How to convert bytearray into integer?
jacky.chao.wang at gmail.com
Tue Aug 17 02:57:45 CEST 2010
Thanks for your reply. Agree and I'll use your suggestions. Thanks!
On Aug 17, 3:36 am, Mark Dickinson <dicki... at gmail.com> wrote:
> On Aug 16, 8:08 pm, Jacky <jacky.chao.w... at gmail.com> wrote:
> > Hi Thomas,
> > Thanks for your comments! Please check mine inline.
> > On Aug 17, 1:50 am, Thomas Jollans <tho... at jollybox.de> wrote:
> > > On Monday 16 August 2010, it occurred to Jacky to exclaim:
> > > > Hi there,
> > > > Recently I'm facing a problem to convert 4 bytes on an bytearray into
> > > > an 32-bit integer. So far as I can see, there're 3 ways:
> > > > a) using struct module,
> > > Yes, that's what it's for, and that's what you should be using.
> > My concern is that struct may need to parse the format string,
> > construct the list, and de-reference index=0 for this generated list
> > to get the int out.
> > There should be some way more efficient?
> Well, you can improve on the struct solution by using the
> struct.Struct class to avoid parsing the format string repeatedly:
> >>> import struct
> >>> S = struct.Struct('<I')
> >>> S.unpack_from(buffer(bytearray([1,2,3,4,5])))
> This doesn't make a huge difference on my machine (OS X 10.6.4, 64-bit
> build of Python 2.6) though; it's probably more effective for long
> format strings. Adding:
> def test_struct2(buf, offset, S=struct.Struct('<I')):
> return S.unpack_from(buf, offset)
> to your test code, I see a speedup of around 8% over your test_struct.
> By the way, you may want to consider using an explicit byte-order/size
> marker in your format string; i.e., use '<I' instead of 'I'. This
> forces a 4-byte little-endian interpretation, regardless of the
> platform you're running Python on.
> Mark- Hide quoted text -
> - Show quoted text -
More information about the Python-list