, October 5, 2012 8:54 AM
> Subject: Re: [Tutor] rounding up to the nearest multiple of 8
>
> On Thu, Oct 04, 2012 at 05:26:13PM -0400, eryksun wrote:
>> On T
On Thu, Oct 04, 2012 at 05:26:13PM -0400, eryksun wrote:
> On Thu, Oct 4, 2012 at 4:04 PM, Joel Goldstick
> wrote:
> >
> my_string = "123"
> pad = 8 - len(my_string) % 8
> my_string = my_string + " " * pad
> my_string
> > '123 '
>
> If len(my_string) is already a multiple
On Thu, Oct 4, 2012 at 4:04 PM, Joel Goldstick wrote:
>
my_string = "123"
pad = 8 - len(my_string) % 8
my_string = my_string + " " * pad
my_string
> '123 '
If len(my_string) is already a multiple of 8, the above sets pad to 8:
>>> s = "12345678"
>>> pad = 8 - len(
On Thu, Oct 4, 2012 at 3:47 PM, Albert-Jan Roskam wrote:
> Hi,
>
> The function below works, but it's such a kludge! Is there a way to make this
> more elegant?
> As said in the docstring, I want to round up a given integer to the nearest
> multiple of 8. I was thinking
> of something like math.
Hi,
The function below works, but it's such a kludge! Is there a way to make this
more elegant?
As said in the docstring, I want to round up a given integer to the nearest
multiple of 8. I was thinking
of something like math.ceil.
def _getPadding(charlen):
""" Helper function to replace n
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