Re: [Tutor] Encoding
2010/3/7 spir denis.s...@gmail.com
Oh, right. And, if i'm not wrong B is an UTF8 string decoded to unicode
(due
to the coding: statement at the top of the file) and re-encoded to latin1
Si! :-)
Ahah. Ok, Grazie!
One more question: Amazon SimpleDB only accepts UTF8.
So, let's say i have to put into an image file:
filestream = file.read()
filetoput = filestream.encode('utf-8')
Do you think this is ok?
Oh, of course everything url-encoded then
Giorgio
Denis
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[Tutor] recursive generator
Hello,
Is it possible at all to have a recursive generator? I think at a iterator for
a recursive data structure (here, a trie). The following code does not work: it
only yields a single value. Like if child.__iter__() were never called.
def __iter__(self):
''' Iteration on (key,value) pairs. '''
print '*',
if self.holdsEntry:
yield (self.key,self.value)
for child in self.children:
print ,
child.__iter__()
print ,
raise StopIteration
With the debug prints in code above, for e in t: print e outputs:
* ('', 0)
print len(t.children) # -- 9
Why is no child.__iter__() executed at all? I imagine this can be caused by the
special feature of a generator recording current execution point. (But then, is
it at all possible to have a call in a generator? Or does the issue only appear
whan the callee is a generator itself?) Else, the must be an obvious error in
my code.
Denis
--
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spir.wikidot.com
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Re: [Tutor] recursive generator
On Sun, 7 Mar 2010 11:58:05 pm spir wrote: Hello, Is it possible at all to have a recursive generator? Yes. def recursive_generator(n): ... if n == 0: ... return ... yield n ... for i in recursive_generator(n-1): ... yield i ... it = recursive_generator(5) it.next() 5 it.next() 4 it.next() 3 it.next() 2 it.next() 1 it.next() Traceback (most recent call last): File stdin, line 1, in module StopIteration I think at a iterator for a recursive data structure (here, a trie). The following code does not work: it only yields a single value. Like if child.__iter__() were never called. def __iter__(self): ''' Iteration on (key,value) pairs. ''' print '*', if self.holdsEntry: yield (self.key,self.value) for child in self.children: print , child.__iter__() print , raise StopIteration __iter__ should be an ordinary function, not a generator. Something like this should work: # Untested. def __iter__(self): ''' Iteration on (key,value) pairs. ''' def inner(): print '*', # Side effects bad... if self.holdsEntry: yield (self.key,self.value) for child in self.children: print , child.__iter__() print , raise StopIteration return inner() This means that your class won't *itself* be an iterator, but calling iter() on it will return a generator object, which of course is an iterator. If you want to make your class an iterator itself, then you need to follow the iterator protocol. __iter__ must return the instance itself, and next must return (not yield) the next iteration. class MyIterator(object): def __init__(self, n): self.value = n def next(self): if self.value == 0: raise StopIteration self.value //= 2 return self.value def __iter__(self): return self See the discussion in the docs about the iterator protocol: http://docs.python.org/library/stdtypes.html#iterator-types Why is no child.__iter__() executed at all? I imagine this can be caused by the special feature of a generator recording current execution point. That's exactly what generators do: when they reach a yield, execution is halted, but the state of the generator is remembered. Then when you call the next() method, execution resumes. (But then, is it at all possible to have a call in a generator? Or does the issue only appear whan the callee is a generator itself?) Else, the must be an obvious error in my code. I don't understand what you mean. -- Steven D'Aprano ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Encoding
On Sun, 7 Mar 2010 13:23:12 +0100
Giorgio anothernetfel...@gmail.com wrote:
One more question: Amazon SimpleDB only accepts UTF8.
[...]
filestream = file.read()
filetoput = filestream.encode('utf-8')
No! What is the content of the file? Do you think it can be a pure python
representation of a unicode text?
uContent = inFile.read().decode(***format***)
process, if any
outFile.write(uContent.encode('utf-8'))
input --decode-- process --encode-- output
This gives me an idea: when working with unicode, it would be cool to have an
optional format parameter for file.read() and write. So, the above would be:
uContent = inFile.read(***format***)
process, if any
outFile.write(uContent, 'utf-8')
Or, maybe even better, the format could be given as third parameter of file
open(); then any read or write operation would directly convert from/to the
said format. What do you all think?
denis
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Re: [Tutor] Encoding
Or, maybe even better, the format could be given as third parameter of file
open(); then any read or write operation would directly convert from/to the
said format. What do you all think?
See the codecs.open() command as an alternative to open().
With all the hassles of encoding, I'm puzzled why anyone would use the
regular open() for anything but binary operations.
Malcolm
- Original message -
From: spir denis.s...@gmail.com
To: Python tutor tutor@python.org
Date: Sun, 7 Mar 2010 14:29:11 +0100
Subject: Re: [Tutor] Encoding
On Sun, 7 Mar 2010 13:23:12 +0100
Giorgio anothernetfel...@gmail.com wrote:
One more question: Amazon SimpleDB only accepts UTF8.
[...]
filestream = file.read()
filetoput = filestream.encode('utf-8')
No! What is the content of the file? Do you think it can be a pure
python representation of a unicode text?
uContent = inFile.read().decode(***format***)
process, if any
outFile.write(uContent.encode('utf-8'))
input --decode-- process --encode-- output
This gives me an idea: when working with unicode, it would be cool to
have an optional format parameter for file.read() and write. So, the
above would be:
uContent = inFile.read(***format***)
process, if any
outFile.write(uContent, 'utf-8')
Or, maybe even better, the format could be given as third parameter of
file open(); then any read or write operation would directly convert
from/to the said format. What do you all think?
denis
--
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spir.wikidot.com
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Re: [Tutor] recursive generator
Steven D'Aprano, 07.03.2010 14:27: On Sun, 7 Mar 2010 11:58:05 pm spir wrote: def __iter__(self): ''' Iteration on (key,value) pairs. ''' print '*', if self.holdsEntry: yield (self.key,self.value) for child in self.children: print , child.__iter__() print , raise StopIteration __iter__ should be an ordinary function, not a generator. Something like this should work: # Untested. def __iter__(self): ''' Iteration on (key,value) pairs. ''' def inner(): print '*', # Side effects bad... if self.holdsEntry: yield (self.key,self.value) for child in self.children: print , child.__iter__() print , raise StopIteration return inner() That's just an unnecessarily redundant variation on the above. It's perfectly ok if __iter__() is a generator method. Stefan ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] __iter__: one obvious way to do it
Hello, Below 6 working way to implement __iter__ for a container here simulated with a plain inner list. Sure, the example is a bit artificial ;-) 1. __iter__ returns a generator _expression_ 2. __iter__ *is* a generator 3. __iter__ returns a generator (this one is a bit weird, i guess) 4. __iter__ returns self, its own iterator via next() 5. __iter__ returns an external iterator object 6. __iter__ returns iter() of a collection built just on time (this one is really contrived) Also, one can always traverse the collection (already existing or built then) itself if it not quasi-infinite (no __iter__ at all). There should be one-- and preferably only one --obvious way to do it http://www.python.org/dev/peps/pep-0020/ I understand some ways fit given use cases better -- or worse. But it's difficult to find the best one, or even a proper one in a given case. Also, generation and iteration are rather abstract notions. And it's too much variety for me. I am lost in this field. I would enjoy a commented and examplified overview. Denis -- la vita e estrany spir.wikidot.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] recursive generator
sorry, forgot to forward this to the list.
On Sun, Mar 7, 2010 at 1:58 PM, spir denis.s...@gmail.com wrote:
Hello,
Is it possible at all to have a recursive generator? I think at a iterator
for a recursive data structure (here, a trie). The following code does not
work: it only yields a single value. Like if child.__iter__() were never
called.
def __iter__(self):
''' Iteration on (key,value) pairs. '''
print '*',
if self.holdsEntry:
yield (self.key,self.value)
for child in self.children:
print ,
child.__iter__()
print ,
raise StopIteration
With the debug prints in code above, for e in t: print e outputs:
* ('', 0)
print len(t.children) # -- 9
Why is no child.__iter__() executed at all? I imagine this can be caused by
the special feature of a generator recording current execution point. (But
then, is it at all possible to have a call in a generator? Or does the issue
only appear whan the callee is a generator itself?) Else, the must be an
obvious error in my code.
Denis
remember that child.__iter__ returns a generator object. Merely
calling the function won't do anything. To actually get elements from
a generator object, you need to call next() on the returned iterator,
or iterate over it in another way (e.g. a for loop). I would write
this method more or less like so:
from itertools import chain
def __iter__(self):
this_entry = [(self.key, self.value)] is self.holds_entry else []
return chain(this_entry, *[iter(c) for c in self.children])
(Warning! untested! I'm pretty sure chain will work like this, but you
might have to do a little tweaking)
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Re: [Tutor] __iter__: one obvious way to do it
On Mon, 8 Mar 2010 02:07:41 am spir wrote: [sorry, forgot the code] Hello, Below 6 working way to implement __iter__ for a container here simulated with a plain inner list. Sure, the example is a bit artificial ;-) 1. __iter__ returns a generator _expression_ def __iter__(self): return (pair(n) for n in self.items) Seems perfectly acceptable to me. That's syntactic sugar for the next one: 2. __iter__ *is* a generator def __iter__(self): for n in self.items: yield pair(n) raise StopIteration As Stefan pointed out, __iter__ can be a generator, so that's okay too. However, the StopIteration at the end is redundant: generators automatically raise StopIteration when they fall off the end at the end of the code. So this is equivalent to the above: def __iter__(self): for n in self.items: yield pair(n) 3. __iter__ returns a generator (this one is a bit weird, i guess) def __iter__(self): return self.pairs() There's nothing weird about it. It's the difference between writing code directly inline, and calling it indirectly: def f(): return [1, 2, 3] versus: def indirect(): return [1, 2, 3] def f(): return indirect() If you have a good reason for the indirection, it is perfectly fine. E.g. you might have a class with multiple iterators: class X: def iter_width(self): Iterate left-to-right pass def iter_depth(self): Iterate top-to-bottom pass def iter_spiral(self): pass def __iter__(self): # Default. return self.iter_width() 4. __iter__ returns self, its own iterator via next() def __iter__(self): self.i=0 return self That's how you would make the class an iterator directly. 5. __iter__ returns an external iterator object def __iter__(self): return Iter(self) Built-ins such as lists, dicts, tuples and sets use that strategy: iter([1,2,3]) listiterator object at 0xb7d08a4c iter(dict(a=1,b=2)) dictionary-keyiterator object at 0xb7d08fe0 6. __iter__ returns iter() of a collection built just on time (this one is really contrived) def __iter__(self): return iter(tuple([pair(n) for n in self.items])) Not contrived, but inefficient. First you build a list, all at once, using a list comprehension. So much for lazy iteration, but sometimes you have good reason for this (see below). Then you copy everything in the list into a tuple. Why? Then you create an iterator from the tuple. If you remove the intermediate tuple, it is a reasonable approach for ensuring that you can modify the original object without changing any iterators made from it. In other words, __iter__ returns a *copy* of the data in self. But the easiest way to do that: def __iter__(self): return iter([pair(n) for n in self.items]) No need to make a tuple first. Also, one can always traverse the collection (already existing or built then) itself if it not quasi-infinite (no __iter__ at all). The point of __iter__ is to have a standard way to traverse data structures, so you can traverse them with for-loops. Otherwise, every data structure needs a different method: for item in tree.traverse(): for item in mapping.next_key(): for item in sequence.get_next_item(): for item in obj.walker(): There should be one-- and preferably only one --obvious way to do it http://www.python.org/dev/peps/pep-0020/ This doesn't mean that there should be *only* one way to do something. It means that the should be one OBVIOUS way to do it. -- Steven D'Aprano ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] recursive generator
On 7 March 2010 12:58, spir denis.s...@gmail.com wrote:
Hello,
Is it possible at all to have a recursive generator? I think at a iterator
for a recursive data structure (here, a trie). The following code does not
work: it only yields a single value. Like if child.__iter__() were never
called.
def __iter__(self):
''' Iteration on (key,value) pairs. '''
print '*',
if self.holdsEntry:
yield (self.key,self.value)
for child in self.children:
print ,
child.__iter__()
print ,
raise StopIteration
With the debug prints in code above, for e in t: print e outputs:
* ('', 0)
print len(t.children) # -- 9
Why is no child.__iter__() executed at all? I imagine this can be caused by
the special feature of a generator recording current execution point. (But
then, is it at all possible to have a call in a generator? Or does the issue
only appear whan the callee is a generator itself?) Else, the must be an
obvious error in my code.
Denis
--
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You are calling child.__iter__(), but it's return value is being thrown away.
What you want to be doing is something like
def __iter__(self):
if self.holdsEntry:
yield self.entry
for child in self.children:
print
for val in child: #implicit call to child.__iter__()
yield val
print
Then, when the child.__iter__() is called, the returned iterator is
iterated, and the values are passed up the call stack. There's
probably a more terse way of doing this using itertools, but I think
this is probably more readable.
Hope this clears things up (a little, anyway...)
--
Rich Roadie Rich Lovely
Just because you CAN do something, doesn't necessarily mean you SHOULD.
In fact, more often than not, you probably SHOULDN'T. Especially if I
suggested it.
10 re-discover BASIC
20 ???
30 PRINT Profit
40 GOTO 10
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Re: [Tutor] __iter__: one obvious way to do it
On Sun, Mar 7, 2010 at 5:37 PM, Steven D'Aprano st...@pearwood.info wrote: There should be one-- and preferably only one --obvious way to do it http://www.python.org/dev/peps/pep-0020/ This doesn't mean that there should be *only* one way to do something. It means that the should be one OBVIOUS way to do it. And in this case, if you consider the thing to do to be traversal of some collection, the one obvious way to do it is through iteration. So, the iterator protocol is a perfect example of this part of the zen; the flexibility you have in implementing iteration is necessary to allow it to be the one way. Hugo ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] recursive generator
On Sun, 7 Mar 2010 17:20:07 +0100
Hugo Arts hugo.yo...@gmail.com wrote:
On Sun, Mar 7, 2010 at 1:58 PM, spir denis.s...@gmail.com wrote:
Hello,
Is it possible at all to have a recursive generator? I think at a iterator
for a recursive data structure (here, a trie). The following code does not
work: it only yields a single value. Like if child.__iter__() were never
called.
def __iter__(self):
''' Iteration on (key,value) pairs. '''
print '*',
if self.holdsEntry:
yield (self.key,self.value)
for child in self.children:
print ,
child.__iter__()
print ,
raise StopIteration
With the debug prints in code above, for e in t: print e outputs:
* ('', 0)
print len(t.children) # -- 9
Why is no child.__iter__() executed at all? I imagine this can be caused by
the special feature of a generator recording current execution point. (But
then, is it at all possible to have a call in a generator? Or does the
issue only appear whan the callee is a generator itself?) Else, the must be
an obvious error in my code.
Denis
remember that child.__iter__ returns a generator object. Merely
calling the function won't do anything. To actually get elements from
a generator object, you need to call next() on the returned iterator,
or iterate over it in another way (e.g. a for loop). I would write
this method more or less like so:
from itertools import chain
def __iter__(self):
this_entry = [(self.key, self.value)] is self.holds_entry else []
return chain(this_entry, *[iter(c) for c in self.children])
(Warning! untested! I'm pretty sure chain will work like this, but you
might have to do a little tweaking)
@ Hugo Steven
Thank you very much. I'll study your proposals as soon as I can, then tell you
about the results.
In the meanwhile, I (recursively) constructed the collection of entries and
returned iter(collection). [This works, indeed, but I also want do know how to
properly build a recursive iterator.]
Denis
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Re: [Tutor] Encoding
Giorgio wrote:
2010/3/7 spir denis.s...@gmail.com
One more question: Amazon SimpleDB only accepts UTF8.
So, let's say i have to put into an image file:
Do you mean a binary file with image data, such as a jpeg? In that
case, an emphatic - NO. not even close.
filestream = file.read()
filetoput = filestream.encode('utf-8')
Do you think this is ok?
Oh, of course everything url-encoded then
Giorgio
Encoding binary data with utf-8 wouldn't make any sense, even if you did
have the right semantics for a text file.
Next problem, 'file' is a built-in keyword. So if you write what you
describe, you're trying to call a non-static function with a class
object, which will error.
Those two lines don't make any sense by themselves. Show us some
context, and we can more sensibly comment on them. And try not to use
names that hide built-in keywords, or Python stdlib names.
If you're trying to store binary data in a repository that only permits
text, it's not enough to pretend to convert it to UTF-8. You need to do
some other escaping, such as UUENCODE, that transforms the binary data
into something resembling text. Then you may or may not need to encode
that text with utf-8, depending on its character set.
DaveA
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Re: [Tutor] Encoding
2010/3/7 Dave Angel da...@ieee.org Those two lines don't make any sense by themselves. Show us some context, and we can more sensibly comment on them. And try not to use names that hide built-in keywords, or Python stdlib names. Hi Dave, I'm considering Amazon SimpleDB as an alternative to PGSQL, but i need to store blobs. Amazon's FAQs says that: Q: What kind of data can I store? You can store any UTF-8 string data in Amazon SimpleDB. Please refer to the Amazon Web Services Customer Agreement http://aws.amazon.com/agreement for details. This is the problem. Any idea? DaveA Giorgio -- -- AnotherNetFellow Email: anothernetfel...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] pgdb and console output
OS = RHEL 5.4 32Bit Python = 2.4.3 #! /usr/bin/env python import pgdb, sys db = pgdb.connect (dsn='192.168.0.1:familydata', user='postgres', password='') cursor = db.cursor () cursor.execute (select * from names) rows = cursor.fetchall () for i in (rows): print i #viewtable (db) #sys.stdout.write() cursor.close () ###END The code as is will display the data to the console. I have read the db API 2.0 at python.org. The problem is viewtable (db) will not work or sys.stdout.write () to write the table data to the console screen. What am I doing wrong here or what do I need to do different? Can anyone point me to a book specificly for pgdb, python and postgre? Last thing is can I define the classes in a separate files and reference them? Kinda like code reuse in .Net? I new to Python so bear with me. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] pgdb and console output
hithere there pyli...@gmail.com wrote #! /usr/bin/env python import pgdb, sys db = pgdb.connect (dsn='192.168.0.1:familydata', user='postgres', password='') cursor = db.cursor () cursor.execute (select * from names) rows = cursor.fetchall () for i in (rows): print i #viewtable (db) No idea what viewtable does, never seen it before... The only case I found on Google was in a tutorial which defined viewtable() as a function, it wasn't in the pgdb module... #sys.stdout.write() This writes nothing to stdout. Apparently successfully from your comment below. Try: sys.stdout.write(str(i)) The code as is will display the data to the console. I have read the db API 2.0 at python.org. The problem is viewtable (db) will not work or sys.stdout.write () to write the table data to the console screen. What am I doing wrong here or what do I need to do different? The print is writing to the console I assume? That is the normal method. Can anyone point me to a book specificly for pgdb, python and postgre? It doesn't look lie you need a postgres specific boook but just to go through the standard Python tutorial. It should make most things clear. Last thing is can I define the classes in a separate files and reference them? Kinda like code reuse in .Net? I new to Python so bear with me. Yes, just define them in a file then import that file as a module. The standard tutorial explains that too. -- Alan Gauld Author of the Learn to Program web site http://www.alan-g.me.uk/ ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Encoding
Giorgio wrote: 2010/3/7 Dave Angel da...@ieee.org Those two lines don't make any sense by themselves. Show us some context, and we can more sensibly comment on them. And try not to use names that hide built-in keywords, or Python stdlib names. Hi Dave, I'm considering Amazon SimpleDB as an alternative to PGSQL, but i need to store blobs. Amazon's FAQs says that: Q: What kind of data can I store? You can store any UTF-8 string data in Amazon SimpleDB. Please refer to the Amazon Web Services Customer Agreement http://aws.amazon.com/agreement for details. This is the problem. Any idea? DaveA Giorgio You still didn't provide the full context. Are you trying to do store binary data, or not? Assuming you are, you could do the UUENCODE suggestion I made. Or use base64: base64.encodestring(/s/) wlll turn binary data into (larger) binary data, also considered a string. The latter is ASCII, so it's irrelevant whether it's considered utf-8 or otherwise. You store the resulting string in your database, and use base64.decodestring(s) to reconstruct your original. There's 50 other ways, some more efficient, but this may be the simplest. DaveA ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] Printing time without if statement
Hi, I have an event-based simulator written in Python (of course). It takes a while to run, and I want to have messages printed every so often to the screen, indicating the simulation time that has passed. Currently, every event that occurs follows the following code snippet: # initilize printstep = 10 nextprint = 10 # for each event if current_time nextprint: print 'time past: ' + str(nextprint) nextprint += printstep This seems stupid to me - why should I have to check each round if the time has been passed? I was thinking that there might be a simpler way, using threading, but I have zero experience with threading, so need your help. The idea was to have a simple thread that waited X time (CPU cycles, say) and then read the current_time variable and printed it. Any simple (simpler?) solutions to this? Thanks Elisha ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Printing time without if statement
On Sun, Mar 07, 2010 at 11:38:49PM -0500, Elisha Rosensweig wrote: Hi, I have an event-based simulator written in Python (of course). It takes a while to run, and I want to have messages printed every so often to the screen, indicating the simulation time that has passed. Currently, every event that occurs follows the following code snippet: It depends on how your system simulates the events and their scheduling, but you might look at the sched module in the standard library. Maybe you would be able to use that to schedule your simulated events and also schedule periodic time output too. -- Steve Willoughby| Using billion-dollar satellites st...@alchemy.com | to hunt for Tupperware. ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Encoding
Giorgio, 05.03.2010 14:56: What i don't understand is why: s = uciao è ciao is converting a string to unicode, decoding it from the specified encoding but t = ciao è ciao t = unicode(t) That should do exactly the same instead of using the specified encoding always assume that if i'm not telling the function what the encoding is, i'm using ASCII. Is this a bug? Did you read the Unicode tutorial at the link I posted? Here's the link again: http://www.amk.ca/python/howto/unicode Stefan ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] Passing nested structures to fcntl.ioctl
Hello everyone,
I have some code that's calling fcntl.ioctl. I need to pass a nested
structure as the 3rd argument of the function. Something like this
typedef struct coordinates{
int x;
int y;
} coordinates;
typedef struct point{
int amp;
coordinates* coord;
} point;
How would I do this? I need to allocate a point structure and then a
coordinate structure and link them up together via the coord member of
point, fill them up with values and pass it to the fcntl.ioctl. The
actual IOCTL implementation will do the indirection to read out the
data.
I can't use ctypes since I'm working on an embedded system with an ARM
core that doesn't have ctypes. The struct module doesn't seem to help
due to the indirection.
Any ideas? I'm personally planning to write a small C extension that
creates the structure based on a specification in memory and then
passes that to the IOCTL call directly but that looks like some work. I
was also told to take a look at Cython which basically seems to be an
easy way of doing what I'm planning. Is there a quicker/easier way?
Thanks.
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Re: [Tutor] Printing time without if statement
On Mon, 8 Mar 2010 03:38:49 pm Elisha Rosensweig wrote: Hi, I have an event-based simulator written in Python (of course). It takes a while to run, and I want to have messages printed every so often to the screen, indicating the simulation time that has passed. Currently, every event that occurs follows the following code snippet: [...] This seems stupid to me - why should I have to check each round if the time has been passed? I was thinking that there might be a simpler way, using threading, but I have zero experience with threading, so need your help. The idea was to have a simple thread that waited X time (CPU cycles, say) and then read the current_time variable and printed it. Any simple (simpler?) solutions to this? That's a brilliant idea. I think I will steal it for my own code :) (Why didn't I think of it myself? Probably because I almost never use threads...) Anyway, here's something that should help: import time import threading class Clock(threading.Thread): def __init__(self, *args, **kwargs): super(Clock, self).__init__(*args, **kwargs) self.finished = False def start(self): print Clock %s started at %s % (self.getName(), time.ctime()) super(Clock, self).start() def run(self): while 1: if self.finished: break print Clock %s still alive at %s % ( self.getName(), time.ctime()) time.sleep(2) print Clock %s quit at %s % (self.getName(), time.ctime()) def quit(self): print Clock %s asked to quit at %s % ( self.getName(), time.ctime()) self.finished = True def do_work(): clock = Clock(name=clock-1) clock.start() # Start processing something hard. for i in xrange(8): print Processing %d... % i # Simulate real work with a sleep. time.sleep(0.75) clock.quit() And in action: do_work() Clock clock-1 started at Mon Mar 8 17:40:42 2010 Processing 0... Clock clock-1 still alive at Mon Mar 8 17:40:43 2010 Processing 1... Processing 2... Clock clock-1 still alive at Mon Mar 8 17:40:45 2010 Processing 3... Processing 4... Processing 5... Clock clock-1 still alive at Mon Mar 8 17:40:47 2010 Processing 6... Processing 7... Clock clock-1 still alive at Mon Mar 8 17:40:49 2010 Clock clock-1 asked to quit at Mon Mar 8 17:40:49 2010 Clock clock-1 quit at Mon Mar 8 17:40:51 2010 There's a bit of a display artifact in the interactive interpreter, when the final quit message is printed: the interpreter doesn't notice it needs to redraw the prompt. But other than that, it should be fine. -- Steven D'Aprano ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
