Page 3 - String to Int conversion speed

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#31

November 9th, 2012, 07:55 PM

KING-OLE

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Quote:

Originally Posted by b49P23TIvg

The division part of your program might be faster using divmod. Then again, you might end up storing a bunch of really long unused quotients. Don't know.

Code:

>>> help(divmod)
Help on built-in function divmod in module __builtin__:

divmod(...)
    divmod(x, y) -> (quotient, remainder)
    
    Return the tuple ((x-x%y)/y, x%y).  Invariant: div*y + mod == x.

Thanks for the divmod tip, but in this case it won't help, as I don't want to i/p unless p is a factor of i, in which case there won't be a remainder.

#32

November 10th, 2012, 10:39 AM

KING-OLE

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Update:

Found a faster way of doing it.

Instead of creating strings and converting them to integers, I multiply the last integer to get the new:

Code:


  oldone=i
  bigone=10**(10**(n-1))

  for cnt in range (1,10):
    i*=bigone
    i+=oldone

Results:

Using the i=str('1'**n) solution takes about 12 minutes for 1 <= n <= 7:

Code:


07:53:01:65 - Start
08:05:00.80 - Finish

New solution, taking the previous i and multiplying it with 10**(10**(n-1)) and adding i back 9 times is a lot faster, taking only about 3 minutes and 5 seconds:

Code:


08:39:47:89 - Start
08:42:52.49 - Finish

Modifying the new solution to div i with the prime factor found, as well as searching the primes reversed (high to low) is about the same, about 1 second faster than the previous:

Code:


08:45:44.06 - Start
08:48:49.00 - Finish

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