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On 05/02/15 18:46, DaveA wrote:
> You don't need much class understanding at all for this. Something
> like:
>
> class Job: def __init__(self, retain, srcpath, suffix, syncpath,
> snappath, ...): self.retain = retain = datetime.timedelta (days =
> r
On February 5, 2015 12:33:56 PM EST, Bob Williams
wrote:
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>
>On 05/02/15 14:59, DaveA wrote:
>>
>>
Sorry my last message was in html, and indentation trashed . I just installed
k9 on my tablet and hadn't yet found the setting for text-only. Le
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On 05/02/15 14:59, DaveA wrote:
>
>
> On February 5, 2015 8:27:29 AM EST, Bob Williams
> wrote:
>> Hi,
>>
>> My script is running under Python 3.4.1 on a 64bit openSUSE
>> linux system. It is a backup script making calls to rsync and
>> btrfs-tools
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On 05/02/15 13:57, Mark Lawrence wrote:
> On 05/02/2015 13:27, Bob Williams wrote:
>
>>
>> I would like to reduce all those repeated calls to do_sync() in
>> main(), for example, to one by putting the *_srcpath and
>> *_*syncpath variables into list
On February 5, 2015 8:27:29 AM EST, Bob Williams
wrote:
>Hi,
>
>My script is running under Python 3.4.1 on a 64bit openSUSE linux
>system. It is a backup script making calls to rsync and btrfs-tools,
>and backing up several different paths. Here is the script, my question
>follows below:
>
>**C
On 05/02/2015 13:27, Bob Williams wrote:
I would like to reduce all those repeated calls to do_sync() in main(), for
example, to one by putting the *_srcpath and *_*syncpath variables into lists
(eg. source_list and sync_list) and using a for loop to get the first item out
of each list, then
Hi,
My script is running under Python 3.4.1 on a 64bit openSUSE linux system. It is
a backup script making calls to rsync and btrfs-tools, and backing up several
different paths. Here is the script, my question follows below:
**Code**
import datetime
import glob
import os, os.path
import subpro
> Do you understand how that works?
Yep. It's crystal clear now. Thank you.
It took a while till I got it, though ;-)
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On Wed, 27 Nov 2013 17:08:11 +0100, Rafael Knuth
wrote:
for x in range(2, n):
if n % x == 0:
print(n, 'equals', x, '*', n//x)
#3 Round:
n = 4
x = 3
When n is 4, the inner loop will test 2, then 3. But since n% x is
zero for values 4 and 2, it will print, then b
Rafael Knuth wrote:
>I am trying to figure out how exactly variables in nested loops are
>generated, and don't get it 100% right yet. Here's my code:
Maybe it's easier if you look at a simpler example like:
for i in range(4):
for j in range(4):
print("i: {}, j: {}".format(i, j))
Do
On Wed, Nov 27, 2013 at 11:08 AM, Rafael Knuth wrote:
> Hej there,
>
> I am trying to figure out how exactly variables in nested loops are
> generated, and don't get it 100% right yet. Here's my code:
>
> for n in range(2, 10):
> for x in range(2, n):
> if n % x == 0:
> pri
Hej there,
I am trying to figure out how exactly variables in nested loops are
generated, and don't get it 100% right yet. Here's my code:
for n in range(2, 10):
for x in range(2, n):
if n % x == 0:
print(n, 'equals', x, '*', n//x)
break
else:
print
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