Wow, thank you Ian for this very detailed answer, and thank you for taking the
time for that! Much appreciated!
If I get this right, I have to somehow fix the value of a_list during the loop,
like when you use the classic default value argument trick with lambdas (damn, I
should have thought of
Le 14/03/19 à 10:45, Peter Otten a écrit :
> Pierre Reinbold wrote:
>
>> Wow, thank you Ian for this very detailed answer, and thank you for taking
>> the time for that! Much appreciated!
>>
>> If I get this right, I have to somehow fix the value of a_list during the
>> loop, like when you use
Pierre Reinbold wrote:
> Wow, thank you Ian for this very detailed answer, and thank you for taking
> the time for that! Much appreciated!
>
> If I get this right, I have to somehow fix the value of a_list during the
> loop, like when you use the classic default value argument trick with
>
Wow, thank you Ian for this very detailed answer, and thank you for taking the
time for that! Much appreciated!
If I get this right, I have to somehow fix the value of a_list during the loop,
like when you use the classic default value argument trick with lambdas (damn, I
should have thought of
You're basically running into this:
https://docs.python.org/3/faq/programming.html#why-do-lambdas-defined-in-a-loop-with-different-values-all-return-the-same-result
To see why, let's try disassembling your function. I'm using Python 3.5
here, but it shouldn't make much of a difference.
py>
Dear all,
I want to implement a function computing the Cartesian product if the elements
of a list of lists, but using generator expressions. I know that it is already
available in itertools but it is for the sake of understanding how things work.
I already have a working recursive version, and
Neal Becker wrote:
I have an object that expects to call a callable to get a value:
class obj:
def __init__ (self, gen):
self.gen = gen
def __call__ (self):
return self.gen()
As written, there is no need for this obj class, it just adds a pointless
layer of indirection.
On Wed, Feb 4, 2015 at 6:23 AM, Neal Becker ndbeck...@gmail.com wrote:
class rpt:
def __init__ (self, value, rpt):
self.value = value; self.rpt = rpt
def __call__ (self):
for i in range (self.rpt):
yield self.value
Note that this class is just reimplementing
Neal Becker wrote:
I have an object that expects to call a callable to get a value:
class obj:
def __init__ (self, gen):
self.gen = gen
def __call__ (self):
return self.gen()
As written that looks a bit like
if boolean_expression == True: ...
as you could replace
inst =
I have an object that expects to call a callable to get a value:
class obj:
def __init__ (self, gen):
self.gen = gen
def __call__ (self):
return self.gen()
Now I want gen to be a callable that repeats N times. I'm thinking, this
sounds perfect for yield
class rpt:
def __init__
On Thu, Feb 5, 2015 at 12:23 AM, Neal Becker ndbeck...@gmail.com wrote:
Now I want gen to be a callable that repeats N times. I'm thinking, this
sounds perfect for yield
class rpt:
def __init__ (self, value, rpt):
self.value = value; self.rpt = rpt
def __call__ (self):
for i
On Wed, Feb 4, 2015 at 9:32 AM, Chris Angelico ros...@gmail.com wrote:
On Thu, Feb 5, 2015 at 12:23 AM, Neal Becker ndbeck...@gmail.com wrote:
Now I want gen to be a callable that repeats N times. I'm thinking, this
sounds perfect for yield
class rpt:
def __init__ (self, value,
On Wed, Feb 4, 2015 at 10:19 AM, Joel Goldstick joel.goldst...@gmail.com
wrote:
On Wed, Feb 4, 2015 at 9:32 AM, Chris Angelico ros...@gmail.com wrote:
On Thu, Feb 5, 2015 at 12:23 AM, Neal Becker ndbeck...@gmail.com wrote:
Now I want gen to be a callable that repeats N times. I'm
I see. Thanks for the clarification.
On Dec 22, 12:35 am, Steven D'Aprano steve
+comp.lang.pyt...@pearwood.info wrote:
On Wed, 21 Dec 2011 21:45:13 -0800, GZ wrote:
Now the question here is this:
def h():
if condition=true:
#I would like to return an itereator with zero
Hi,
I am wondering what would be the best way to return an iterator that
has zero items.
I just noticed the following two are different:
def f():
pass
def g():
if 0: yield 0
pass
for x in f(): print x
Traceback (most recent call last):
File string, line 1, in fragment
TypeError:
On Wed, 21 Dec 2011 21:45:13 -0800, GZ wrote:
Hi,
I am wondering what would be the best way to return an iterator that has
zero items.
return iter([])
I just noticed the following two are different:
def f():
pass
That creates a function that does nothing, and then returns None
On Thu, Dec 22, 2011 at 4:45 PM, GZ zyzhu2...@gmail.com wrote:
def h():
if condition=true:
#I would like to return an itereator with zero length
else:
for ...: yield x
Easy. Just 'return' in the conditional.
def h():
if condition:
return
On Wed, 21 Dec 2011 21:45:13 -0800, GZ wrote:
Now the question here is this:
def h():
if condition=true:
#I would like to return an itereator with zero length
else:
for ...: yield x
In other words, when certain condition is met, I want to yield nothing.
How to
On Wed, 22 Dec 2010 15:49:31 -0800, Dan Stromberg wrote:
def generator():
i = 0
while True:
yield i
i += 1
Shorter version:
from itertools import count as generator
--
http://mail.python.org/mailman/listinfo/python-list
So I have a generator, either as a free function or in a class and I
want to generate objects that are initialized from the generated things.
def generator():
for whatever:
yield something
class Object():
def __init__(self):
self.data = # the next
On 12/22/2010 3:15 PM Victor Eijkhout said...
So I have a generator, either as a free function or in a class and I
want to generate objects that are initialized from the generated things.
def generator():
for whatever:
yield something
class Object():
def
On Wed, Dec 22, 2010 at 3:15 PM, Victor Eijkhout s...@sig.for.address wrote:
So I have a generator, either as a free function or in a class and I
want to generate objects that are initialized from the generated things.
def generator():
for whatever:
yield something
Dan Stromberg drsali...@gmail.com wrote:
You likely want a class variable:
Sounds like an elegant solution. Thanks!
Victor.
--
Victor Eijkhout -- eijkhout at tacc utexas edu
--
http://mail.python.org/mailman/listinfo/python-list
On Fri, Sep 19, 2008 at 9:51 PM, Steven D'Aprano
[EMAIL PROTECTED] wrote:
Extending len() to support iterables sounds like a good idea, except that
it's not.
Here are two iterables:
def yes(): # like the Unix yes command
while True:
yield y
def rand(total):
Return random
Steven D'Aprano:
Extending len() to support iterables sounds like a good idea, except that it's
not.
Python language lately has shifted toward more and more usage of lazy
iterables (see range lazy by default, etc). So they are now quite
common. So extending len() to make it act like leniter()
On Mon, 22 Sep 2008 04:21:12 -0700, bearophileHUGS wrote:
Steven D'Aprano:
Extending len() to support iterables sounds like a good idea, except
that it's not.
Python language lately has shifted toward more and more usage of lazy
iterables (see range lazy by default, etc). So they are now
Steven D'Aprano:
I'm sorry, I don't recognise leniter(). Did I miss something?
I have removed the docstring/doctests:
def leniter(iterator):
if hasattr(iterator, __len__):
return len(iterator)
nelements = 0
for _ in iterator:
nelements += 1
return nelements
it
Simon Mullis [EMAIL PROTECTED] writes:
l = [ 1.1.1.1, 1.2.2.2, 1.2.2.3, 1.3.1.2, 1.3.4.5]
...
dict_of_counts = dict([(v[0:3], count) for v in l])
Untested:
def major_version(version_string):
convert '1.2.3.2' to '1.2'
return '.'.join(version_string.split('.')[:2])
Gerard flanagan wrote:
George Sakkis wrote:
..
Note that this works correctly only if the versions are already sorted
by major version.
Yes, I should have mentioned it. Here's a fuller example below. There's
maybe better ways of sorting version numbers, but this is what I do.
Indeed,
Boris Borcic wrote:
Gerard flanagan wrote:
George Sakkis wrote:
..
Note that this works correctly only if the versions are already sorted
by major version.
Yes, I should have mentioned it. Here's a fuller example below.
There's maybe better ways of sorting version numbers, but this is
Gerard flanagan:
data.sort()
datadict = \
dict((k, len(list(g))) for k,g in groupby(data, lambda s:
'.'.join(s.split('.',2)[:2])))
That code may run correctly, but it's quite unreadable, while good
Python programmers value high readability. So the right thing to do is
to split that line
On Sep 19, 2:01 pm, [EMAIL PROTECTED] wrote:
Gerard flanagan:
data.sort()
datadict = \
dict((k, len(list(g))) for k,g in groupby(data, lambda s:
'.'.join(s.split('.',2)[:2])))
That code may run correctly, but it's quite unreadable, while good
Python programmers value high
On Fri, 19 Sep 2008 17:00:56 -0700, MRAB wrote:
Extending len() to support iterables sounds like a good idea, except
that it could be misleading when:
len(file(path))
returns the number of lines and /not/ the length in bytes as you might
first think!
Extending len() to support
MRAB:
except that it could be misleading when:
len(file(path))
returns the number of lines and /not/ the length in bytes as you might
first think! :-)
Well, file(...) returns an iterable of lines, so its len is the number
of lines :-)
I think I am able to always remember this fact.
Hi,
Let's say I have an arbitrary list of minor software versions of an
imaginary software product:
l = [ 1.1.1.1, 1.2.2.2, 1.2.2.3, 1.3.1.2, 1.3.4.5]
I'd like to create a dict with major_version : count.
(So, in this case:
dict_of_counts = { 1.1 : 1,
1.2 : 2,
Simon Mullis napisał(a):
Something like:
dict_of_counts = dict([(v[0:3], count) for v in l])
I can't seem to figure out how to get count, as I cannot do x += 1
or x++ as x may or may not yet exist, and I haven't found a way to
create default values.
It seems to me that the count you're
On Sep 18, 10:54 am, Simon Mullis [EMAIL PROTECTED] wrote:
Hi,
Let's say I have an arbitrary list of minor software versions of an
imaginary software product:
l = [ 1.1.1.1, 1.2.2.2, 1.2.2.3, 1.3.1.2, 1.3.4.5]
I'd like to create a dict with major_version : count.
(So, in this case:
On Sep 18, 10:54 am, Simon Mullis [EMAIL PROTECTED] wrote:
Hi,
Let's say I have an arbitrary list of minor software versions of an
imaginary software product:
l = [ 1.1.1.1, 1.2.2.2, 1.2.2.3, 1.3.1.2, 1.3.4.5]
I'd like to create a dict with major_version : count.
(So, in this case:
Simon Mullis wrote:
Hi,
Let's say I have an arbitrary list of minor software versions of an
imaginary software product:
l = [ 1.1.1.1, 1.2.2.2, 1.2.2.3, 1.3.1.2, 1.3.4.5]
I'd like to create a dict with major_version : count.
(So, in this case:
dict_of_counts = { 1.1 : 1,
On Sep 18, 10:54 am, Simon Mullis [EMAIL PROTECTED] wrote:
Hi,
Let's say I have an arbitrary list of minor software versions of an
imaginary software product:
l = [ 1.1.1.1, 1.2.2.2, 1.2.2.3, 1.3.1.2, 1.3.4.5]
I'd like to create a dict with major_version : count.
(So, in this case:
On Sep 18, 11:43 am, Gerard flanagan [EMAIL PROTECTED] wrote:
Simon Mullis wrote:
Hi,
Let's say I have an arbitrary list of minor software versions of an
imaginary software product:
l = [ 1.1.1.1, 1.2.2.2, 1.2.2.3, 1.3.1.2, 1.3.4.5]
I'd like to create a dict with major_version :
Haha!
Thanks for all of the suggestions... (I love this list!)
SM
2008/9/18 [EMAIL PROTECTED]:
On Sep 18, 10:54 am, Simon Mullis [EMAIL PROTECTED] wrote:
Hi,
Let's say I have an arbitrary list of minor software versions of an
imaginary software product:
l = [ 1.1.1.1, 1.2.2.2, 1.2.2.3,
George Sakkis wrote:
On Sep 18, 11:43 am, Gerard flanagan [EMAIL PROTECTED] wrote:
Simon Mullis wrote:
Hi,
Let's say I have an arbitrary list of minor software versions of an
imaginary software product:
l = [ 1.1.1.1, 1.2.2.2, 1.2.2.3, 1.3.1.2, 1.3.4.5]
I'd like to create a dict with
Mathias Panzenboeck wrote:
Robert Kern wrote:
Timothy Wu wrote:
Hi,
Using generator recursively is not doing what I expect:
def test_gen(x):
yield x
x = x - 1
if x != 0:
test_gen(x)
The only thing that the last line does is *create* a new generator
object. You
Hi,
Using generator recursively is not doing what I expect:
def test_gen(x):
yield x
x = x - 1
if x != 0:
test_gen(x)
for item in test_gen(3):
print item
This gives me a single number 3 and not printing 2 and 1 as I would expect.
What is wrong??
Timothy
--
Timothy Wu wrote:
Hi,
Using generator recursively is not doing what I expect:
def test_gen(x):
yield x
x = x - 1
if x != 0:
test_gen(x)
The only thing that the last line does is *create* a new generator object. You
need to actually iterate over it and yield its
On 11/26/06, Robert Kern [EMAIL PROTECTED] wrote:
The only thing that the last line does is *create* a new generator object.
You
need to actually iterate over it and yield its values. E.g.
In [2]: def test_gen(x):
...: yield x
...: x -= 1
...: if x != 0:
...:
Robert Kern wrote:
Timothy Wu wrote:
Hi,
Using generator recursively is not doing what I expect:
def test_gen(x):
yield x
x = x - 1
if x != 0:
test_gen(x)
The only thing that the last line does is *create* a new generator object. You
need to actually iterate over
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