Re: When to use assert

[Steven D'Aprano]

Chris Angelico wrote:

> On Wed, Oct 22, 2014 at 12:44 PM, Steven D'Aprano
>  wrote:
>> def do_something(instance_or_id):
>> instance = Model.get(instance_or_id)
>> assert isinstance(instance, Model)
>> # Code that assumes that instance is an object of type Model
>>
>>
>> That means that the logic for what is acceptable as a Model is all in one
>> place, namely the Model.get method, and callers don't need to care about
>> the pre-condition "argument is a Model or an integer ID", they only need
>> to care about the post-condition "result of get() is a Model".
> 
> And at that point, the assertion is redundant, on par with:
> 
> a = len(seq)
> assert isinstance(a, int)
> 
> because you shouldn't have to assert what's part of a function's
> guarantee.

That depends on how well you trust the function, how paranoid you are, and
whether you wish to include a checked comment, among other considerations.
I'd prefer to write the above than:

a = len(seq)
# At this point, a is an int.

because comments inside code that aren't checked are technically known
as "lies" . Ha ha only serious.

I wouldn't write such a comment for len() since I would expect anyone
reading the code to know what len() does, but the same doesn't necessarily
apply for every function call I make.

Checking the post-condition of a built-in like len() is too paranoid for my
tastes, as len() enforces the rule that __len__() methods return a
non-negative integer, and there are millions of lines of Python code around
the world calling len(). Somebody surely have noticed by now if len()
violated that post-condition. But for a function from my own code base,
where I might only have dozens of users (oh to have that many!) and fewer
unit tests than perhaps I ought to, I might not be quite so confident that
the assertion was redundant. To err is human, and so there are occasions
when it is appropriate to trust but verify.

As I wrote:

This is why assert can be divisive. Since we can vary in our 
confidence about the correctness of code, one person's useful 
assertion may be another person's useless runtime test.

http://import-that.dreamwidth.org/676.html


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Re: Permissions on files installed by pip?

[Adam Funk]

On 2014-10-17, Jean-Michel Pichavant wrote:

> - Original Message -
>> From: "Adam Funk" 
>> To: python-list@python.org
>> Sent: Thursday, 16 October, 2014 9:29:46 PM
>> Subject: Permissions on files installed by pip?
>> 
>> I've been using the python-nltk package on Ubuntu, but I need ntlk
>> 3.0
>> now.  I used 'sudo aptitude purge python-nltk' to get rid of my
>> existing installation, & followed instructions on the nltk website
>> [1]
>> starting at step 4 (since I already have python-pip & python-numpy
>> packages installed).
>> 
>> $ sudo pip install -U
>> 
>> I couldn't get it to work, until I realized that the permissions &
>> ownership on /usr/local/lib/python2.7/dist-packages were 'drwx--S---
>> root staff'.  A 'chmod -R a+rX' on that directory seems to have fixed
>> it.  Is it normal for sudo pip install to set the permissions that
>> way, or did I do something wrong?
>
> On debian wheezy:
>
> ls -al /usr/local/lib/python2.7/dist-packages  
>
> drwxrwsr-x 5 root staff 4.0K Jun 30 15:16 ./
>
> I'm not sure pip is responsible for this anyway, so my money goes on "you did 
> something wrong" :)

Probably something to do with the way I have sudo set up then.
Thanks.


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is war.   --- Henry Miller
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Re: When to use assert

[Chris Angelico]

On Thu, Oct 23, 2014 at 2:49 AM, Steven D'Aprano
 wrote:
> Chris Angelico wrote:
>
>> And at that point, the assertion is redundant, on par with:
>>
>> a = len(seq)
>> assert isinstance(a, int)
>>
>> because you shouldn't have to assert what's part of a function's
>> guarantee.
>
> That depends on how well you trust the function, how paranoid you are, and
> whether you wish to include a checked comment, among other considerations.
> I'd prefer to write the above than:
>
> a = len(seq)
> # At this point, a is an int.
>
> because comments inside code that aren't checked are technically known
> as "lies" . Ha ha only serious.
> ...
> Checking the post-condition of a built-in like len() is too paranoid for my
> tastes...
>
> As I wrote:
>
> This is why assert can be divisive. Since we can vary in our
> confidence about the correctness of code, one person's useful
> assertion may be another person's useless runtime test.

I agree that the assert is preferable to the comment. But maybe my
level of paranoia is just lower than most people's, as I wouldn't
bother checking the post-conditions of pretty much anything. Do you
put these assertions every time you call the function? Every time
you're depending on its return value? At what point should you stop
writing Python code and start using a language with a lot more
boilerplate and checking (I believe Haskell fits that, though I'm not
overly familiar with the language)?

This is the job of a test suite. You don't pepper your code with
assertions to the effect that "I just pushed something onto my queue,
it should now have this item in it"; you create a test case for it,
and verify your function there. In the rest of the code, you trust
that your test suite passes, and don't waste time with assertions.

Or is that insufficiently paranoid?

ChrisA
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Re: (test) ? a:b

[Dan Stromberg]

On Wed, Oct 22, 2014 at 2:05 AM,   wrote:
> Em quarta-feira, 22 de outubro de 2014 06h29min55s UTC-2, ast  escreveu:
>> Hello
>>
>>
>>
>> Is there in Python something like:
>>
>>
>>
>> j = (j >= 10) ? 3 : j+1;
>>
>>
>>
>> as in C language ?
>>
>>
>>
>> thx
>
> without not:
> j = [j+1, 3][j>=10]
> with not:
> j = [3, j+1][not (j>=10)]

This is not very readable, and eagerly evaluates the two list values.
A proper ternary operator will not evaluate the unused
candidate-result. This matters a little for performance, but matters
more if one or both of the candidate results have side-effects.

It's better to use "j = 3 if j >= 10 else j + 1".

What you've suggested here was one of the ternary operator workarounds
before a true ternary operator was introduced in 2.5.

I don't use Python's ternary operator much though - I tend to find
if+else more clear, and it shows up well in a debugger. The ternary
operator tends to lead to long one-liners (especially if nested!),
which are usually best avoided.

I'd even go so far as to say that most uses of a ternary operator
probably should be a small function.  It's slower, but it lends itself
to DRY and abstraction better.
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Re: When to use assert

[Dan Stromberg]

On Wed, Oct 22, 2014 at 9:01 AM, Chris Angelico  wrote:
> On Thu, Oct 23, 2014 at 2:49 AM, Steven D'Aprano
>  wrote:
>> Chris Angelico wrote:
> I agree that the assert is preferable to the comment. But maybe my
> level of paranoia is just lower than most people's, as I wouldn't
> bother checking the post-conditions of pretty much anything. Do you
> put these assertions every time you call the function? Every time
> you're depending on its return value? At what point should you stop
> writing Python code and start using a language with a lot more
> boilerplate and checking (I believe Haskell fits that, though I'm not
> overly familiar with the language)?

I like to use assertions and "if cond: raise ValueError('foo')" a lot.

I think Eiffel may be the poster-child for a language with
pre-conditions, post-conditions and assertions.

I think you're in good company - a lot of developers don't use assertions much.

I like assertions, because they tend to stop bugs pretty quickly.  If
you have 3 functions, one calling another calling another, assertions
in each can keep you from having to backtrack among them when
debugging, instead going directly to the problem's source.

> This is the job of a test suite.

Test suites are great, and I can't really question your reliance on
them.  I love having lots of automated tests.  But for the reason I
described above, I still like having lots of assertions.

> You don't pepper your code with
> assertions to the effect that "I just pushed something onto my queue,
> it should now have this item in it"; you create a test case for it,
> and verify your function there. In the rest of the code, you trust
> that your test suite passes, and don't waste time with assertions.

I wouldn't test that a value was added to a queue immediately after
adding it.  That's excessive, and may even require an abstraction
violation.

But if, for example, I have a string with 3 known-good values, I'll
if/elif/elif/else, and make the else raise an AssertionError.  The
assertion should never fire, but if the code changes, it could, and if
there's a typo somewhere, it could then too.

> Or is that insufficiently paranoid?

With good tests, you're probably fine.
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Re: (test) ? a:b

[Steven D'Aprano]

Ned Batchelder wrote:

> On 10/22/14 5:05 AM, busca...@gmail.com wrote:

>> without not:
>> j = [j+1, 3][j>=10]
>> with not:
>> j = [3, j+1][not (j>=10)]
>>
> 
> Why on earth would you recommend this outdated hack, when there's a true
> conditional operator?
> 
>  j = 3 if j >= 10 else j+1

I think that's a bit harsh. Especially since this appears to have been
Buscacio's first post here. Hopefully not his(?) last post!

The old (b, a)[condition] idiom is not outdated for anyone supporting Python
2.4, and I wouldn't call it a hack. Indexing into a sequence with a bool is
basic to Python's semantics: True and False are ints equal to 1 and 0
respectively. It's also a technique easily extensible to more than two
values:

'01TX'[n % 4]

is in my opinion more readable than:

i = n % 4
'0' if i == 0 else '1' if i == 1 else 'T' if i == 3 else 'X'


> Of course, many people feel like the conditional operator isn't worth
> the squished-up unreadability, but if someone asks for a conditional
> operator, at least show them one!

The advantage of the `x if cond else y` operator is that it is a
short-cutting operator, it doesn't evaluate either x or y unless needed.
But for many cases that's not important, and in those cases I won't say I
prefer the old (y, x)[cond] idiom, but neither do I dislike it.


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Re: Python 3.4.1 and blitzdb issue

[Juan Christian]

Oh yes, and here is what the call to the API returns:

{"adult":false,"also_known_as":["George Walton Lucas Jr.
"],"biography":"Arguably the most important film innovator in the history
of the medium, George Lucas continually \"pushed the envelope\" of
filmmaking technology since his early days as a student at U.S.C.
Considered a wunderkind by his contemporaries, he had a much harder time
communicating his vision to studio executives, whose meddling managed to
compromise each of his first three feature directing efforts in some way.
The monumental success of \"Star Wars\" (1977) ushered in the era of the
\"summer blockbuster,\" which, despite the later popularity of low budget
independent films, was still the prevailing mentality powering the
Hollywood engine.\n\nThough he set the tone and established the
expectations which influenced studios to devote the bulk of their resources
to films designed to blast off into hyperspace for spectacular profits, it
was doubtful that a film as revolutionary as \"Star Wars\" was in its day
could get made in the later blockbuster assembly line climate of the new
millennium.","birthday":"1944-05-14","deathday":"","homepage":"","id":1,"imdb_id":"nm184","name":"George
Lucas","place_of_birth":"Modesto - California -
USA","popularity":2.185575,"profile_path":"/rJ1zvSeZfge0mHtLnzJn4Mkw18S.jpg"}

On Wed, Oct 22, 2014 at 2:34 PM, Juan Christian 
wrote:

> Testing code:
>
> CODE -
>
> #!/usr/bin/env
>
> import requests
> from blitzdb import Document, FileBackend
>
>
> API_URL = 'http://api.themoviedb.org/3'
> API_KEY = 'ddf30289'
>
>
> class Actor(Document):
> pass
>
>
> def get_actor(_id):
> r = requests.get('{}/person/{}?api_key={}'.format(API_URL, str(_id),
> API_KEY))
> return r.json()
>
>
> actor_1 = Actor(get_actor(1))
> actor_2 = Actor(get_actor(2))
>
> backend = FileBackend("db.blitz")
> actor_1.save(backend)
> actor_2.save(backend)
>
> print(backend.get(Actor,{'imdb_id' : 'nm184'}))
> print('\n')
> print(backend.get(Actor,{'imdb_id' : 'nm434'}))
>
>
> OUTPUT -
>
> Warning: cjson could not be imported, CJsonSerializer will not be
> available.
> Traceback (most recent call last):
>   File ".\uff.py", line 27, in 
> print(backend.get(Actor,{'imdb_id' : 'nm184'}))
>   File "C:\Python34\lib\site-packages\blitzdb\backends\file\backend.py",
> line 456, in get
> raise cls.DoesNotExist
> blitzdb.document.DoesNotExist: DoesNotExist(Actor)
>
>
> QUESTION -
>
> Why the output says that Actor doesn't exists when I already added it here
> 'actor_1.save(backend)' and 'actor_2.save(backend)'
>
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Re: Matplotlib: getting a figure to show without plt.show()

[Peter Pearson]

On Wed, 22 Oct 2014 12:38:02 +0200, Peter Otten wrote:
> Peter Pearson wrote:
[snip]
>> def callback(event):
>> global n, first
>> fig = plt.figure(2)
>> fig.clear()
>> plt.plot([0,1],[0,n])
>> n += 1  # (Pretending something changes from one plot to the next.)
>> if first:
>> first = False
>> plt.show()
>> else:
>> fig.canvas.draw()
>>
>> Can someone tell me the right way?
[snip]

> def callback(event):
> global n, fig2
>
> if fig2 is None:
> fig2 = plt.figure(2)
>
> fig2.clear()
> fig2.gca().plot([0, .5, 1], [0, 1/n, 1])
> fig2.show()

Thank you for pointing out two useful things:

 * I like fig2.gca().plot better than plt.plot, since it makes it
   clear that I'm talking about fig2 and not some other part of the
   vast and inscrutable plt kingdom.

 * fig2.show is different from plt.show, and in fact this solves my
   problem exactly, since fig2.show returns immediately.

By the way, I will dispense with your "if fig2 is None" test.  In real
life, I say something like fig2 = plt.figure("Packet Sizes").  If a
figure named "Packet Sizes" doesn't exist, plt.figure creates it,
otherwise it returns a pointer to the existing figure -- *and* it titles
the figure's window with "Packet Sizes" rather than "2".

[snip]
> As I'm not an expert for matplotlib you might also post your question on the 
> matplotlib mailing list.

Sound advice.  Next time.  Thanks again.


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Python 3.4.1 and blitzdb issue

[Juan Christian]

Testing code:

CODE -

#!/usr/bin/env

import requests
from blitzdb import Document, FileBackend


API_URL = 'http://api.themoviedb.org/3'
API_KEY = 'ddf30289'


class Actor(Document):
pass


def get_actor(_id):
r = requests.get('{}/person/{}?api_key={}'.format(API_URL, str(_id),
API_KEY))
return r.json()


actor_1 = Actor(get_actor(1))
actor_2 = Actor(get_actor(2))

backend = FileBackend("db.blitz")
actor_1.save(backend)
actor_2.save(backend)

print(backend.get(Actor,{'imdb_id' : 'nm184'}))
print('\n')
print(backend.get(Actor,{'imdb_id' : 'nm434'}))


OUTPUT -

Warning: cjson could not be imported, CJsonSerializer will not be available.
Traceback (most recent call last):
  File ".\uff.py", line 27, in 
print(backend.get(Actor,{'imdb_id' : 'nm184'}))
  File "C:\Python34\lib\site-packages\blitzdb\backends\file\backend.py",
line 456, in get
raise cls.DoesNotExist
blitzdb.document.DoesNotExist: DoesNotExist(Actor)


QUESTION -

Why the output says that Actor doesn't exists when I already added it here
'actor_1.save(backend)' and 'actor_2.save(backend)'
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Re: (test) ? a:b

[Chris Angelico]

On Thu, Oct 23, 2014 at 3:28 AM, Steven D'Aprano
 wrote:
> It's also a technique easily extensible to more than two
> values:
>
> '01TX'[n % 4]
>
> is in my opinion more readable than:
>
> i = n % 4
> '0' if i == 0 else '1' if i == 1 else 'T' if i == 3 else 'X'

That's true when it's fundamentally arithmetic. But part of that
readability difference is the redundancy in the second one. What if it
weren't so redundant?

'Negative' if x < 0 else 'Low' if x < 10 else 'Mid' if x < 20 else 'High'

You can't easily turn that into a dict lookup, nor indexing. It's
either a chained if/elif tree or nested if/else expressions, which
come to the same thing. So I'd say all the techniques have their
advantages and disadvantages. The most important thing to note is
where they differ in semantics, like the short-circuiting of if/else.

ChrisA
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Re: Python 3.4.1 and blitzdb issue

[Stéphane Wirtel]

Maybe ask on the project on github. Andreas is a good guy and will reply 
asap.



On 22 Oct 2014, at 18:34, Juan Christian wrote:


Testing code:

CODE -

#!/usr/bin/env

import requests
from blitzdb import Document, FileBackend


API_URL = 'http://api.themoviedb.org/3'
API_KEY = 'ddf30289'


class Actor(Document):
 pass


def get_actor(_id):
 r = requests.get('{}/person/{}?api_key={}'.format(API_URL, str(_id),
API_KEY))
 return r.json()


actor_1 = Actor(get_actor(1))
actor_2 = Actor(get_actor(2))

backend = FileBackend("db.blitz")
actor_1.save(backend)
actor_2.save(backend)

print(backend.get(Actor,{'imdb_id' : 'nm184'}))
print('\n')
print(backend.get(Actor,{'imdb_id' : 'nm434'}))


OUTPUT -

Warning: cjson could not be imported, CJsonSerializer will not be 
available.

Traceback (most recent call last):
File ".\uff.py", line 27, in 
 print(backend.get(Actor,{'imdb_id' : 'nm184'}))
File "C:\Python34\lib\site-packages\blitzdb\backends\file\backend.py",
line 456, in get
 raise cls.DoesNotExist
blitzdb.document.DoesNotExist: DoesNotExist(Actor)


QUESTION -

Why the output says that Actor doesn't exists when I already added it 
here

'actor_1.save(backend)' and 'actor_2.save(backend)'
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Re: (-1)**1000

[Terry Reedy]

On 10/22/2014 4:27 AM, ast wrote:

Hello

If i am writing (-1)**1000 on a python program, will the
interpreter do (-1)*(-1)*...*(-1) or something clever ?


The answer depends on the implementation.


In fact i have (-1)**N with N an integer potentially big.

I do some tests that suggest that Python is clever


You probably mean "CPython is clever".  Other implementations may or may 
not have the same optimizations.




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Terry Jan Reedy

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Re: (test) ? a:b

[Michael Torrie]

On 10/22/2014 05:45 AM, Mark Lawrence wrote:
>>> without not:
>>> j = [j+1, 3][j>=10]
>>> with not:
>>> j = [3, j+1][not (j>=10)]
>>>
>>
>> Oh it's a trick !
>> thx
> 
> IMHO it's just dreadful.  Why people insist on messing around like this 
> I really don't know, it just drives me nuts.

This actually was the standard idiom used by many python programs before
Python 2.5.  But I agree.  Don't do this anymore! Python has a ternary
if expression. Also the ternary if expression does, I believe
short-circuit logic, so the non-chosen path is not calculated.  This
hack does not. Could lead to interesting bugs depending on your assumptions.

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Re: (test) ? a:b

[alister]

On Thu, 23 Oct 2014 03:28:48 +1100, Steven D'Aprano wrote:

>> Why on earth would you recommend this outdated hack, when there's a
>> true conditional operator?
>> 
>>  j = 3 if j >= 10 else j+1
> 
> I think that's a bit harsh. Especially since this appears to have been
> Buscacio's first post here. Hopefully not his(?) last post!
> 
> The old (b, a)[condition] idiom is not outdated for anyone supporting
> Python 2.4, and I wouldn't call it a hack. Indexing into a sequence with
> a bool is basic to Python's semantics: True and False are ints equal to
> 1 and 0 respectively. It's also a technique easily extensible to more
> than two values:
> 
> '01TX'[n % 4]
> 
> is in my opinion more readable than:
> 
> i = n % 4 '0' if i == 0 else '1' if i == 1 else 'T' if i == 3 else
> 'X'

chained ternary operations are evil no mater what style or language they 
are written in as they rapidly become confusing & unreadable

"Readability counts"

in my opinion they are better written as nested if statements 

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Telling the truth when you don't mean to.
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Re: (test) ? a:b

[BartC]

"Rustom Mody"  wrote in message 
news:7d2ea3c1-504e-4f5c-8338-501b1483d...@googlegroups.com...

On Wednesday, October 22, 2014 5:01:08 PM UTC+5:30, Ned Batchelder wrote:

On 10/22/14 5:05 AM, buscacio wrote:
> Em quarta-feira, 22 de outubro de 2014 06h29min55s UTC-2, ast 
> escreveu:

>> Hello
>> Is there in Python something like:
>> j = (j >= 10) ? 3 : j+1;
>> as in C language ?
>> thx
> without not:
> j = [j+1, 3][j>=10]
> with not:
> j = [3, j+1][not (j>=10)]



Why on earth would you recommend this outdated hack, when there's a true
conditional operator?



To learn a bit about the interchangeability of control and data 
structures?


[Just playing devil's advocate]


But it doesn't do the same thing.

Comparing:

x = cond ? f() : g();   # C version

with

x = [f(), g()] [cond]

the latter evaluates both f() and g() instead of just one. Apart from being 
inefficient, it can have unintended side-effects.


--
Bartc



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Re: (test) ? a:b

[Matthew Ruffalo]

On 10/22/2014 12:40 PM, Chris Angelico wrote:
> That's true when it's fundamentally arithmetic. But part of that
> readability difference is the redundancy in the second one. What if it
> weren't so redundant?
>
> 'Negative' if x < 0 else 'Low' if x < 10 else 'Mid' if x < 20 else 'High'
>
> You can't easily turn that into a dict lookup, nor indexing. It's
> either a chained if/elif tree or nested if/else expressions, which
> come to the same thing.
No, you can't turn that into a dict lookup, but this is one of the
canonical use cases for the bisect module:

>>> from bisect import bisect
>>> breakpoints = [0, 10, 20]
>>> labels = ['Negative', 'Low', 'Mid', 'High']
>>> values = [-5, 5, 15, 25]
>>> [labels[bisect(breakpoints, value)] for value in values]
['Negative', 'Low', 'Mid', 'High']

It's also worth noting that using bisect is O(log(n)) instead of O(n),
but if you're going to hit a point where the asymptotic behavior matters
I'm sure you will have long since abandoned a manually-written if/elif
chain.

MMR...

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Re: (test) ? a:b

[Chris Angelico]

On Thu, Oct 23, 2014 at 5:27 AM, Matthew Ruffalo  wrote:
> On 10/22/2014 12:40 PM, Chris Angelico wrote:
>> That's true when it's fundamentally arithmetic. But part of that
>> readability difference is the redundancy in the second one. What if it
>> weren't so redundant?
>>
>> 'Negative' if x < 0 else 'Low' if x < 10 else 'Mid' if x < 20 else 'High'
>>
>> You can't easily turn that into a dict lookup, nor indexing. It's
>> either a chained if/elif tree or nested if/else expressions, which
>> come to the same thing.
> No, you can't turn that into a dict lookup, but this is one of the
> canonical use cases for the bisect module:
>
 from bisect import bisect
 breakpoints = [0, 10, 20]
 labels = ['Negative', 'Low', 'Mid', 'High']
 values = [-5, 5, 15, 25]
 [labels[bisect(breakpoints, value)] for value in values]
> ['Negative', 'Low', 'Mid', 'High']
>
> It's also worth noting that using bisect is O(log(n)) instead of O(n),
> but if you're going to hit a point where the asymptotic behavior matters
> I'm sure you will have long since abandoned a manually-written if/elif
> chain.

Indeed. If the performance is making any difference, something's gone
wrong. I'm seeing that as not significantly clearer than the chained
if/else statement or expression: you need a list of breakpoints and a
list of results, which need to be kept in sync. But there's probably
no ideal solution to this.

ChrisA
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setuptools + data_files = 2

[luc2]

hello, would you know how to make data_files work in setuptools ?
i can't figure out how to put datas in the generated .tar.gz

$ find .
./hello
./hello/__init__.py
./share
./share/test_file.txt
./setup.py

$ cat ./hello/__init__.py
def hello():
print( 'hello' )

$ cat ./share/test_file.txt
this is a test

$ cat ./hello/setup.py
from setuptools import setup, find_packages
setup(
name = "Hello",
version = "0.1",
packages = find_packages(),
data_files = [
( 'share', ['share/test_file.txt'] )
]
)

$ python setup.py sdist

$ tar tvf dist/Hello-0.1.tar.gz | grep test_file.txt

$ echo "no test_file.txt :-("
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Re: (test) ? a:b

[Ned Batchelder]

On 10/22/14 12:28 PM, Steven D'Aprano wrote:

Ned Batchelder wrote:


On 10/22/14 5:05 AM, busca...@gmail.com wrote:



without not:
j = [j+1, 3][j>=10]
with not:
j = [3, j+1][not (j>=10)]



Why on earth would you recommend this outdated hack, when there's a true
conditional operator?

  j = 3 if j >= 10 else j+1


I think that's a bit harsh. Especially since this appears to have been
Buscacio's first post here. Hopefully not his(?) last post!


You are right, it sounds a bit harsh. Sorry.

--
Ned Batchelder, http://nedbatchelder.com

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I am out of trial and error again Lists

[Seymore4Head]

def nametonumber(name):
lst=[""]
for x,y in enumerate (name):
lst=lst.append(y)
print (lst)
return (lst)
a=["1-800-getcharter"]
print (nametonumber(a))#18004382427837

  
The syntax for when to use a () and when to use [] still throws me a
curve.

For now, I am trying to end up with a list that has each character in
"a" as a single item.

I get:
None
None
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Re: I am out of trial and error again Lists

[Joel Goldstick]

On Wed, Oct 22, 2014 at 4:30 PM, Seymore4Head
 wrote:
> def nametonumber(name):
> lst=[""]
> for x,y in enumerate (name):
> lst=lst.append(y)
> print (lst)
> return (lst)
> a=["1-800-getcharter"]
> print (nametonumber(a))#18004382427837
>
>
> The syntax for when to use a () and when to use [] still throws me a
> curve.
() is tuples which are immutable which means that the items can't be
changed.  [] is list which means that each item can be changed.
Tuples are useful because they can be used as keys in dictionaries and
are guarantied not to change.  Lists are useful because they can be
updated.

What you are doing confuses me.  You don't use x, which is the enumerated value.

FIrst lst should be lst = [] .  You don't need to set the first
element in the list to an empty string.  You just want to establish
that you have an empty list called lst
Second, you don't need lst = lst.append(y) because you can just say
lst.append(y).  This will append the y value to the end of the list.
As to converting letters to the corresponding numbers on a phone
keypad, you don't show you code here for that
>
> For now, I am trying to end up with a list that has each character in
> "a" as a single item.
>
> I get:
> None
> None
> --
> https://mail.python.org/mailman/listinfo/python-list



-- 
Joel Goldstick
http://joelgoldstick.com
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Wed, 22 Oct 2014 16:57:00 -0400, Joel Goldstick
 wrote:

>On Wed, Oct 22, 2014 at 4:30 PM, Seymore4Head
> wrote:
>> def nametonumber(name):
>> lst=[""]
>> for x,y in enumerate (name):
>> lst=lst.append(y)
>> print (lst)
>> return (lst)
>> a=["1-800-getcharter"]
>> print (nametonumber(a))#18004382427837
>>
>>
>> The syntax for when to use a () and when to use [] still throws me a
>> curve.
>() is tuples which are immutable which means that the items can't be
>changed.  [] is list which means that each item can be changed.
>Tuples are useful because they can be used as keys in dictionaries and
>are guarantied not to change.  Lists are useful because they can be
>updated.
>
>What you are doing confuses me.  You don't use x, which is the enumerated 
>value.
>
>FIrst lst should be lst = [] .  You don't need to set the first
>element in the list to an empty string.  You just want to establish
>that you have an empty list called lst
>Second, you don't need lst = lst.append(y) because you can just say
>lst.append(y).  This will append the y value to the end of the list.
>As to converting letters to the corresponding numbers on a phone
>keypad, you don't show you code here for that
>>
>> For now, I am trying to end up with a list that has each character in
>> "a" as a single item.
>>
>> I get:
>> None
>> None
>> --
>> https://mail.python.org/mailman/listinfo/python-list

The lst=lst.append(y)
Was the mistake I never could see.

I am using enumerate just for practice.  To me that is just as easy as
typing len(something) and it seems more flexible.

and...the reason I don't show the code for the conversions is that
I haven't got that far yet.  :)

Thank you
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Re: I am out of trial and error again Lists

[Mark Lawrence]

On 22/10/2014 21:57, Joel Goldstick wrote:

On Wed, Oct 22, 2014 at 4:30 PM, Seymore4Head
 wrote:

def nametonumber(name):
 lst=[""]
 for x,y in enumerate (name):
 lst=lst.append(y)
 print (lst)
 return (lst)
a=["1-800-getcharter"]
print (nametonumber(a))#18004382427837


The syntax for when to use a () and when to use [] still throws me a
curve.

() is tuples which are immutable which means that the items can't be
changed.  [] is list which means that each item can be changed.
Tuples are useful because they can be used as keys in dictionaries and
are guarantied not to change.  Lists are useful because they can be
updated.



This is wrong, commas define tuples.

type mytest.py
a = 1, 2
print(type(a))

c:\Users\Mark\MyPython>mytest.py



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what you can do for our language.

Mark Lawrence

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Re: I am out of trial and error again Lists

[sohcahtoa82]

On Wednesday, October 22, 2014 2:06:35 PM UTC-7, Seymore4Head wrote:
> On Wed, 22 Oct 2014 16:57:00 -0400, Joel Goldstick
>  wrote:
> 
> >On Wed, Oct 22, 2014 at 4:30 PM, Seymore4Head
> > wrote:
> >> def nametonumber(name):
> >> lst=[""]
> >> for x,y in enumerate (name):
> >> lst=lst.append(y)
> >> print (lst)
> >> return (lst)
> >> a=["1-800-getcharter"]
> >> print (nametonumber(a))#18004382427837
> >>
> >>
> >> The syntax for when to use a () and when to use [] still throws me a
> >> curve.
> >() is tuples which are immutable which means that the items can't be
> >changed.  [] is list which means that each item can be changed.
> >Tuples are useful because they can be used as keys in dictionaries and
> >are guarantied not to change.  Lists are useful because they can be
> >updated.
> >
> >What you are doing confuses me.  You don't use x, which is the enumerated 
> >value.
> >
> >FIrst lst should be lst = [] .  You don't need to set the first
> >element in the list to an empty string.  You just want to establish
> >that you have an empty list called lst
> >Second, you don't need lst = lst.append(y) because you can just say
> >lst.append(y).  This will append the y value to the end of the list.
> >As to converting letters to the corresponding numbers on a phone
> >keypad, you don't show you code here for that
> >>
> >> For now, I am trying to end up with a list that has each character in
> >> "a" as a single item.
> >>
> >> I get:
> >> None
> >> None
> >> --
> >> https://mail.python.org/mailman/listinfo/python-list
> 
> The lst=lst.append(y)
> Was the mistake I never could see.
> 
> I am using enumerate just for practice.  To me that is just as easy as
> typing len(something) and it seems more flexible.
> 
> and...the reason I don't show the code for the conversions is that
> I haven't got that far yet.  :)
> 
> Thank you

I'm still confused as to why you're using enumerate.  Using it when you don't 
need to for "practice" just seems strange.  You don't even need to use 
len(something) in your case.  You should just be using 'for y in name:' if you 
don't need that x.  Enumerate is essentially just a shortcut to zipping a range 
based on the length.  For example...

for x, y in enumerate(name):

is equivalent to:

for x, y in zip(range(len(name)), name):

And both are pointless if you're not using the x.

Also, is there a reason why you're defining 'a' to be a list with a single 
string value, rather than just defining it as a string?  It seems like you 
should probably just have:

a = "1-800-getcharter"
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Wed, 22 Oct 2014 14:35:18 -0700 (PDT), sohcahto...@gmail.com wrote:

>On Wednesday, October 22, 2014 2:06:35 PM UTC-7, Seymore4Head wrote:
>> On Wed, 22 Oct 2014 16:57:00 -0400, Joel Goldstick
>>  wrote:
>> 
>> >On Wed, Oct 22, 2014 at 4:30 PM, Seymore4Head
>> > wrote:
>> >> def nametonumber(name):
>> >> lst=[""]
>> >> for x,y in enumerate (name):
>> >> lst=lst.append(y)
>> >> print (lst)
>> >> return (lst)
>> >> a=["1-800-getcharter"]
>> >> print (nametonumber(a))#18004382427837
>> >>
>> >>
>> >> The syntax for when to use a () and when to use [] still throws me a
>> >> curve.
>> >() is tuples which are immutable which means that the items can't be
>> >changed.  [] is list which means that each item can be changed.
>> >Tuples are useful because they can be used as keys in dictionaries and
>> >are guarantied not to change.  Lists are useful because they can be
>> >updated.
>> >
>> >What you are doing confuses me.  You don't use x, which is the enumerated 
>> >value.
>> >
>> >FIrst lst should be lst = [] .  You don't need to set the first
>> >element in the list to an empty string.  You just want to establish
>> >that you have an empty list called lst
>> >Second, you don't need lst = lst.append(y) because you can just say
>> >lst.append(y).  This will append the y value to the end of the list.
>> >As to converting letters to the corresponding numbers on a phone
>> >keypad, you don't show you code here for that
>> >>
>> >> For now, I am trying to end up with a list that has each character in
>> >> "a" as a single item.
>> >>
>> >> I get:
>> >> None
>> >> None
>> >> --
>> >> https://mail.python.org/mailman/listinfo/python-list
>> 
>> The lst=lst.append(y)
>> Was the mistake I never could see.
>> 
>> I am using enumerate just for practice.  To me that is just as easy as
>> typing len(something) and it seems more flexible.
>> 
>> and...the reason I don't show the code for the conversions is that
>> I haven't got that far yet.  :)
>> 
>> Thank you
>
>I'm still confused as to why you're using enumerate.  Using it when you don't 
>need to for "practice" just seems strange.  You don't even need to use 
>len(something) in your case.  You should just be using 'for y in name:' if you 
>don't need that x.  Enumerate is essentially just a shortcut to zipping a 
>range based on the length.  For example...
>
>for x, y in enumerate(name):
>
>is equivalent to:
>
>for x, y in zip(range(len(name)), name):
>
>And both are pointless if you're not using the x.
>
>Also, is there a reason why you're defining 'a' to be a list with a single 
>string value, rather than just defining it as a string?  It seems like you 
>should probably just have:
>
>a = "1-800-getcharter"


The lst=lst part was throwing an error I didn't understand.  That was
the only reason I added brackets around the 'a'.  I was getting in to
trial and error stuff.   Mostly error.

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Re: I am out of trial and error again Lists

[Seymore4Head]

On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
 wrote:

One more question.
if y in str(range(10)
Why doesn't that work.
I commented it out and just did it "long hand"

def nametonumber(name):
lst=[]
nx=[]
for x in (name):
lst.append(x)
for y in (lst):
#if y in str(range(10)):
if y in "1234567890":
nx.append(y)
if y in " -()":
nx.append(y)
if y in "abc":
nx.append("2")
if y in "def":
nx.append("3")
if y in "ghi":
nx.append("4")
if y in "jkl":
nx.append("5")
if y in "mno":
nx.append("6")
if y in "pqrs":
nx.append("7")
if y in "tuv":
nx.append("8")
if y in "wxyz":
nx.append("9")
number="".join(str(e) for e in nx)
return (number)
a="1-800-getcharter"
print (nametonumber(a))#1800 438 2427 837
a="1-800-leo laporte"
print (nametonumber(a))
a="1 800 callaprogrammer"
print (nametonumber(a))

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Re: I am out of trial and error again Lists

[Seymore4Head]

BTW I know I didn't check for Caps yet.

On Wed, 22 Oct 2014 18:30:17 -0400, Seymore4Head
 wrote:

>On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
> wrote:
>
>One more question.
>if y in str(range(10)
>Why doesn't that work.
>I commented it out and just did it "long hand"
>
>def nametonumber(name):
>lst=[]
>nx=[]
>for x in (name):
>lst.append(x)
>for y in (lst):
>#if y in str(range(10)):
>if y in "1234567890":
>nx.append(y)
>if y in " -()":
>nx.append(y)
>if y in "abc":
>nx.append("2")
>if y in "def":
>nx.append("3")
>if y in "ghi":
>nx.append("4")
>if y in "jkl":
>nx.append("5")
>if y in "mno":
>nx.append("6")
>if y in "pqrs":
>nx.append("7")
>if y in "tuv":
>nx.append("8")
>if y in "wxyz":
>nx.append("9")
>number="".join(str(e) for e in nx)
>return (number)
>a="1-800-getcharter"
>print (nametonumber(a))#1800 438 2427 837
>a="1-800-leo laporte"
>print (nametonumber(a))
>a="1 800 callaprogrammer"
>print (nametonumber(a))
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Re: I am out of trial and error again Lists

[Denis McMahon]

On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head wrote:

> def nametonumber(name):
> lst=[""]
> for x,y in enumerate (name):
> lst=lst.append(y)
> print (lst)
> return (lst)
> a=["1-800-getcharter"]
> print (nametonumber(a))#18004382427837
> 
> 
> The syntax for when to use a () and when to use [] still throws me a
> curve.
> 
> For now, I am trying to end up with a list that has each character in
> "a" as a single item.
> 
> I get:
> None None

First of all, an empty list is created with:

emptylist = []

whereas

x = [""]

creates a list containing one element, that element being an empty 
string. Not the same thing!

Did you try stepping through your code line by line in the interpreter to 
see what happened at each step?

note that append is a method of a list object, it has no return value, 
the original list is modified in place.

>>> l = ["a","b","c"]   # declare a list
>>> l.append( "d" ) # use the append method
>>> l   # show the list
['a', 'b', 'c', 'd']

So your line:

lst = lst.append(y)

should be:

lst.append(y)

Finally, did you really intend to pass a single element list into the 
function, or did you intend to pass a string into the function?

There is a difference between:

a=["1-800-getcharter"] 

which creates a single element list, the one element is the string "1-800-
getcharter", and:

a="1-800-getcharter"

which creates a string variable with the value "1-800-getcharter"

when you pass a list containing a string to your function, enumerate will 
look at each list element, so if your list contains one string, enumerate 
will return the pair 0, the_string, so the string gets appended to your 
empty list as a single item.

The code I think you wanted to write is as follows:

def nametonumber(name):
lst=[]
for x,y in enumerate(name):
lst.append(y)
return lst

a="1-800-getcharter"
print ( nametonumber(a) )

I suggests that you study very carefully the differences between this and 
your original code until you understand the reason and effect of every 
difference, as only by doing so will you discover the misconceptions 
which you seem to be operating under, and until you can get some of those 
straightened out, you're not going to make a lot of progress.

Try running the original code and my suggested alternative line by line 
in the interpreter, and examining the state of relevant variables after 
each line of execution.

Here's a code file with both your original code and my modified code with 
comprehensive print statements inserted for debugging. By referencing the 
debugging statements back to the code, you should be able to determine 
exactly where in your original code the value of "none" comes from.

### code starts

print ( "original code" )

def nametonumber(name):
print ("a) name =", name)
lst=[""]
print ( "b) lst = ", lst )
for x,y in enumerate (name):
print ( "c) x = ", x, "; y = ", y, "; lst = ", lst )
lst=lst.append(y)
print ( "d) lst = ", lst )
print (lst)
return (lst)

a=["1-800-getcharter"]
print ( "e) a = ", a )
print (nametonumber(a))

print ( "modified code" )

def nametonumber2(name):
print ("f) name =", name)
lst=[]
print ( "g) lst = ", lst )
for x,y in enumerate(name):
print ( "h) x = ", x, "; y = ", y, "; lst = ", lst )
lst.append(y)
print ( "i) lst = ", lst )
return lst

a="1-800-getcharter"
print ( "j) a = ", a )
print ( nametonumber2(a) )

### code ends

If you run the above code exactly as it is, you should see in the output 
how the enumeration reacts according to the different data it is given to 
enumerate, and also where lst is assigned the value none.

As I said above, getting your head round why this is happening is 
essential, and I really do suggest that you slow down and try and 
understand these basic concepts, because at the moment it seems you are 
striving to attempt more and more complicated things without 
understanding the basics upon which they are constructed, and like many 
other similar newsgroups, it's been my experience in the past that there 
is limited tolerance here for people who repeatedly make the same basic 
errors without learning from them.

-- 
Denis McMahon, denismfmcma...@gmail.com
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Re: I am out of trial and error again Lists

[Mark Lawrence]

On 22/10/2014 23:30, Seymore4Head wrote:

On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
 wrote:

One more question.
if y in str(range(10)
Why doesn't that work.


Invalid syntax, it should obviously be:-

if y in str(range(10)):

OTOH if you've simply mistyped above what did you expect to happen and 
what actually happened?  Give us the actual code that you ran and the 
full traceback.


--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence

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Re: I am out of trial and error again Lists

[MRAB]

On 2014-10-22 23:30, Seymore4Head wrote:

On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
 wrote:

One more question.
if y in str(range(10)
Why doesn't that work.


In what way doesn't it work?

If you want to know what it returns, print it out.


I commented it out and just did it "long hand"

def nametonumber(name):
 lst=[]
 nx=[]
 for x in (name):


You don't need to put the parens around name...


 lst.append(x)
 for y in (lst):


...or around lst.

You're iterating over a string, putting its characters into a list, and
then iterating over that list. You're doing more work than you need to!


 #if y in str(range(10)):
 if y in "1234567890":
 nx.append(y)
 if y in " -()":
 nx.append(y)
 if y in "abc":
 nx.append("2")
 if y in "def":
 nx.append("3")
 if y in "ghi":
 nx.append("4")
 if y in "jkl":
 nx.append("5")
 if y in "mno":
 nx.append("6")
 if y in "pqrs":
 nx.append("7")
 if y in "tuv":
 nx.append("8")
 if y in "wxyz":
 nx.append("9")
 number="".join(str(e) for e in nx)


The list nx already contains strings, so you don't need str here.


 return (number)


You don't need the parens here either.


a="1-800-getcharter"
print (nametonumber(a))#1800 438 2427 837
a="1-800-leo laporte"
print (nametonumber(a))
a="1 800 callaprogrammer"
print (nametonumber(a))



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Re: Problem with Android Build [SOLVED]

[Cyd Haselton]

On Tue, Oct 21, 2014 at 1:57 PM, Chris Angelico  wrote:

> On Wed, Oct 22, 2014 at 5:53 AM, Cyd Haselton  wrote:
> > I forgot to add...I also removed and/or commented out lines referencing
> > Modules/pwdmodule.o.
>
> Sounds like the normal sort of work involved in porting to a new
> platform. I've done a few of those kinds of jobs - ported Pike to
> OS/2, and to MinGW (there is an official Pike for Windows, but I
> wanted to use MinGW rather than MSVC), and there's a lot of fiddling
> around to be done!
>
> ChrisA
>

This problem is fixed. I'd previously removed Makefile commands to build
posixmodule.o because of an undeclared reference to I_PUSH.  After picking
through the source and history, I added them back in and instead added an
#ifndef __ANDROID__ around the function that used I_PUSH.

Ran a make clean...then make, and the newly built python was able to find
sysconfig and build pybuilddir.txt

Running into a different error, which I will post separate from this one.
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Wed, 22 Oct 2014 22:43:14 + (UTC), Denis McMahon
 wrote:

>On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head wrote:
>
>> def nametonumber(name):
>> lst=[""]
>> for x,y in enumerate (name):
>> lst=lst.append(y)
>> print (lst)
>> return (lst)
>> a=["1-800-getcharter"]
>> print (nametonumber(a))#18004382427837
>> 
>> 
>> The syntax for when to use a () and when to use [] still throws me a
>> curve.
>> 
>> For now, I am trying to end up with a list that has each character in
>> "a" as a single item.
>> 
>> I get:
>> None None
>
>First of all, an empty list is created with:
>
>emptylist = []
>
>whereas
>
>x = [""]
>
>creates a list containing one element, that element being an empty 
>string. Not the same thing!
>
>Did you try stepping through your code line by line in the interpreter to 
>see what happened at each step?
>
>note that append is a method of a list object, it has no return value, 
>the original list is modified in place.
>
 l = ["a","b","c"]   # declare a list
 l.append( "d" ) # use the append method
 l   # show the list
>['a', 'b', 'c', 'd']
>
>So your line:
>
>lst = lst.append(y)
>
>should be:
>
>lst.append(y)
>
>Finally, did you really intend to pass a single element list into the 
>function, or did you intend to pass a string into the function?
>
Those string errors were desperate attempts to fix the "append" error
I didn't understand.  

>There is a difference between:
>
>a=["1-800-getcharter"] 
>
>which creates a single element list, the one element is the string "1-800-
>getcharter", and:
>
>a="1-800-getcharter"
>
>which creates a string variable with the value "1-800-getcharter"
>
>when you pass a list containing a string to your function, enumerate will 
>look at each list element, so if your list contains one string, enumerate 
>will return the pair 0, the_string, so the string gets appended to your 
>empty list as a single item.
>
>The code I think you wanted to write is as follows:
>
>def nametonumber(name):
>lst=[]
>for x,y in enumerate(name):
>lst.append(y)
>return lst
>
>a="1-800-getcharter"
>print ( nametonumber(a) )
>
>I suggests that you study very carefully the differences between this and 
>your original code until you understand the reason and effect of every 
>difference, as only by doing so will you discover the misconceptions 
>which you seem to be operating under, and until you can get some of those 
>straightened out, you're not going to make a lot of progress.
>
>Try running the original code and my suggested alternative line by line 
>in the interpreter, and examining the state of relevant variables after 
>each line of execution.
>
>Here's a code file with both your original code and my modified code with 
>comprehensive print statements inserted for debugging. By referencing the 
>debugging statements back to the code, you should be able to determine 
>exactly where in your original code the value of "none" comes from.
>
>### code starts
>
>print ( "original code" )
>
>def nametonumber(name):
>print ("a) name =", name)
>lst=[""]
>print ( "b) lst = ", lst )
>for x,y in enumerate (name):
>print ( "c) x = ", x, "; y = ", y, "; lst = ", lst )
>lst=lst.append(y)
>print ( "d) lst = ", lst )
>print (lst)
>return (lst)
>
>a=["1-800-getcharter"]
>print ( "e) a = ", a )
>print (nametonumber(a))
>
>print ( "modified code" )
>
>def nametonumber2(name):
>print ("f) name =", name)
>lst=[]
>print ( "g) lst = ", lst )
>for x,y in enumerate(name):
>print ( "h) x = ", x, "; y = ", y, "; lst = ", lst )
>lst.append(y)
>print ( "i) lst = ", lst )
>return lst
>
>a="1-800-getcharter"
>print ( "j) a = ", a )
>print ( nametonumber2(a) )
>
>### code ends
>
>If you run the above code exactly as it is, you should see in the output 
>how the enumeration reacts according to the different data it is given to 
>enumerate, and also where lst is assigned the value none.
>
>As I said above, getting your head round why this is happening is 
>essential, and I really do suggest that you slow down and try and 
>understand these basic concepts, because at the moment it seems you are 
>striving to attempt more and more complicated things without 
>understanding the basics upon which they are constructed, and like many 
>other similar newsgroups, it's been my experience in the past that there 
>is limited tolerance here for people who repeatedly make the same basic 
>errors without learning from them.

Thanks a lot for all your suggestions.  I haven't learned to use the
interpreter yet.  I do plan on learning to use it.
The problem with that at the moment is that I have enrolled in an
online computer class.  They use Codeskulptor.
Codeskulptor code is not compatible to standard Python.  When I finish
the class I do plan on using Python 3 and will learn the Python 3
stuff.
Codeskulptor has a Viz mode.  I have tried using it, but so

Re: I am out of trial and error again Lists

[Seymore4Head]

On Wed, 22 Oct 2014 23:55:57 +0100, Mark Lawrence
 wrote:

>On 22/10/2014 23:30, Seymore4Head wrote:
>> On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
>>  wrote:
>>
>> One more question.
>> if y in str(range(10)
>> Why doesn't that work.
>
>Invalid syntax, it should obviously be:-
>
>if y in str(range(10)):
>
>OTOH if you've simply mistyped above what did you expect to happen and 
>what actually happened?  Give us the actual code that you ran and the 
>full traceback.

I don't get an error.  It just doesn't print the numbers correctly.
http://imgur.com/a/2loQV
You can click on the images and make them larger.
Thanks
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Mark Lawrence]

On 23/10/2014 00:26, Seymore4Head wrote:

On Wed, 22 Oct 2014 23:55:57 +0100, Mark Lawrence
 wrote:


On 22/10/2014 23:30, Seymore4Head wrote:

On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
 wrote:

One more question.
if y in str(range(10)
Why doesn't that work.


Invalid syntax, it should obviously be:-

if y in str(range(10)):

OTOH if you've simply mistyped above what did you expect to happen and
what actually happened?  Give us the actual code that you ran and the
full traceback.


I don't get an error.  It just doesn't print the numbers correctly.
http://imgur.com/a/2loQV
You can click on the images and make them larger.
Thanks



I suggest you try str(range(10)) from the interactive prompt and see 
exactly what you get, as it's nothing like what you expect :)


--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence

--
https://mail.python.org/mailman/listinfo/python-list

The “trials” in “trial and error” should be as simple as possible (was: I am out of trial and error again Lists)

[Ben Finney]

Seymore4Head  writes:

> Those string errors were desperate attempts to fix the "append" error
> I didn't understand.

It's normal when learning to get one's code into a mess.

But, when trying to trouble-shoot, please adopt the habit of
*simplifying* the examples, to better understand them.

At the least, when presenting code here for asking questions, ensure
you've make a simple-as-possible example showing the problem.

In other words: rip out any complicated parts to see if they're
relevant; if the same problem happens, that part wasn't relevant to the
example, and you should omit it when presenting it here.

You'll find that this will get you more sensible answers, As a bonus, it
will often lead to you understanding the problem enough to solve it!

-- 
 \  “At my lemonade stand I used to give the first glass away free |
  `\  and charge five dollars for the second glass. The refill |
_o__)contained the antidote.” —Emo Philips |
Ben Finney

-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Seymore4Head]

On Thu, 23 Oct 2014 00:44:01 +0100, Mark Lawrence
 wrote:

>On 23/10/2014 00:26, Seymore4Head wrote:
>> On Wed, 22 Oct 2014 23:55:57 +0100, Mark Lawrence
>>  wrote:
>>
>>> On 22/10/2014 23:30, Seymore4Head wrote:
 On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
  wrote:

 One more question.
 if y in str(range(10)
 Why doesn't that work.
>>>
>>> Invalid syntax, it should obviously be:-
>>>
>>> if y in str(range(10)):
>>>
>>> OTOH if you've simply mistyped above what did you expect to happen and
>>> what actually happened?  Give us the actual code that you ran and the
>>> full traceback.
>>
>> I don't get an error.  It just doesn't print the numbers correctly.
>> http://imgur.com/a/2loQV
>> You can click on the images and make them larger.
>> Thanks
>>
>
>I suggest you try str(range(10)) from the interactive prompt and see 
>exactly what you get, as it's nothing like what you expect :)

I see that now.  So there may not be a short hand solution.

Thanks
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Steven D'Aprano]

Seymore4Head wrote:

> Those string errors were desperate attempts to fix the "append" error
> I didn't understand.

Ah, the good ol' "make random changes to the code until the error goes away"
technique. You know that it never works, right?

Start by *reading the error message*, assuming you're getting an error
message. I'm the first person to admit that Python's error messages are not
always as clear as they should be, especially syntax errors, but still
there is a lot of information that can be gleamed from most error messages.
Take this attempt to use append:

py> mylist.append(23)
Traceback (most recent call last):
  File "", line 1, in 
NameError: name 'mylist' is not defined

That tells me that I have forgotten to define a variable mylist. So I fix
that:

py> mylist = 23
py> mylist.append(23)
Traceback (most recent call last):
  File "", line 1, in 
AttributeError: 'int' object has no attribute 'append'


That tells me that I can't append to a int. After googling for "Python
append" I learn that I can append to a list, so I try again:

py> mylist = []
py> mylist.append(23)
py> print(mylist)
[23]


Success!

If you are familiar with other programming languages, it might help to think
of append() as being like a procedure in Pascal, for example. You call
append() with an argument, but don't expect a return result.

Technically, *all* functions and methods in Python return something, even if
just the special value None, which can lead to "Gotchas!" like this one:

py> mylist = mylist.append(42)  # Don't do this!
py> print(mylist)  # I expect [23, 42] but get None instead.
None

Oops. One of the small annoyances of Python is that there is no way to tell
ahead of time, except by reading the documentation, whether something is a
proper function that returns a useful value, or a procedure-like function
that returns None. That's just something you have to learn.

The interactive interpreter is your friend. Learn to experiment at the
interactive interpreter -- you do know how to do that, don't you? If not,
ask. At the interactive interpreter, if a function or method returns a
value, it will be printed, *except for None*. So a function that doesn't
print anything might be procedure-like, and one which does print something
might not be:

py> mylist = [1, 5, 2, 6, 4, 3]
py> sorted(mylist)  # proper function returns a value
[1, 2, 3, 4, 5, 6]
py> mylist.sort()  # procedure-like function returns None
py> print(mylist)  # and modifies the list in place
[1, 2, 3, 4, 5, 6]


-- 
Steven

-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Seymore4Head]

On Thu, 23 Oct 2014 11:05:08 +1100, Steven D'Aprano
 wrote:

>Seymore4Head wrote:
>
>> Those string errors were desperate attempts to fix the "append" error
>> I didn't understand.
>
>Ah, the good ol' "make random changes to the code until the error goes away"
>technique. You know that it never works, right?
>
>Start by *reading the error message*, assuming you're getting an error
>message. I'm the first person to admit that Python's error messages are not
>always as clear as they should be, especially syntax errors, but still
>there is a lot of information that can be gleamed from most error messages.
>Take this attempt to use append:
>
>py> mylist.append(23)
>Traceback (most recent call last):
>  File "", line 1, in 
>NameError: name 'mylist' is not defined
>
>That tells me that I have forgotten to define a variable mylist. So I fix
>that:
>
>py> mylist = 23
>py> mylist.append(23)
>Traceback (most recent call last):
>  File "", line 1, in 
>AttributeError: 'int' object has no attribute 'append'
>
>
>That tells me that I can't append to a int. After googling for "Python
>append" I learn that I can append to a list, so I try again:
>
>py> mylist = []
>py> mylist.append(23)
>py> print(mylist)
>[23]
>
>
>Success!
>
>If you are familiar with other programming languages, it might help to think
>of append() as being like a procedure in Pascal, for example. You call
>append() with an argument, but don't expect a return result.
>
>Technically, *all* functions and methods in Python return something, even if
>just the special value None, which can lead to "Gotchas!" like this one:
>
>py> mylist = mylist.append(42)  # Don't do this!
>py> print(mylist)  # I expect [23, 42] but get None instead.
>None
>
>Oops. One of the small annoyances of Python is that there is no way to tell
>ahead of time, except by reading the documentation, whether something is a
>proper function that returns a useful value, or a procedure-like function
>that returns None. That's just something you have to learn.
>
>The interactive interpreter is your friend. Learn to experiment at the
>interactive interpreter -- you do know how to do that, don't you? If not,
>ask. At the interactive interpreter, if a function or method returns a
>value, it will be printed, *except for None*. So a function that doesn't
>print anything might be procedure-like, and one which does print something
>might not be:
>
>py> mylist = [1, 5, 2, 6, 4, 3]
>py> sorted(mylist)  # proper function returns a value
>[1, 2, 3, 4, 5, 6]
>py> mylist.sort()  # procedure-like function returns None
>py> print(mylist)  # and modifies the list in place
>[1, 2, 3, 4, 5, 6]

I am going to get around to learning the interpreter soon.

Thanks
-- 
https://mail.python.org/mailman/listinfo/python-list

the ressurection of ZOPE for web domination? bluebream and caveman the answer?

[johannes falcone]

i loved the rant about how zope would have all these features, and then some 
other python framework would come on with like 1 and act like its the bomb, and 
zope was like we been doing that and more for X years

those who dont study zope are doomed to repeat it!!!

is zope scoffing at drupal? botle? pyramid ?
-- 
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Re: (test) ? a:b

[Steven D'Aprano]

Michael Torrie wrote:

> On 10/22/2014 05:45 AM, Mark Lawrence wrote:
 without not:
 j = [j+1, 3][j>=10]
 with not:
 j = [3, j+1][not (j>=10)]

>>>
>>> Oh it's a trick !
>>> thx
>> 
>> IMHO it's just dreadful.  Why people insist on messing around like this
>> I really don't know, it just drives me nuts.
> 
> This actually was the standard idiom used by many python programs before
> Python 2.5.  But I agree.  Don't do this anymore! Python has a ternary
> if expression. Also the ternary if expression does, I believe
> short-circuit logic, so the non-chosen path is not calculated.  This
> hack does not. Could lead to interesting bugs depending on your
> assumptions.

Working code doesn't suddenly become non-working code just because Python
adds a second way to do something. The standard idiom

(value_if_false, value_if_true)[condition]

worked from (at least) Python 1.5 to Python 2.4, and it continues to work
today. There's nothing wrong with it: it does what it does, nothing more,
nothing less, and the only hashish part of this is that bools *are* ints,
with True == 1 and False == 0. That was heavily debated back in 2.3 or
thereabouts when bools were first introduced, and the decision made then
wasn't reverted in Python 3, so you can take it as By Design and not a
fluke of history. I think it is fair to say that Guido likes it that bools
are ints.

The older idiom isn't *exactly* the same as the ternary if operator, since
that short-circuits, but for many purposes short-circuiting is not
important or needed. If you don't need short-circuiting, or you dislike the
order of `value_if_true if condition else value_if_false`, or you need to
support Python 2.4 or older, or *simply because you like it*, there is
nothing wrong with using the older `sequence[flag]` idiom.

(It amuses me that not that many years ago the general attitude here was
that ternary if was an abomination that no right-thinking person should
ever use because the order of terms isn't identical to C, and now the
attitude seems to be that anyone *not* using ternary if is a heretic who
deserves to be set on fire :-)



-- 
Steven

-- 
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Re: I am out of trial and error again Lists

[MRAB]

On 2014-10-23 01:02, Seymore4Head wrote:

On Thu, 23 Oct 2014 00:44:01 +0100, Mark Lawrence
 wrote:


On 23/10/2014 00:26, Seymore4Head wrote:

On Wed, 22 Oct 2014 23:55:57 +0100, Mark Lawrence
 wrote:


On 22/10/2014 23:30, Seymore4Head wrote:

On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
 wrote:

One more question.
if y in str(range(10)
Why doesn't that work.


Invalid syntax, it should obviously be:-

if y in str(range(10)):

OTOH if you've simply mistyped above what did you expect to happen and
what actually happened?  Give us the actual code that you ran and the
full traceback.


I don't get an error.  It just doesn't print the numbers correctly.
http://imgur.com/a/2loQV
You can click on the images and make them larger.
Thanks



I suggest you try str(range(10)) from the interactive prompt and see
exactly what you get, as it's nothing like what you expect :)


I see that now.  So there may not be a short hand solution.


range(10) will yield the integers 0-9.

You can iterate over them, turning each into a string, and then join
them together:

''.join(str(n) for n in range(10))

--
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[MRAB]

On 2014-10-23 01:10, Seymore4Head wrote:

On Thu, 23 Oct 2014 11:05:08 +1100, Steven D'Aprano
 wrote:


Seymore4Head wrote:


Those string errors were desperate attempts to fix the "append" error
I didn't understand.


Ah, the good ol' "make random changes to the code until the error goes away"
technique. You know that it never works, right?

Start by *reading the error message*, assuming you're getting an error
message. I'm the first person to admit that Python's error messages are not
always as clear as they should be, especially syntax errors, but still
there is a lot of information that can be gleamed from most error messages.
Take this attempt to use append:

py> mylist.append(23)
Traceback (most recent call last):
 File "", line 1, in 
NameError: name 'mylist' is not defined

That tells me that I have forgotten to define a variable mylist. So I fix
that:

py> mylist = 23
py> mylist.append(23)
Traceback (most recent call last):
 File "", line 1, in 
AttributeError: 'int' object has no attribute 'append'


That tells me that I can't append to a int. After googling for "Python
append" I learn that I can append to a list, so I try again:

py> mylist = []
py> mylist.append(23)
py> print(mylist)
[23]


Success!

If you are familiar with other programming languages, it might help to think
of append() as being like a procedure in Pascal, for example. You call
append() with an argument, but don't expect a return result.

Technically, *all* functions and methods in Python return something, even if
just the special value None, which can lead to "Gotchas!" like this one:

py> mylist = mylist.append(42)  # Don't do this!
py> print(mylist)  # I expect [23, 42] but get None instead.
None

Oops. One of the small annoyances of Python is that there is no way to tell
ahead of time, except by reading the documentation, whether something is a
proper function that returns a useful value, or a procedure-like function
that returns None. That's just something you have to learn.

The interactive interpreter is your friend. Learn to experiment at the
interactive interpreter -- you do know how to do that, don't you? If not,
ask. At the interactive interpreter, if a function or method returns a
value, it will be printed, *except for None*. So a function that doesn't
print anything might be procedure-like, and one which does print something
might not be:

py> mylist = [1, 5, 2, 6, 4, 3]
py> sorted(mylist)  # proper function returns a value
[1, 2, 3, 4, 5, 6]
py> mylist.sort()  # procedure-like function returns None
py> print(mylist)  # and modifies the list in place
[1, 2, 3, 4, 5, 6]


I am going to get around to learning the interpreter soon.


Why wait?

You're trying to learn the language _now_, and checking things
interactively will help you.

--
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Re: I am out of trial and error again Lists

[alex23]

On 23/10/2014 10:02 AM, Seymore4Head wrote:

On Thu, 23 Oct 2014 00:44:01 +0100, Mark Lawrence
 wrote:

One more question.
if y in str(range(10)
Why doesn't that work.

I suggest you try str(range(10)) from the interactive prompt and see
exactly what you get, as it's nothing like what you expect :)


I see that now.  So there may not be a short hand solution.


There are two 'short hand solutions' to do what you want here:

import string

if y in string.digits:


Or even simpler:

if y.isdigit():
...

--
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Seymore4Head]

On Thu, 23 Oct 2014 02:31:57 +0100, MRAB 
wrote:

>On 2014-10-23 01:10, Seymore4Head wrote:
>> On Thu, 23 Oct 2014 11:05:08 +1100, Steven D'Aprano
>>  wrote:
>>
>>>Seymore4Head wrote:
>>>
 Those string errors were desperate attempts to fix the "append" error
 I didn't understand.
>>>
>>>Ah, the good ol' "make random changes to the code until the error goes away"
>>>technique. You know that it never works, right?
>>>
>>>Start by *reading the error message*, assuming you're getting an error
>>>message. I'm the first person to admit that Python's error messages are not
>>>always as clear as they should be, especially syntax errors, but still
>>>there is a lot of information that can be gleamed from most error messages.
>>>Take this attempt to use append:
>>>
>>>py> mylist.append(23)
>>>Traceback (most recent call last):
>>>  File "", line 1, in 
>>>NameError: name 'mylist' is not defined
>>>
>>>That tells me that I have forgotten to define a variable mylist. So I fix
>>>that:
>>>
>>>py> mylist = 23
>>>py> mylist.append(23)
>>>Traceback (most recent call last):
>>>  File "", line 1, in 
>>>AttributeError: 'int' object has no attribute 'append'
>>>
>>>
>>>That tells me that I can't append to a int. After googling for "Python
>>>append" I learn that I can append to a list, so I try again:
>>>
>>>py> mylist = []
>>>py> mylist.append(23)
>>>py> print(mylist)
>>>[23]
>>>
>>>
>>>Success!
>>>
>>>If you are familiar with other programming languages, it might help to think
>>>of append() as being like a procedure in Pascal, for example. You call
>>>append() with an argument, but don't expect a return result.
>>>
>>>Technically, *all* functions and methods in Python return something, even if
>>>just the special value None, which can lead to "Gotchas!" like this one:
>>>
>>>py> mylist = mylist.append(42)  # Don't do this!
>>>py> print(mylist)  # I expect [23, 42] but get None instead.
>>>None
>>>
>>>Oops. One of the small annoyances of Python is that there is no way to tell
>>>ahead of time, except by reading the documentation, whether something is a
>>>proper function that returns a useful value, or a procedure-like function
>>>that returns None. That's just something you have to learn.
>>>
>>>The interactive interpreter is your friend. Learn to experiment at the
>>>interactive interpreter -- you do know how to do that, don't you? If not,
>>>ask. At the interactive interpreter, if a function or method returns a
>>>value, it will be printed, *except for None*. So a function that doesn't
>>>print anything might be procedure-like, and one which does print something
>>>might not be:
>>>
>>>py> mylist = [1, 5, 2, 6, 4, 3]
>>>py> sorted(mylist)  # proper function returns a value
>>>[1, 2, 3, 4, 5, 6]
>>>py> mylist.sort()  # procedure-like function returns None
>>>py> print(mylist)  # and modifies the list in place
>>>[1, 2, 3, 4, 5, 6]
>>
>> I am going to get around to learning the interpreter soon.
>>
>Why wait?
>
>You're trying to learn the language _now_, and checking things
>interactively will help you.

Because most of the practice I am getting is not using Python.  I use
Codeskulptor.  

OK.Now is as good a time as ever.

Thanks
-- 
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Thu, 23 Oct 2014 11:37:27 +1000, alex23  wrote:

>On 23/10/2014 10:02 AM, Seymore4Head wrote:
>> On Thu, 23 Oct 2014 00:44:01 +0100, Mark Lawrence
>>  wrote:
>> One more question.
>> if y in str(range(10)
>> Why doesn't that work.
>>> I suggest you try str(range(10)) from the interactive prompt and see
>>> exactly what you get, as it's nothing like what you expect :)
>>
>> I see that now.  So there may not be a short hand solution.
>
>There are two 'short hand solutions' to do what you want here:
>
> import string
>
> if y in string.digits:
> 
>
>Or even simpler:
>
> if y.isdigit():
> ...
Thanks
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Seymore4Head]

On Wed, 22 Oct 2014 21:35:19 -0400, Seymore4Head
 wrote:

>On Thu, 23 Oct 2014 02:31:57 +0100, MRAB 
>wrote:
>
>>On 2014-10-23 01:10, Seymore4Head wrote:
>>> On Thu, 23 Oct 2014 11:05:08 +1100, Steven D'Aprano
>>>  wrote:
>>>
Seymore4Head wrote:

> Those string errors were desperate attempts to fix the "append" error
> I didn't understand.

Ah, the good ol' "make random changes to the code until the error goes away"
technique. You know that it never works, right?

Start by *reading the error message*, assuming you're getting an error
message. I'm the first person to admit that Python's error messages are not
always as clear as they should be, especially syntax errors, but still
there is a lot of information that can be gleamed from most error messages.
Take this attempt to use append:

py> mylist.append(23)
Traceback (most recent call last):
  File "", line 1, in 
NameError: name 'mylist' is not defined

That tells me that I have forgotten to define a variable mylist. So I fix
that:

py> mylist = 23
py> mylist.append(23)
Traceback (most recent call last):
  File "", line 1, in 
AttributeError: 'int' object has no attribute 'append'


That tells me that I can't append to a int. After googling for "Python
append" I learn that I can append to a list, so I try again:

py> mylist = []
py> mylist.append(23)
py> print(mylist)
[23]


Success!

If you are familiar with other programming languages, it might help to think
of append() as being like a procedure in Pascal, for example. You call
append() with an argument, but don't expect a return result.

Technically, *all* functions and methods in Python return something, even if
just the special value None, which can lead to "Gotchas!" like this one:

py> mylist = mylist.append(42)  # Don't do this!
py> print(mylist)  # I expect [23, 42] but get None instead.
None

Oops. One of the small annoyances of Python is that there is no way to tell
ahead of time, except by reading the documentation, whether something is a
proper function that returns a useful value, or a procedure-like function
that returns None. That's just something you have to learn.

The interactive interpreter is your friend. Learn to experiment at the
interactive interpreter -- you do know how to do that, don't you? If not,
ask. At the interactive interpreter, if a function or method returns a
value, it will be printed, *except for None*. So a function that doesn't
print anything might be procedure-like, and one which does print something
might not be:

py> mylist = [1, 5, 2, 6, 4, 3]
py> sorted(mylist)  # proper function returns a value
[1, 2, 3, 4, 5, 6]
py> mylist.sort()  # procedure-like function returns None
py> print(mylist)  # and modifies the list in place
[1, 2, 3, 4, 5, 6]
>>>
>>> I am going to get around to learning the interpreter soon.
>>>
>>Why wait?
>>
>>You're trying to learn the language _now_, and checking things
>>interactively will help you.
>
>Because most of the practice I am getting is not using Python.  I use
>Codeskulptor.  
>
>OK.Now is as good a time as ever.
>
>Thanks

Now I remember why...nothing happens
http://i.imgur.com/MIRpqzY.jpg

If I click on the shell window, I can get the grayed options to show
up for one turn.
I hit step and everything goes gray again.

http://i.imgur.com/NtMdmU1.jpg

Not a very fruitful exercise.  :(
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Re: I am out of trial and error again Lists

[Larry Hudson]

On 10/22/2014 05:02 PM, Seymore4Head wrote:

On Thu, 23 Oct 2014 00:44:01 +0100, Mark Lawrence
 wrote:


  (This is in reference to the line:  if y in str(range(10)):)


I suggest you try str(range(10)) from the interactive prompt and see
exactly what you get, as it's nothing like what you expect :)


I see that now.  So there may not be a short hand solution.



Yes there is.

That statement would work in Py2 as-is, but in Py3 it needs to be changed to:

if y in list(range(10)):

This give you a list not a string, but that's actually what you want here.  If you _really_ want 
a string, use join():  "".join(list(range(10)))


--  Also, from another post:  ---

> Thanks a lot for all your suggestions.  I haven't learned to use the
> interpreter yet.  I do plan on learning to use it.

You are making yourself work several hundred times harder if you don't learn to use the 
interactive mode!


> The problem with that at the moment is that I have enrolled in an
> online computer class.  They use Codeskulptor.
> Codeskulptor code is not compatible to standard Python.

I would be GREATLY bothered by the fact that it is not standard Python -- it sounds like a poor 
course.  You also seems to imply that you are only using those on-line tools and don't have a 
real Python installed on your own system.  If so, that is an extremely bad choice.  And if so, 
stop everything until you do install it NOW (definitely Python 3).  Then you can play with 
things on your own system IN THE INTERACTIVE MODE!!!  That's when you will start actually learning.


 -=- Larry -=-

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Re: I am out of trial and error again Lists

[Seymore4Head]

On Wed, 22 Oct 2014 19:58:24 -0700, Larry Hudson 
wrote:

>On 10/22/2014 05:02 PM, Seymore4Head wrote:
>> On Thu, 23 Oct 2014 00:44:01 +0100, Mark Lawrence
>>  wrote:
>
>  (This is in reference to the line:  if y in str(range(10)):)
>
>>> I suggest you try str(range(10)) from the interactive prompt and see
>>> exactly what you get, as it's nothing like what you expect :)
>>
>> I see that now.  So there may not be a short hand solution.
>>
>
>Yes there is.
>
>That statement would work in Py2 as-is, but in Py3 it needs to be changed to:
>
>if y in list(range(10)):
>
>This give you a list not a string, but that's actually what you want here.  If 
>you _really_ want 
>a string, use join():  "".join(list(range(10)))
>
>--  Also, from another post:  ---
>
> > Thanks a lot for all your suggestions.  I haven't learned to use the
> > interpreter yet.  I do plan on learning to use it.
>
>You are making yourself work several hundred times harder if you don't learn 
>to use the 
>interactive mode!
>
> > The problem with that at the moment is that I have enrolled in an
> > online computer class.  They use Codeskulptor.
> > Codeskulptor code is not compatible to standard Python.
>
>I would be GREATLY bothered by the fact that it is not standard Python -- it 
>sounds like a poor 
>course.  You also seems to imply that you are only using those on-line tools 
>and don't have a 
>real Python installed on your own system.  If so, that is an extremely bad 
>choice.  And if so, 
>stop everything until you do install it NOW (definitely Python 3).  Then you 
>can play with 
>things on your own system IN THE INTERACTIVE MODE!!!  That's when you will 
>start actually learning.
>
>  -=- Larry -=-

OK  Good advice.  I already have Py3 installed.  Interactive mode is
the next item on my do list.

Thanks
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Re: I am out of trial and error again Lists

[Rustom Mody]

On Thursday, October 23, 2014 8:28:39 AM UTC+5:30, Larry Hudson wrote:

> --  Also, from another post:  ---
>  > Thanks a lot for all your suggestions.  I haven't learned to use the
>  > interpreter yet.  I do plan on learning to use it.
> You are making yourself work several hundred times harder if you don't learn 
> to use the 
> interactive mode!

+100

That a noob takes the longest path out of noob-land --- well thats the 
fun of life!

That experienced people are not peremptory enough
-- yeah "My way or Highway" is rude but necessary at times -- is unfortunate.
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Re: (test) ? a:b

[Gregory Ewing]

Chris Angelico wrote:

I've seen much MUCH worse... where multiple conditional
expressions get combined arithmetically, and then the result used
somewhere...


In the days of old-school BASIC it was common to
exploit the fact that boolean expressions evaluated
to 0 or 1 (or -1, depending on your dialect :) to
achieve conditional expressions or other tricks.

Probably forgiveable, given that there were no real
conditional expressions, and every byte of your 48K
or less of RAM was precious...

--
Greg
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Re: (test) ? a:b

[Vito De Tullio]

Dennis Lee Bieber wrote:

>> x = [f(), g()] [cond]
>>
>>the latter evaluates both f() and g() instead of just one. Apart from
>>being inefficient, it can have unintended side-effects.
> 
> Ah, but what would
> 
> x = [f, g][cond]()
> 
> produce?


headache

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By ZeD

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Re: I am out of trial and error again Lists

[Terry Reedy]

On 10/22/2014 10:58 PM, Larry Hudson wrote:


This give you a list not a string, but that's actually what you want
here.  If you _really_ want a string, use join():  "".join(list(range(10)))


Interactive mode makes it really easy to test before posting.

>>> "".join(list(range(10)))

Traceback (most recent call last):
  File "", line 1, in 
"".join(list(range(10)))
TypeError: sequence item 0: expected string, int found

>>> "".join(str(i) for i in range(10))
'0123456789'

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(test) ? a:b

[ast]

Hello

Is there in Python something like:

j = (j >= 10) ? 3 : j+1;

as in C language ?

thx
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(-1)**1000

[ast]

Hello

If i am writing (-1)**1000 on a python program, will the
interpreter do (-1)*(-1)*...*(-1) or something clever ?

In fact i have (-1)**N with N an integer potentially big.

I do some tests that suggest that Python is clever

thx

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Re: (test) ? a:b

[Jean-Michel Pichavant]

- Original Message -
> From: "ast" 
> To: python-list@python.org
> Sent: Wednesday, 22 October, 2014 10:29:43 AM
> Subject: (test) ? a:b
> 
> Hello
> 
> Is there in Python something like:
> 
> j = (j >= 10) ? 3 : j+1;
> 
> as in C language ?
> 
> thx

j = 3 if j >=10 else j+1

Cheers

JM


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Re: (-1)**1000

[Jean-Michel Pichavant]

- Original Message -
> From: "ast" 
> To: python-list@python.org
> Sent: Wednesday, 22 October, 2014 10:27:34 AM
> Subject: (-1)**1000
> 
> Hello
> 
> If i am writing (-1)**1000 on a python program, will the
> interpreter do (-1)*(-1)*...*(-1) or something clever ?
> 
> In fact i have (-1)**N with N an integer potentially big.
> 
> I do some tests that suggest that Python is clever
> 
> thx

Python will yield the correct results. That is the most clever thing to do.
If you really worried about execution speed (I assume that what your question 
implies), Python may not be the language you need.

However, know that there are these modules "numpy" and "scipy" which are used 
by the scientific community which provide a python interface (it's a python 
module) but most of the heavy lifting is done in C (you can embed C in python 
code). 

For instance
http://docs.scipy.org/doc/numpy/reference/generated/numpy.power.html

Use this module if speed is what you're looking for.

JM


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Re: (-1)**1000

[Chris Angelico]

On Wed, Oct 22, 2014 at 7:27 PM, ast  wrote:
> If i am writing (-1)**1000 on a python program, will the
> interpreter do (-1)*(-1)*...*(-1) or something clever ?
>
> In fact i have (-1)**N with N an integer potentially big.

Exponentiation is far more efficient than the naive implementation of
iterated multiplication. Any modern programming language on any modern
CPU architecture should be able to handle this kind of thing. But even
the naive approach is likely to be fast enough.

>>> x=1
>>> for i in range(100): x*=-1

I had to go as far as a million iterations before this, implemented
purely in Python with absolutely no optimization, demonstrated a
visible pause (of about a quarter second) on my not-exactly-new
Windows laptop. My Linux desktop, with a rather hotter CPU, has no
trouble with a million, so I'd have to go higher to get a pause out of
it.

And actually, about half of that time is spent in the loop - replacing
the assignment with "pass" still leaves half the iteration time.

Poor performance is a crime. Python is innocent until proven guilty.
And the burden of proof is seldom met.

ChrisA
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Re: (test) ? a:b

[buscacio]

Em quarta-feira, 22 de outubro de 2014 06h29min55s UTC-2, ast  escreveu:
> Hello
> 
> 
> 
> Is there in Python something like:
> 
> 
> 
> j = (j >= 10) ? 3 : j+1;
> 
> 
> 
> as in C language ?
> 
> 
> 
> thx

without not:
j = [j+1, 3][j>=10]
with not:
j = [3, j+1][not (j>=10)]

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Re: (-1)**1000

[Peter Otten]

ast wrote:

> If i am writing (-1)**1000 on a python program, will the
> interpreter do (-1)*(-1)*...*(-1) or something clever ?
> 
> In fact i have (-1)**N with N an integer potentially big.
> 
> I do some tests that suggest that Python is clever

Let's see:

$ python3
Python 3.4.0 (default, Apr 11 2014, 13:05:11) 
[GCC 4.8.2] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import dis
>>> def f():
... return (-1)**1000
... 
>>> dis.dis(f)
  2   0 LOAD_CONST   4 (1)
  3 RETURN_VALUE

So yes, CPython replaces the expression (-1)**1000 with its value during 
compilation.

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Re: (test) ? a:b

[Chris Angelico]

On Wed, Oct 22, 2014 at 8:05 PM,   wrote:
> Em quarta-feira, 22 de outubro de 2014 06h29min55s UTC-2, ast  escreveu:
>> Hello
>>
>>
>>
>> Is there in Python something like:
>>
>>
>>
>> j = (j >= 10) ? 3 : j+1;
>>
>>
>>
>> as in C language ?
>>
>>
>>
>> thx
>
> without not:
> j = [j+1, 3][j>=10]
> with not:
> j = [3, j+1][not (j>=10)]

There's a distinct semantic difference there, though. Compare these:

/* C */
int x = (y != 0) ? 65536/y : 0;

# Python
x = 65536/y if y else 0

# Python, your way
x = [0, 65536/y][y!=0]

Do you see where the problem is?

Plus, subscripting a literal list is far less readable than the
ternary operator.

Also: Please can you avoid Google Groups, or if you must use it,
please at least clean up the excessive blank lines before you post.
I've left them so you can see what we have to cope with. Thanks!

ChrisA
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Re: (test) ? a:b

[Mark Lawrence]

On 22/10/2014 10:05, busca...@gmail.com wrote:

Em quarta-feira, 22 de outubro de 2014 06h29min55s UTC-2, ast  escreveu:

Hello



Is there in Python something like:



j = (j >= 10) ? 3 : j+1;



as in C language ?



thx


without not:
j = [j+1, 3][j>=10]
with not:
j = [3, j+1][not (j>=10)]



The death penalty should be reintroduced into the UK for two crimes, 
writing code like the above and using google groups.


--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence

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Re: (test) ? a:b

[ast]

 a écrit dans le message de 
news:7839376e-fc27-4299-ae63-4ddf17ef9...@googlegroups.com...

Em quarta-feira, 22 de outubro de 2014 06h29min55s UTC-2, ast  escreveu:

Hello



Is there in Python something like:



j = (j >= 10) ? 3 : j+1;



as in C language ?



thx


without not:
j = [j+1, 3][j>=10]
with not:
j = [3, j+1][not (j>=10)]



Oh it's a trick !
thx 


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Re: (-1)**1000

[ast]

"Chris Angelico"  a écrit dans le message de 
news:mailman.15058.1413968065.18130.python-l...@python.org...

On Wed, Oct 22, 2014 at 7:27 PM, ast  wrote:

If i am writing (-1)**1000 on a python program, will the
interpreter do (-1)*(-1)*...*(-1) or something clever ?

In fact i have (-1)**N with N an integer potentially big.


Exponentiation is far more efficient than the naive implementation of
iterated multiplication.


In the very particular case of (-1)**N,  I belive that Python check
the odd or even parity of N and provides the result accordingly.

I tried:

(-1)**10

1

(-1)**11

-1

and it is instantaneous


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Re: (test) ? a:b

[Chris Angelico]

On Wed, Oct 22, 2014 at 8:16 PM, Mark Lawrence  wrote:
>> without not:
>> j = [j+1, 3][j>=10]
>> with not:
>> j = [3, j+1][not (j>=10)]
>>
>
> The death penalty should be reintroduced into the UK for two crimes, writing
> code like the above and using google groups.

No no no. Code like that doesn't deserve death, just community
service. I've seen much MUCH worse... where multiple conditional
expressions get combined arithmetically, and then the result used
somewhere... I also may have been guilty of same, myself, though I'm
going to plead the internet's equivalent of the Fifth Amendment to the
US Constitution and not incriminate myself by showing the code...

ChrisA
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Re: (-1)**1000

[Peter Otten]

ast wrote:

> 
> "Chris Angelico"  a écrit dans le message de
> news:mailman.15058.1413968065.18130.python-l...@python.org...
>> On Wed, Oct 22, 2014 at 7:27 PM, ast  wrote:
>>> If i am writing (-1)**1000 on a python program, will the
>>> interpreter do (-1)*(-1)*...*(-1) or something clever ?
>>>
>>> In fact i have (-1)**N with N an integer potentially big.
>>
>> Exponentiation is far more efficient than the naive implementation of
>> iterated multiplication.
> 
> In the very particular case of (-1)**N,  I belive that Python check
> the odd or even parity of N and provides the result accordingly.
> 
> I tried:
 (-1)**10
> 1
 (-1)**11
> -1
> 
> and it is instantaneous

Not instantaneous once you defeat the peephole optimizer by introducing a 
variable:

$ python3 -m timeit '(-1)**101'
1000 loops, best of 3: 0.0356 usec per loop
$ python3 -m timeit -s'a = 101' '(-1)**a'
10 loops, best of 3: 3.23 usec per loop

When you increase the exponent you might discern a pattern:

$ python3 -m timeit -s 'a = 10**10' '(-1)**a'
100 loops, best of 3: 1.42 usec per loop
$ python3 -m timeit -s 'a = 10**100' '(-1)**a'
10 loops, best of 3: 11.6 usec per loop
$ python3 -m timeit -s 'a = 10**1000' '(-1)**a'
1 loops, best of 3: 101 usec per loop
$ python3 -m timeit -s 'a = 10**1' '(-1)**a'
1000 loops, best of 3: 992 usec per loop

That looks like log(a) while a parity check takes constant time:

$ python3 -m timeit -s 'a = 10**10' 'a & 1'
1000 loops, best of 3: 0.124 usec per loop
$ python3 -m timeit -s 'a = 10**100' 'a & 1'
1000 loops, best of 3: 0.124 usec per loop
$ python3 -m timeit -s 'a = 10**1000' 'a & 1'
1000 loops, best of 3: 0.122 usec per loop


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Re: Matplotlib: getting a figure to show without plt.show()

[Peter Otten]

Peter Pearson wrote:

> I'm using Matplotlib to present a "control" window with clickable
> buttons, and to plot things in another window when you click buttons,
> in the style of the code below.  I'm ashamed of the stinky way I
> use "first" to call plt.show the first time data are plotted but then
> to call fig.canvas.draw for subsequent data plots.  

> def callback(event):
> global n, first
> fig = plt.figure(2)
> fig.clear()
> plt.plot([0,1],[0,n])
> n += 1  # (Pretending something changes from one plot to the next.)
> if first:
> first = False
> plt.show()
> else:
> fig.canvas.draw()

> Can someone tell me the right way?

I don't see what's wrong with doing something different the first time 
around.

> If I call plt.show every time, then after about 40 plots I get
> "RuntimeError: maximum recursion depth exceeded while calling a Python
> object", which makes sense because I'm getting one layer deeper in
> callbacks with every plot (plt.show doesn't return).  But if I call
> fig.canvas.draw every time, the window for the data plot never appears
> on my screen.

If your backend uses tkinter, then the problem may be that plt.show() starts 
a new mainloop. Over here this causes the script to hang on termination. 
Through try-and-error I came up with

import matplotlib.pyplot as plt
from matplotlib.widgets import Button

def callback(event):
global n, fig2

if fig2 is None:
fig2 = plt.figure(2)

fig2.clear()
fig2.gca().plot([0, .5, 1], [0, 1/n, 1])
fig2.show()

n += 1 

n = 1
fig2 = None

plt.figure(1)

quit_button = Button(plt.axes([.1, .3, .4, .2]), "Quit")
quit_button.on_clicked(lambda x: plt.close("all"))

plot_button = Button(plt.axes([.1, .1, .4, .2]), "Plot")
plot_button.on_clicked(callback)

plt.show()

If you don't mind that the second window shows up immediately you can modify 
that to

import matplotlib.pyplot as plt
from matplotlib.widgets import Button

def callback(event):
global n

fig2.clear()
fig2.gca().plot([0, .5, 1], [0, 1/n, 1])
fig2.show()

n += 1 

n = 1

plt.figure(1)

quit_button = Button(plt.axes([.1, .3, .4, .2]), "Quit")
quit_button.on_clicked(lambda x: plt.close("all"))

plot_button = Button(plt.axes([.1, .1, .4, .2]), "Plot")
plot_button.on_clicked(callback)

fig2 = plt.figure(2)

plt.show()

and thus avoid the special case. 

As I'm not an expert for matplotlib you might also post your question on the 
matplotlib mailing list.

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Re: (-1)**1000

[Michiel Overtoom]

On Oct 22, 2014, at 12:29, Peter Otten wrote:

> That looks like log(a) while a parity check takes constant time:
> $ python3 -m timeit -s 'a = 10**10' 'a & 1'


Do you mean 'parity' as in http://en.wikipedia.org/wiki/Parity_bit ?  Because a 
parity bit denotes whether the *number* of '1' bits is even or odd, not the 
value of the least significant bit.

Greetings,

-- 
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- RiderOfGiraffes

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Re: (-1)**1000

[Tim Chase]

On 2014-10-22 12:29, Peter Otten wrote:
> That looks like log(a) while a parity check takes constant time:
> 
> $ python3 -m timeit -s 'a = 10**10' 'a & 1'
> 1000 loops, best of 3: 0.124 usec per loop
> $ python3 -m timeit -s 'a = 10**100' 'a & 1'
> 1000 loops, best of 3: 0.124 usec per loop
> $ python3 -m timeit -s 'a = 10**1000' 'a & 1'
> 1000 loops, best of 3: 0.122 usec per loop

Just for the record, this is a one-bit even/odd check (which is
useful & fast in this sign-of-large-exponent case), not a parity check
(which typically counts the number of "1" bits, adds the parity bit,
and asserts the result is even)

-tkc



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Re: Flush stdin

[Marko Rauhamaa]

Dan Stromberg :

> On Mon, Oct 20, 2014 at 9:41 PM, Marko Rauhamaa  wrote:
>> Terminal devices support line buffering on write.
> Yes, though that's not the only place it's useful.
>
>> Line buffering on read is an illusion created by higher-level libraries.
>> The low-level read function reads in blocks of bytes.
>
> Actually, doesn't line buffering sometimes exist inside an OS kernel?
> stty/termios/termio/sgtty relate here, for *ix examples.  Supporting
> code: http://stromberg.dnsalias.org/~strombrg/ttype/  It turns on
> character-at-a-time I/O in the tty driver via a variety of methods for
> portability.  I wrote it in C before I took an interest in Python.

I was being sloppy in my TTY terminology. A TTY device is running inside
the kernel and thus "writes" by copying bytes from its kernel buffer
into the user space when the user space process calls read(2).


Marko
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Re: (-1)**1000

[Peter Otten]

Michiel Overtoom wrote:

> 
> On Oct 22, 2014, at 12:29, Peter Otten wrote:
> 
>> That looks like log(a) while a parity check takes constant time:
>> $ python3 -m timeit -s 'a = 10**10' 'a & 1'
> 
> 
> Do you mean 'parity' as in http://en.wikipedia.org/wiki/Parity_bit ? 
> Because a parity bit denotes whether the *number* of '1' bits is even or
> odd, not the value of the least significant bit.

No, I meant the lsb. The OP introduced the term 'parity'; not sure if that 
was erroneous, too, or if there is an angle to the problem that escapes me.

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Re: (test) ? a:b

[Ned Batchelder]

On 10/22/14 5:05 AM, busca...@gmail.com wrote:

Em quarta-feira, 22 de outubro de 2014 06h29min55s UTC-2, ast  escreveu:

Hello



Is there in Python something like:



j = (j >= 10) ? 3 : j+1;



as in C language ?



thx


without not:
j = [j+1, 3][j>=10]
with not:
j = [3, j+1][not (j>=10)]



Why on earth would you recommend this outdated hack, when there's a true 
conditional operator?


j = 3 if j >= 10 else j+1

Of course, many people feel like the conditional operator isn't worth 
the squished-up unreadability, but if someone asks for a conditional 
operator, at least show them one!


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403 forbidden error

[diyaraik]

Hai,

  Could anyone please help me to resolve 403 forbidden error while logging into 
an application.

Following is the error details:

Traceback (most recent call last):
  File "./example6.py", line 18, in 
response = urllib2.urlopen(req)
  File "/usr/lib/python2.7/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
  File "/usr/lib/python2.7/urllib2.py", line 406, in open
response = meth(req, response)
  File "/usr/lib/python2.7/urllib2.py", line 519, in http_response
'http', request, response, code, msg, hdrs)
  File "/usr/lib/python2.7/urllib2.py", line 444, in error
return self._call_chain(*args)
  File "/usr/lib/python2.7/urllib2.py", line 378, in _call_chain
result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 527, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: FORBIDDEN


Sorry if the question is not relevant as im new to python.


Regards,
Diya
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Re: (test) ? a:b

[Mark Lawrence]

On 22/10/2014 10:27, Chris Angelico wrote:

On Wed, Oct 22, 2014 at 8:16 PM, Mark Lawrence  wrote:

without not:
j = [j+1, 3][j>=10]
with not:
j = [3, j+1][not (j>=10)]



The death penalty should be reintroduced into the UK for two crimes, writing
code like the above and using google groups.


No no no. Code like that doesn't deserve death, just community
service. I've seen much MUCH worse... where multiple conditional
expressions get combined arithmetically, and then the result used
somewhere... I also may have been guilty of same, myself, though I'm
going to plead the internet's equivalent of the Fifth Amendment to the
US Constitution and not incriminate myself by showing the code...

ChrisA



Perhaps you're correct.  Is there anything worse than looking at a 
dreadful piece of code that makes no sense at all and knowing that you'd 
written it six months earlier?


--
My fellow Pythonistas, ask not what our language can do for you, ask
what you can do for our language.

Mark Lawrence

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Re: (test) ? a:b

[Marko Rauhamaa]

Ned Batchelder :

> Why on earth would you recommend this outdated hack, when there's a
> true conditional operator? [...] if someone asks for a conditional
> operator, at least show them one!

No good deed goes unpunished.


Marko
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Re: (test) ? a:b

[Mark Lawrence]

On 22/10/2014 10:14, ast wrote:


 a écrit dans le message de
news:7839376e-fc27-4299-ae63-4ddf17ef9...@googlegroups.com...

Em quarta-feira, 22 de outubro de 2014 06h29min55s UTC-2, ast  escreveu:

Hello



Is there in Python something like:



j = (j >= 10) ? 3 : j+1;



as in C language ?



thx


without not:
j = [j+1, 3][j>=10]
with not:
j = [3, j+1][not (j>=10)]



Oh it's a trick !
thx


IMHO it's just dreadful.  Why people insist on messing around like this 
I really don't know, it just drives me nuts.


Also would you please access this list via 
https://mail.python.org/mailman/listinfo/python-list or read and action 
this https://wiki.python.org/moin/GoogleGroupsPython to prevent us 
seeing double line spacing and single line paragraphs, thanks.


--
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what you can do for our language.

Mark Lawrence

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Problems with selenium 2 and python 3.4.1

[novozhiloffvadim]

Hi all, i have a little problem. I have a simple automation to fill login form 
fields. Actually, it passes good, but there's the problem. I need to see actual 
output in my console after the script filled fields, like "Logged in 
successfully" or "Username not found". I tried many stuff, but nothing worked 
this way, my last try was while loop and it works great, but only when I have 
positive result. I wrote a second condition, but when I type incorrect data, it 
drives me crazy to see all these errors in my console. So here's the code and 
part of output, any help or thoughts would be appreciated.


from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import Select
from selenium.common.exceptions import NoSuchElementException

baseurl = "http://www.somesite/login";
email = input("Type an email: ")
password = input("Type a password: ")

xpaths = { 'loginBox' : "//input[@id='session_email']",
   'passwordBox' : "//input[@id='session_password']",
   'submitButton' :   "//input[@class='ufs-but']",
   'success' : "//div[@class='flash-message success']",
   'error' : "//span[@class='form_error']"
 }


mydriver = webdriver.Firefox()
mydriver.get(baseurl)


mydriver.find_element_by_xpath(xpaths['loginBox']).send_keys(email)


mydriver.find_element_by_xpath(xpaths['passwordBox']).send_keys(password)


mydriver.find_element_by_xpath(xpaths['submitButton']).click()


while mydriver.find_element_by_xpath(xpaths['success']):
print("Success")
if mydriver.find_element_by_xpath(xpaths['error']):
print("No")


And there's what I got when I try to interrupt an error:

File "ab.py", line 32, in 
while mydriver.find_element_by_xpath(xpaths['success']):
  File 
"/usr/local/lib/python3.4/site-packages/selenium-2.43.0-py3.4.egg/selenium/webdriver/remote/webdriver.py",
 line 230, in find_element_by_xpath
return self.find_element(by=By.XPATH, value=xpath)
  File 
"/usr/local/lib/python3.4/site-packages/selenium-2.43.0-py3.4.egg/selenium/webdriver/remote/webdriver.py",
 line 662, in find_element
{'using': by, 'value': value})['value']
  File 
"/usr/local/lib/python3.4/site-packages/selenium-2.43.0-py3.4.egg/selenium/webdriver/remote/webdriver.py",
 line 173, in execute
self.error_handler.check_response(response)
  File 
"/usr/local/lib/python3.4/site-packages/selenium-2.43.0-py3.4.egg/selenium/webdriver/remote/errorhandler.py",
 line 166, in check_response
raise exception_class(message, screen, stacktrace)
selenium.common.exceptions.NoSuchElementException: Message: 'Unable to locate 
element: {"method":"xpath","selector":"//div[@class=\'flash-message 
success\']"}' ; Stacktrace: 
at FirefoxDriver.prototype.findElementInternal_ 
(file:///tmp/tmpjax8kj1u/extensions/fxdri...@googlecode.com/components/driver-component.js:9618:26)
at FirefoxDriver.prototype.findElement 
(file:///tmp/tmpjax8kj1u/extensions/fxdri...@googlecode.com/components/driver-component.js:9627:3)
at DelayedCommand.prototype.executeInternal_/h 
(file:///tmp/tmpjax8kj1u/extensions/fxdri...@googlecode.com/components/command-processor.js:11612:16)
at DelayedCommand.prototype.executeInternal_ 
(file:///tmp/tmpjax8kj1u/extensions/fxdri...@googlecode.com/components/command-processor.js:11617:7)
at DelayedCommand.prototype.execute/< 
(file:///tmp/tmpjax8kj1u/extensions/fxdri...@googlecode.com/components/command-processor.js:11559:5)
 

As I said, successfull result ain't a problem. Hope to get any help.
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Re: 403 forbidden error

[Chris Angelico]

On Wed, Oct 22, 2014 at 10:36 PM,   wrote:
>   Could anyone please help me to resolve 403 forbidden error while logging 
> into an application.

That comes down tot he server you're talking to. Maybe your
username/password is wrong, or maybe you need to send back a cookie,
or something. If you do some web searches, you should be able to find
some info about HTTP; the better you understand the protocol, the more
you'll understand of what urllib2 is saying.

ChrisA
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Re: (test) ? a:b

[Rustom Mody]

On Wednesday, October 22, 2014 5:01:08 PM UTC+5:30, Ned Batchelder wrote:
> On 10/22/14 5:05 AM, buscacio wrote:
> > Em quarta-feira, 22 de outubro de 2014 06h29min55s UTC-2, ast  escreveu:
> >> Hello
> >> Is there in Python something like:
> >> j = (j >= 10) ? 3 : j+1;
> >> as in C language ?
> >> thx
> > without not:
> > j = [j+1, 3][j>=10]
> > with not:
> > j = [3, j+1][not (j>=10)]

> Why on earth would you recommend this outdated hack, when there's a true 
> conditional operator?


To learn a bit about the interchangeability of control and data structures?

[Just playing devil's advocate]

Doesn't change the fact that as a practice it should not be done
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Re: (-1)**1000

[Ian Kelly]

On Wed, Oct 22, 2014 at 5:02 AM, Peter Otten <__pete...@web.de> wrote:
> Michiel Overtoom wrote:
>
>>
>> On Oct 22, 2014, at 12:29, Peter Otten wrote:
>>
>>> That looks like log(a) while a parity check takes constant time:
>>> $ python3 -m timeit -s 'a = 10**10' 'a & 1'
>>
>>
>> Do you mean 'parity' as in http://en.wikipedia.org/wiki/Parity_bit ?
>> Because a parity bit denotes whether the *number* of '1' bits is even or
>> odd, not the value of the least significant bit.
>
> No, I meant the lsb. The OP introduced the term 'parity'; not sure if that
> was erroneous, too, or if there is an angle to the problem that escapes me.

Since the OP just wrote "parity", not "parity bit", I would assume
they meant as in http://en.wikipedia.org/wiki/Parity_(mathematics)
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Re: (-1)**1000

[Ian Kelly]

On Wed, Oct 22, 2014 at 4:43 AM, Tim Chase
 wrote:
> On 2014-10-22 12:29, Peter Otten wrote:
>> That looks like log(a) while a parity check takes constant time:
>>
>> $ python3 -m timeit -s 'a = 10**10' 'a & 1'
>> 1000 loops, best of 3: 0.124 usec per loop
>> $ python3 -m timeit -s 'a = 10**100' 'a & 1'
>> 1000 loops, best of 3: 0.124 usec per loop
>> $ python3 -m timeit -s 'a = 10**1000' 'a & 1'
>> 1000 loops, best of 3: 0.122 usec per loop
>
> Just for the record, this is a one-bit even/odd check

Which is just a verbose way of writing "parity check", even if that
phrase is usually used in another context.
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Re: Flush stdin

[random832]

On Tue, Oct 21, 2014, at 19:16, Dan Stromberg wrote:
> Actually, doesn't line buffering sometimes exist inside an OS kernel?
> stty/termios/termio/sgtty relate here, for *ix examples.  Supporting
> code: http://stromberg.dnsalias.org/~strombrg/ttype/  It turns on
> character-at-a-time I/O in the tty driver via a variety of methods for
> portability.  I wrote it in C before I took an interest in Python.

Yes, and 90% of the time, when someone says they want to "flush stdin",
what they really want to do is go to the next line after they've
sloppily read part of the line they're on (and the behavior they are
seeing that they object to is that their next read function reads the
rest of the current line). The appropriate course of action in these
cases is to actually read to the next newline and discard the data, not
to do any kind of flush.
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Re: (-1)**1000

[Ned Batchelder]

On 10/22/14 5:27 AM, ast wrote:


"Chris Angelico"  a écrit dans le message de
news:mailman.15058.1413968065.18130.python-l...@python.org...

On Wed, Oct 22, 2014 at 7:27 PM, ast  wrote:

If i am writing (-1)**1000 on a python program, will the
interpreter do (-1)*(-1)*...*(-1) or something clever ?

In fact i have (-1)**N with N an integer potentially big.


Exponentiation is far more efficient than the naive implementation of
iterated multiplication.


In the very particular case of (-1)**N,  I belive that Python check
the odd or even parity of N and provides the result accordingly.

I tried:

(-1)**10

1

(-1)**11

-1

and it is instantaneous




Keep in mind that actually calculating the exponentiation wouldn't do 
10 multiplications anyway: the clever 
way to do integer powers is by squaring based on the binary 
representation of the exponent.  It's explained here: 
http://stackoverflow.com/a/101613/14343


So even if Python is actually calculating the value, it's only doing 75 
multiplications or so.


--
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Re: (test) ? a:b

[alister]

On Wed, 22 Oct 2014 12:41:49 +0100, Mark Lawrence wrote:

> On 22/10/2014 10:27, Chris Angelico wrote:
>> On Wed, Oct 22, 2014 at 8:16 PM, Mark Lawrence
>>  wrote:
 without not:
 j = [j+1, 3][j>=10]
 with not:
 j = [3, j+1][not (j>=10)]


>>> The death penalty should be reintroduced into the UK for two crimes,
>>> writing code like the above and using google groups.
>>
>> No no no. Code like that doesn't deserve death, just community service.
>> I've seen much MUCH worse... where multiple conditional expressions get
>> combined arithmetically, and then the result used somewhere... I also
>> may have been guilty of same, myself, though I'm going to plead the
>> internet's equivalent of the Fifth Amendment to the US Constitution and
>> not incriminate myself by showing the code...
>>
>> ChrisA
>>
>>
> Perhaps you're correct.  Is there anything worse than looking at a
> dreadful piece of code that makes no sense at all and knowing that you'd
> written it six months earlier?

looking a a dreadful piece of unreadable & unfathomable code & knowing 
you wrote it only Yesterday ;-) 



-- 
Chamberlain's Laws:
(1) The big guys always win.
(2) Everything tastes more or less like chicken.
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Re: (test) ? a:b

[Chris Angelico]

On Thu, Oct 23, 2014 at 2:12 AM, alister
 wrote:
>> Perhaps you're correct.  Is there anything worse than looking at a
>> dreadful piece of code that makes no sense at all and knowing that you'd
>> written it six months earlier?
>
> looking a a dreadful piece of unreadable & unfathomable code & knowing
> you wrote it only Yesterday ;-)

Sounds like you have some experience with Perl...

ChrisA
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Re: (test) ? a:b

[alister]

On Thu, 23 Oct 2014 02:18:42 +1100, Chris Angelico wrote:

> On Thu, Oct 23, 2014 at 2:12 AM, alister
>  wrote:
>>> Perhaps you're correct.  Is there anything worse than looking at a
>>> dreadful piece of code that makes no sense at all and knowing that
>>> you'd written it six months earlier?
>>
>> looking a a dreadful piece of unreadable & unfathomable code & knowing
>> you wrote it only Yesterday ;-)
> 
> Sounds like you have some experience with Perl...
> 
> ChrisA

Never met the girl I deny everything.

Actual no experience in perl & my professional programming experience is 
limited to 8 bit assembler (68XX & 8051) 30 years ago. 
c & c++ were too low level for my liking so I started looking at python 
for fun on my Linux box's and was blown away by its smoothness.

-- 
On the other hand, life can be an endless parade of TRANSSEXUAL
QUILTING BEES aboard a cruise ship to DISNEYWORLD if only we let it!!
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Re: Flush stdin

[Marko Rauhamaa]

random...@fastmail.us:

> Yes, and 90% of the time, when someone says they want to "flush
> stdin", what they really want to do is go to the next line after
> they've sloppily read part of the line they're on (and the behavior
> they are seeing that they object to is that their next read function
> reads the rest of the current line). The appropriate course of action
> in these cases is to actually read to the next newline and discard the
> data, not to do any kind of flush.

I'm not sure I have seen that. However, somewhat analogously, there are
linux text utilities that read a number of lines and leave the input
intact. Since you can't really effectively read lines, the utilities
routinely read past the designated endpoint and then seek back to the
end of the line.

For example, consider this script:

seq 2 >test.dat
{
 head -n 5 >/dev/null
 head -n 5
} /dev/null
 head -n 5
}

I get:


1861
1862
1863
1864

because you can't seek back a pipe. The first "head" command has
greedily read in the first 1860 lines and the second one continues where
the first one left off.


Marko
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