Help me write better Code

[sssdevelop]

Thank you so much Terry Jan Reedy. You have given best advice - yup, i am beginner in Python. 
Your reply has done grooming :) 

thx,



On Thursday, July 10, 2014 12:16:48 AM UTC+5:30, Terry Reedy wrote:
> On 7/9/2014 10:27 AM, sssdevelop wrote:
> 
> > Hello,
> 
> >
> 
> > I have working code - but looking for better/improved code. Better coding practices, better algorithm :)
> 
> >
> 
> > Problem: Given sequence of increasing integers, print blocks of consecutive integers.
> 
> 
> 
> > Input: [51, 53, 55, 67, 68, 91, 92, 93, 94, 99]
> 
> > Outout: [67, 68], [91, 92, 93, 94]
> 
> 
> 
> Recommendations:
> 
> 1. If you are just beginning Python, use the current version, now 3.4.
> 
> 
> 
> 2. Separate interface code that gets input and presents output from the 
> 
> function that processes the increasing sequence. The function should not 
> 
> care whether the ints come from a user, file, or calculation.
> 
> 
> 
> 3. Think in terms of iterables and iterators rather than lists (this is 
> 
> clearer in 3.x, where some builtins have been converted). The function 
> 
> should not care what class is used to convey the sequence of numbers. 
> 
> This happens to make it easier to solve the 'pump-priming' problem you 
> 
> stumbled over.
> 
> 
> 
> 4. For designing a loop, what is the loop invariant that you want, that 
> 
> will make writing the code easy. For this problem, "tem is a non-emptly 
> 
> list of consecutive integers". Remember that a list of one int 
> 
> qualifies. Using for loops with the proper iterable solves the other 
> 
> part of the loop invariant: the current item is the next item to be 
> 
> compared to the last item of tem. If tem is always non-empty, that 
> 
> comparison is always possible.
> 
> 
> 
> 5. Remember that Python code is generic unless constrained. What should 
> 
> happen if the function gets non-int numbers, with or without an integer 
> 
> value? What should happen if the sequence is not increasing, but 
> 
> contains consecutive subsequences. For beginning code, one could decide 
> 
> to meet the spec given for input that meets the condition and not care 
> 
> otherwise.  The code below works for any sequence (even infinite) of 
> 
> objects that can be incremented by 1 and compared to the next.
> 
> 
> 
> 6. Write an automated test. For one test, something like this works.
> 
> 
> 
> ci = consec([51, 53, 55, 67, 68, 91, 92, 93, 94, 99])
> 
> print(next(ci) == [67, 68], next(ci) == [91, 92, 93, 94])
> 
> 
> 
> but since you (properly) noted several test cases
> 
> 
> 
> a = [51, 53, 55, 67, 68, 91, 92, 93, 94, 99]
> 
> #a = []
> 
> #a = [10]
> 
> #a = [10, 11, 12, 15]
> 
> 
> 
> I went ahead and used unittest, at the cost of three lines of 
> 
> 'boilerplate' code. I added a case with a final consecutive sequence. 
> 
> Good thing, because it initially failed because I initially forgot to 
> 
> check tem after the loop.
> 
> 
> 
> import unittest
> 
> 
> 
> def consec(iterable):
> 
>      "Yield blocks of consecutive integers as a list."
> 
>      it = iter(iterable)
> 
>      first = next(it)
> 
>      tem = [first]
> 
>      for n in it:
> 
>          # tem is a non-empty list of consecutive ints
> 
>          if n == tem[-1] + 1:
> 
>              tem.append(n)
> 
>          else:
> 
>              if len(tem) >= 2:
> 
>                  yield tem
> 
>              tem = [n]
> 
>      if len(tem) >= 2:
> 
>          yield tem
> 
> 
> 
> class Test(unittest.TestCase):
> 
>      def test_consec(self):
> 
>          def eq(seq, blocks):
> 
>              self.assertEqual(list(consec(seq)), blocks)
> 
>          eq((), [])
> 
>          eq([1], [])
> 
>          eq([1,2,3], [[1,2,3]])  # block at beginning or end
> 
>          eq([-1, 1,2,3, 5], [[1,2,3]])  # block in middle
> 
>          eq((-1, 1,2,3, 5, 7,8,9, 11), [[1,2,3], [7,8,9]])  # 2 blocks
> 
> 
> 
> unittest.main(verbosity=2)
> 
>  >>>
> 
> test_consec (__main__.Test) ... ok
> 
> 
> 
> ----------------------------------------------------------------------
> 
> Ran 1 test in 0.016s
> 
> 
> 
> OK
> 
> 
> 
> -- 
> 
> Terry Jan Reedy