Newbie lost(new info)

[John Roth]

"Angelo Secchi" <secchi at sssup.it> wrote in message
news:mailman.150.1077786326.8594.python-list at python.org...
>
> John and Anthon thanks for your help.
> If I did it correctly this is the outcome of the code you asked me to
> try:
>
> >from binascii import hexlify
> >inf = file('foo','rb')
> >data = inf.read(1970) # 1970 is the exact length of the line
> >for i in range(4):
> >     index = i * 32
> >     print hexlify(data[index: index+32])
>
>
> 3131313131313131312030313032323230313033343130323238353332303031
> 3220202020203032323835333230303132303030303031303030303030314338
> 3930303030303120203939333531302030313130303030312020313220313335
> 313030313032323230313033343133343146ee3bb40000000000000000465de3
>
>
>
> >from binascii import hexlify
> >inf = file('foo','rb')
> >data = inf.read(1970)
> >for i in range(223):
> >     index = i * 4 + 113
> >     print hexlify(data[index: index + 4])
>
>
> 46ee3bb4 (I know that this should be 15612852)
> 00000000 (I know that this should be 0)
> 00000000 (I know that this should be 0)
> 465de39a (I know that this should be 6153114)
> 00000000 (I know that this should be 0)
> 00000000 (I know that this should be 0)
> ...
>
>
> John just to understand what you said, if the file is in IBM propietary
> binary format (EBCDIC ?) I cannot convert it in ASCII using Python?

The character data seems to be in ASCII, so it won't be any
problem.

I'm going to have to go onto
the IBM site later today to check the floating point formats; someone who
knows IEEE-488 should be able to tell you whether or not those match
standard formats, though.

John Roth
>
> Thanks again
> Angelo
>
>
>
>
>
>
>
> On Wed, 25 Feb 2004 20:18:47 -0500
> "John Roth" <newsgroups at jhrothjr.com> wrote:
>
> >
> > "Anton Vredegoor" <anton at vredegoor.doge.nl> wrote in message
> > news:403d3b33$0$11500$3a628fcd at reader2.nntp.hccnet.nl...
> > > "John Roth" <newsgroups at jhrothjr.com> wrote:
> > >
> > > >For the rest of it, I'd like to see a *real* hex dump in mainframe
> > > >format. From what you've given us so far I'm not certain whether
> > > >the struct module can convert the data for you.
> > >
> > > Probably what John wants to see is the output of something like
> > > this:
> > >
> > > #one 'line' of data (since it's a binary file there is no real line
> > > #ending convention: a line is just a specific number of bytes long,
> > > #it's important to find out how many exactly)
> > >
> > > from binascii import hexlify
> > > inf = file('somefile','rb')
> > > data = inf.read(1005) #a 113 bytes string + 232 4 bytes floats
> > > L = map(hexlify,data)
> > > print L
> > >
> > > Anton
> >
> > Thank you! I wasn't aware of that module. It looks like it
> > should do exactly what's needed.
> >
> > Although this would probably do instead of the last 2 lines
> > (and excuse the fact that it's real ugly code, as well as
> > untested)
> >
> > for i in range(4):
> >     index = i * 32
> >     print hexlify(data[index: index+32])
> > for i in range(232):
> >     index = i * 4 + 113
> >     print hexlify(data[index: index + 4])
> >
> > John Roth
> > >
> > >
> >
> >
> > -- 
> > http://mail.python.org/mailman/listinfo/python-list
>
>
> --
> ========================================================
>  Angelo Secchi                     PGP Key ID:EA280337
> ========================================================
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