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Subject: Re: Python, convert an integer into an index?
To: python-list@python.org
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From: Lorenzo Sutton <lorenzofsutton@gmail.com>
Date: Thu, 24 Sep 2015 18:13:14 +0200
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On 23/09/2015 17:32, Denis McMahon wrote:
> On Tue, 22 Sep 2015 14:43:55 -0700, Chris Roberts wrote:
>
>> results = 134523      #(Integer)
>
> This appears to be an integer expressed (presumably) in base 10 with 6
> digits
>
>> Desired:
>> results = [1, 2, 3, 4, 5, 2, 3]   #(INDEX)
>
> This appears to be a python list of 7 elements, with the first and the
> the third through seventh elements corresponding to the first and the
> second through sixth most significant digits respectively of the
> previously discussed integer.
>
> I can't actually see any direct method of creating the list given from
> the number given.
>
> However, if I understand the intent of the question you meant to ask, you
> might find that the following code does something interesting:
>
> x = 9876543210
> y = []
>
> while x > 0:
>      y.append(x % 10)
>      x = int(x / 10)
>
> y = list(reversed(y))
> print y

I like the math approach even if the pythonic list string is quicker...

One 'math' way would also be (avoiding the list reverse, but need to 
import math):

 >>> import math
 >>> result = 1234567
 >>> digits = int(math.log10(result) + 1)
 >>> y = []
 >>> for x in range(digits, 0, -1):
	number = result % (10 ** x) / (10 **(x-1))
	y.append(int(number))

 >>> y
[1, 2, 3, 4, 5, 6, 7]