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Numpy, uint64 and Big O notation.

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  Numpy, uint64 and Big O notation. marksabbath@gmail.com - 2014-10-09 06:08 -0700
    Re: Numpy, uint64 and Big O notation. Skip Montanaro <skip.montanaro@gmail.com> - 2014-10-09 10:39 -0500
    Re: Numpy, uint64 and Big O notation. Skip Montanaro <skip.montanaro@gmail.com> - 2014-10-09 11:31 -0500

#78891 — Numpy, uint64 and Big O notation.

Hey all!

I'm trying to find out the best way to multiply an uint64 (numpy). Could someone help me find the best way to achieve that and where can I find the time and space complexity in a Big O notation?

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#78898

On Thu, Oct 9, 2014 at 8:08 AM,  <marksabbath@gmail.com> wrote:

> I'm trying to find out the best way to multiply an uint64 (numpy). Could someone help me find the best way to achieve that and where can I find the time and space complexity in a Big O notation?

Multiply it by what? This works fine for me:

>>> import numpy
>>> x = numpy.uint64(57)
>>> x
57
>>> type(x)
<type 'numpy.uint64'>
>>> xx = x * x
>>> type(xx)
<type 'numpy.uint64'>
>>> xx
3249
>>> xxx = xx * xx
>>> xxx
10556001
>>> xxxx = xxx * xxx
>>> xxxx
111429157112001
>>> type(xxxx)
<type 'numpy.uint64'>
>>> xxxxx = xxxx * xxxx
>>> type(xxxxx)
<type 'numpy.uint64'>
>>> xxxxx
12238449225650938241

I assume under the covers there must be a Python long somewhere:

>>> int(xxxxx)
12238449225650938241L

I think your question needs to be more concrete.

Skip

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#78903

[Multipart message — attachments visible in raw view] — view raw

(For future reference, when responding to answers, it's worthwhile to
continue to cc python-list.)

On Thu, Oct 9, 2014 at 11:12 AM, Marcos Schratzenstaller <
marksabbath@gmail.com> wrote:
> The numpy has a function which manipulate 64 bits integers, but I couldn't
> find a specific method to multiplicate this kind of objects can not be
> manipulated as the same way, so if python is using the correct function
> (assuning python is detecting the object, calling the correct function) it
> is not a problem, but, where can I find this deep information?

Marcos,

Yes, the builtin help() function serves as reasonable documentation in this
case:

>>> help(numpy.int64)
Help on class int64 in module numpy:

class int64(signedinteger, __builtin__.int)
 |  64-bit integer. Character code 'l'. Python int compatible.
 |
 |  Method resolution order:
 |      int64
 |      signedinteger
 |      integer
 |      number
 |      generic
 |      __builtin__.int
 |      __builtin__.object
 |
 |  Methods defined here:
...
 |  Methods inherited from generic:
 |
 |  __abs__(...)
 |      x.__abs__() <==> abs(x)
 |
 |  __add__(...)
 |      x.__add__(y) <==> x+y
...
 |  __mod__(...)
 |      x.__mod__(y) <==> x%y
 |
 |  __mul__(...)
 |      x.__mul__(y) <==> x*y
...

The __mul__ method implements multiplication. I don't know the specific
implementation in numpy, but it's probably little more than a thin wrapper
around x * y (with perhaps some type and overflow checking).

Skip

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