On Jan 9, 2:41 am, Tim Chase [EMAIL PROTECTED] wrote:
I decided that I was just trying to be too smooth by 1/2 so I fell back to
messages = open(os.path.join(host_path,'messages.txt'), 'wb')
deliveries = open(os.path.join(host_path,'deliveries.txt'), 'wb')
actions =
Paul Hankin wrote:
This can be more cleanly written using locals()
for fn in filenames:
locals()[fn] = open(os.path.join(host_path, fname + '.txt', 'wb')
from the reference manual:
locals()
Update and return a dictionary representing the current
local symbol table.
On Jan 9, 10:02 am, Fredrik Lundh [EMAIL PROTECTED] wrote:
Paul Hankin wrote:
This can be more cleanly written using locals()
for fn in filenames:
locals()[fn] = open(os.path.join(host_path, fname + '.txt', 'wb')
from the reference manual:
locals()
Update and return a
Paul Hankin wrote:
Thanks Fredrik! I learnt something today.
I wonder if there's a reason why it doesn't raise an exception when
you try to write to it? That would seem better to me than having it
sometimes update variables and sometimes not.
probably because it returns a standard
Fredrik == Fredrik Lundh [EMAIL PROTECTED] writes:
Fredrik Seriously, for a limited number of files, the dictionary approach
Fredrik is mostly pointless; you end up replacing
Fredrik foo = open(foo)
Fredrik foo.write(...)
Fredrik with
Fredrik somedict[foo] = open(foo)
Fredrik
BJ == BJ Swope [EMAIL PROTECTED] writes:
I (at least) think the code looks much nicer.
BJ #Referring to files to write in various places...
BJ open_files['deliveries'].write(flat_line)
BJ open_files['deliveries'].write('\n')
If you were doing a lot with the deliveries file at some point, you
You don't need for fn in open_files.keys():, you can just use for fn in
open_files:, but simpler than that is to just use the dictionary values:
for fn in open_files.values():
fn.close()
This can also work for standard variable names:
for f in (messages, deliveries, actions, parts,
BJ Swope wrote:
given a list such as
['messages', 'recipients', 'viruses']
how would I iterate over the list and use the values as variables and
open the variable names a files?
I tried
for outfile in ['messages', 'recipients', 'viruses']:
filename = os.path.join(Host_Path,
On Jan 8, 2008 6:03 AM, Fredrik Lundh [EMAIL PROTECTED] wrote:
BJ Swope wrote:
given a list such as
['messages', 'recipients', 'viruses']
how would I iterate over the list and use the values as variables and
open the variable names a files?
I tried
for outfile in
On Jan 8, 10:03 pm, Fredrik Lundh [EMAIL PROTECTED] wrote:
BJ Swope wrote:
given a list such as
['messages', 'recipients', 'viruses']
how would I iterate over the list and use the values as variables and
open the variable names a files?
I tried
for outfile in ['messages',
BJ Swope wrote:
the code looks ok. please define not working.
Yep, defining not working is always helpful! :)
I want to have all 3 files open at the same time. I will write to each
of the files later in my script but just the last file is open for writing.
to keep more than one
On Jan 8, 2008 6:54 AM, BJ Swope [EMAIL PROTECTED] wrote:
given a list such as
['messages', 'recipients', 'viruses']
how would I iterate over the list and use the values as variables and
open the variable names a files?
I tried
for outfile in ['messages',
BJ Swope wrote:
On Jan 8, 2008 6:03 AM, Fredrik Lundh [EMAIL PROTECTED] wrote:
BJ Swope wrote:
given a list such as
['messages', 'recipients', 'viruses']
how would I iterate over the list and use the values as variables and
open the variable names a files?
I tried
for
On Jan 8, 1:03 pm, Fredrik Lundh [EMAIL PROTECTED] wrote:
BJ Swope wrote:
given a list such as
['messages', 'recipients', 'viruses']
how would I iterate over the list and use the values as variables and
open the variable names a files?
I tried
for outfile in ['messages',
Fredrik Lundh [EMAIL PROTECTED] writes:
BJ Swope wrote:
the code looks ok. please define not working.
Yep, defining not working is always helpful! :)
I want to have all 3 files open at the same time. I will write to
each of the files later in my script but just the last file is open
On Jan 8, 2008 7:22 AM, Hrvoje Niksic [EMAIL PROTECTED] wrote:
Fredrik Lundh [EMAIL PROTECTED] writes:
BJ Swope wrote:
the code looks ok. please define not working.
Yep, defining not working is always helpful! :)
I want to have all 3 files open at the same time. I will write
Hi BJ
Fredrik Lundh [EMAIL PROTECTED] writes:
Or in a dict:
open_files = {}
for fn in ['messages', 'recipients', 'viruses']:
open_files[fn] = open(getfilename(fn), 'w')
I decided that I was just trying to be too smooth by 1/2 so I fell back
to ...
messages =
I decided that I was just trying to be too smooth by 1/2 so I fell back to
messages = open(os.path.join(host_path,'messages.txt'), 'wb')
deliveries = open(os.path.join(host_path,'deliveries.txt'), 'wb')
actions = open(os.path.join(host_path,'actions.txt'), 'wb')
parts =
On Jan 8, 2008 9:34 PM, Terry Jones [EMAIL PROTECTED] wrote:
I think you should revisit this decision. Something like Fredrik's code
is
the way to go. It has multiple advantages:
- It's much shorter.
- It's arguably easier to add/remove to/from.
- It has less risk of error (much less
Terry Jones wrote:
I think you should revisit this decision. Something like Fredrik's code is
the way to go.
He used my suggestion, just for a few more files than he had in his
original post.
Seriously, for a limited number of files, the dictionary approach is
mostly pointless; you end up
given a list such as
['messages', 'recipients', 'viruses']
how would I iterate over the list and use the values as variables and open
the variable names a files?
I tried
for outfile in ['messages', 'recipients', 'viruses']:
filename = os.path.join(Host_Path, outfile)
outfile =
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