Re: I am out of trial and error again Lists

[giacomo boffi]

Rustom Mody rustompm...@gmail.com writes:

 What would you say to a person who
 - Buys a Lambhorgini

I'd say: Don't buy a Lambhorgini from that nice guy you met at a party,
  but buy a Lamborghini by an authorized dealer  ;-)

-- 
I was a kid when Lamborghini launched the Miura!
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Side-effects [was Re: I am out of trial and error again Lists]

[Steven D'Aprano]

Chris Angelico wrote:

 On Sat, Oct 25, 2014 at 4:40 PM, Rustom Mody rustompm...@gmail.com
 wrote:
 Its generally accepted that side-effecting functions are not a good idea
 -- typically a function that returns something and changes global state.
 
 Only in certain circles. Not in Python. There are large numbers of
 functions with side effects (mutator methods like list.append,
 anything that needs lots of state like random.random, everything with
 external effect like I/O, heaps of stuff), and it is most definitely
 not frowned upon.

Hmmm. You might be overstating it a tad. You are absolutely correct that
Python is not a strict functional language with no side-effects.

However, mutator methods on a class don't change global state, they change
the state of an instance. Even random.random and friends don't change
global state, they change a (hidden) instance, and you can create your own
instances when needed. That gives you the convenience of pseudo-global
state while still allowing the flexibility of having separate instances.
Even global variables in Python are global to the module, not global to the
entire application.

So although Python code isn't entirely side-effect free, the general advice
to avoid writing code that relies on global state and operates via
side-effect still applies. To take an extreme example, we do this:

print(len(something))

not this:

LEN_ARGUMENT = something
len()
print(LEN_RESULT)

We *could* write code like that in Python, but as a general rule we don't.
If we did, it would be frowned upon. I would say that Python is a pragmatic
language which uses whatever idiom seems best at the time, but over all it
prefers mutable state for compound objects, prefers immutable state for
scalar objects (like numbers), and discourages the unnecessary use of
global mutable state.


-- 
Steven

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Re: Side-effects [was Re: I am out of trial and error again Lists]

[Chris Angelico]

On Sun, Oct 26, 2014 at 5:12 PM, Steven D'Aprano
steve+comp.lang.pyt...@pearwood.info wrote:
 However, mutator methods on a class don't change global state, they change
 the state of an instance. Even random.random and friends don't change
 global state, they change a (hidden) instance, and you can create your own
 instances when needed. That gives you the convenience of pseudo-global
 state while still allowing the flexibility of having separate instances.
 Even global variables in Python are global to the module, not global to the
 entire application.

True, but they do still change state; I oversimplified a bit, but
certainly there are plenty of functions that change state in some way,
and are not frowned upon.

Module level variables really are global, though. You can't easily and
conveniently reinstantiate a module in Python; if you import random;
random.seed(1234), you can confidently expect that some other module
that uses random.random will be affected. Sure, there are ways around
that, but that's true of any form of global state - for a start, it's
usually process level state, in that spawning a new process will
isolate one from another. The only thing that isn't truly global is
the namespace; I can write x = {} and you can write x = [] and
they don't collide. They're still global (process-level), it's just
that they're foo.x and bar.x and are thus distinct.

 I would say that Python is a pragmatic
 language which uses whatever idiom seems best at the time, but over all it
 prefers mutable state for compound objects, prefers immutable state for
 scalar objects (like numbers), and discourages the unnecessary use of
 global mutable state.

It's not really compound vs scalar, but yes, I agree. Practicality
beats purity, on so many levels. Functions with side effects aren't
considered some sort of weird beast that we have to permit for the
sake of I/O; they're a fundamental part of the language.

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

[Terry Reedy]

On 10/26/2014 1:08 AM, Dennis Lee Bieber wrote:

On Sat, 25 Oct 2014 18:48:59 -0400, Terry Reedy tjre...@udel.edu
declaimed the following:


C:\Users\Wulfraed\Documentspython3


You must have done something extra to make this work on Windows.


Possibly hand-edited my system PATH -- I've got a rather nasty one
(inserting line breaks)...


I was referring to the existence of 'python3' (and 'python2').  I 
presume these are created by the ActiveState installer that you said you 
used.  There are not created by the PSF Windows installer.  It instead 
creates 'py' linked to the new launcher; 'py -2' and 'py -3' launch the 
latest available python 2 or python3.


--
Terry Jan Reedy

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Re: I am out of trial and error again Lists

[Rustom Mody]

On Saturday, October 25, 2014 11:20:03 AM UTC+5:30, Chris Angelico wrote:
 On Sat, Oct 25, 2014 at 4:40 PM, Rustom Mody  wrote:
  Its generally accepted that side-effecting functions are not a good idea
  -- typically a function that returns something and changes global state.
 
 Only in certain circles. Not in Python. There are large numbers of
 functions with side effects (mutator methods like list.append,
 anything that needs lots of state like random.random, everything with
 external effect like I/O, heaps of stuff), and it is most definitely
 not frowned upon.
 
 In Python 3 (or Python 2 with the future directive), print is a
 function, print() an expression. It's not semantically a statement.

Ok
So give me a valid (ie useful) use where instead of the usual
l=[1,2,3]
l.append(4)

we have

foo(l.append(4))

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Re: I am out of trial and error again Lists

[Chris Angelico]

On Sat, Oct 25, 2014 at 4:55 PM, Rustom Mody rustompm...@gmail.com wrote:
 So give me a valid (ie useful) use where instead of the usual
 l=[1,2,3]
 l.append(4)

 we have

 foo(l.append(4))

Given that l.append(4) will always return None, there's not a lot of
point passing that return value to something, unless you're doing this
inside a lambda or something dumb like that. It won't be Pythonic.
Your point?

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

[Larry Hudson]

On 10/24/2014 09:37 AM, Seymore4Head wrote:
snip


import string

Not needed, delete it.


def nametonumber(name):
 lst=[]
 nx=[]
 digit=[]
Not needed.  You create digit as an empty list, them immediately follow by assigning a string to 
it (NOT a _list_ of characters, but an actual string.)



 digit=.join(str(i) for i in range(10))
digit is now the _string_ 0123456789.  Actually, a direct assignment here would be easier and 
shorter;

digit = 0123456789


 for x in name:
 lst.append(x)

lst is now a list of the characters in name.  That works, but you can do the 
same thing with:
   lst = list(name)


 for y in (lst):

Parentheses not needed.


 if y in lst(range(1,10)):
This is the line the traceback is referring to.  lst is a list, but the parentheses tell Python 
you are trying to call it as a function.  That is what the TypeError: 'list' object is not 
callable means.  Also it seems you are now trying to make a list of the digits 1-9 without the 
zero.  Is this what you want?  I don't think so, but if that IS what you want you can do it 
easier with slicing (look it up, VERY useful):  (Remember, digit is already a string of the ten 
digits, with zero as the first character.)

if y in digit[1:]:
Otherwise if you do want zero included you simply need:
if y in digit:


 #if y in 1234567890:
 #if y.isdigit():
 #if y in digit:

All of those should work, with the zero.


 #if y in string.digits:
Ok (I think?), but the string module is effectively obsolete -- most of its capabilities have 
been replaced with newer, better string methods.  It is RARELY useful.  I have never used it 
myself so don't know too much about it.  It is definitely not needed here, and is why I 
suggested deleting the import.



 nx.append(y)

 ...

 -=- Larry -=-

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Re: I am out of trial and error again Lists

[Larry Hudson]

On 10/24/2014 12:07 PM, Seymore4Head wrote:

On Fri, 24 Oct 2014 19:40:39 +0100, Mark Lawrence

snip

How many more times, state what you expect to happen and what actually
happens.  doesn't work is useless.  Please read this http://sscce.org/


Good suggestion.
OK  how is this?
It doesn't print what I expect.


That is NO different from the useless crap you consistently give us!  Tell us EXACTLY WHAT YOU 
EXPECT!!



Does it print what you expect?

Yes it does.  But what I expect is different from what you (erroneously) expect.


name=123-xyz-abc
for x in name:
 if x in range(10):
x is a character (a one-element string).  range(10) is a list of ints.  A string will never 
match an int.  BTW, as it is used here, range(10) is for Py2, for Py3 it needs to be 
list(range(10)).



 print (Range,(x))
 if x in str(range(10)):
Once again, find out what str(range(10)) actually is.  It is NOT what you think it is.  And I'll 
reiterate what everyone here keeps telling you:  USE THE INTERACTIVE MODE to see what really 
goes on.  If you keep resisting this you are making your understanding several hundred times 
more difficult.



 print (String range,(x))


Sorry for the harsh tone.  I'm old, and the curmudgeon is starting to come out 
in me.

 -=- Larry -=-

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Re: I am out of trial and error again Lists

[Mark Lawrence]

On 25/10/2014 03:41, Seymore4Head wrote:

On Fri, 24 Oct 2014 19:16:21 -0700, Larry Hudson org...@yahoo.com
wrote:


On 10/24/2014 07:38 AM, Seymore4Head wrote:
snip

I do get the difference.  I don't actually use Python 2.  I use
CodeSkulptor.  I do have Python 3 installed.  Actually I have Python 2
installed but IDLE defaults to Python 3.  So it is a pain to actually
load Python 2.



Exactly HOW are you trying to run Idle?  A default install of Py2 and Py3 in 
Windows should have
also installed Idle for each version.  In my Win7 system, they are BOTH in the 
standard menu,
you should be able to call up either one.

OT:  Side comment:  I rarely use Windows these days, maybe once every two or 
three months -- I
MUCH prefer Linux.  Among other reasons its a far better environment for 
programming.  I only
have one (active) system with Windows installed, and two others with Linux 
only.  Actually make
that three, if you count my Raspberry Pi.   :-)

  -=- Larry -=-

I have a directory of my py files.  I right click on one of the py
files and open with IDLE   Windows XP

If I try to open a py file I have for Python 2 it still opens using
IDLE 3.

I don't have many py 2 files anyway.



How about running your Python 2 version of IDLE and opening your files 
using File-Open or CTRL-O?


--
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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

[Steven D'Aprano]

Rustom Mody wrote:

 On Saturday, October 25, 2014 11:20:03 AM UTC+5:30, Chris Angelico wrote:
 On Sat, Oct 25, 2014 at 4:40 PM, Rustom Mody  wrote:
  Its generally accepted that side-effecting functions are not a good
  idea -- typically a function that returns something and changes global
  state.
 
 Only in certain circles. Not in Python. There are large numbers of
 functions with side effects (mutator methods like list.append,
 anything that needs lots of state like random.random, everything with
 external effect like I/O, heaps of stuff), and it is most definitely
 not frowned upon.
 
 In Python 3 (or Python 2 with the future directive), print is a
 function, print() an expression. It's not semantically a statement.
 
 Ok
 So give me a valid (ie useful) use where instead of the usual
 l=[1,2,3]
 l.append(4)
 
 we have
 
 foo(l.append(4))

Your question seems to be non-sequitor. To me, it doesn't appear to have any
relationship to Chris' comments.

But to answer your question, Ruby has mutator methods like list.append
return the list being mutated, to make it easy to chain multiple calls:

l.append(4).reverse().sort()


That would make it easy and convenient to modify a list immediately pass the
modified list to a function and embed it in an expression:

# Python's way
l.append(4)
values = [1, 2, foo(l), 3]

# Pseudo-Ruby way
values = [1, 2, foo(l.append(4)), 3]


Languages where all values are immutable *by definition* have to return a
new list, since you can't modify the original, hence they too can easily be
embedded in an expression.

Many people try to write something like this:

old_lists = [a, b, c, d]
new_lists = [thelist.append(x) for thelist in old_lists if len(thelist)  5]

only to be unpleasantly surprised to discover than new_lists now contains
None instead of the lists. Having methods return a result rather than
behave like a procedure is a valid (i.e. useful) design choice.


-- 
Steven

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Re: I am out of trial and error again Lists

[Ian Kelly]

On Sat, Oct 25, 2014 at 12:46 AM, Larry Hudson
org...@yahoo.com.dmarc.invalid wrote:
 name=123-xyz-abc
 for x in name:
  if x in range(10):

 x is a character (a one-element string).  range(10) is a list of ints.  A
 string will never match an int.  BTW, as it is used here, range(10) is for
 Py2, for Py3 it needs to be list(range(10)).

The last comment is incorrect. You can do membership tests on a Python
3 range object:

 range(2, 4)
range(2, 4)
 for i in range(5):
...   print(i, i in range(2, 4))
...
0 False
1 False
2 True
3 True
4 False

You can also index them:
 range(50, 100)[37]
87

slice them:
 range(0, 100, 2)[:25]
range(0, 50, 2)

get their length:
 len(range(25, 75))
50

search them:
 range(25, 75).index(45)
20

and generally do most operations that you would expect to be able to
do with an immutable sequence type, with the exceptions of
concatenation and multiplication. And as I observed elsewhere in the
thread, such operations are often more efficient on range objects than
by constructing and operating on the equivalent lists.
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Re: I am out of trial and error again Lists

[Grant Edwards]

On 2014-10-24, Denis McMahon denismfmcma...@gmail.com wrote:
 On Fri, 24 Oct 2014 10:38:31 -0400, Seymore4Head wrote:

 Thanks everyone for your suggestions.

 Try loading the following in codeskulptor:

 http://www.codeskulptor.org/#user38_j6kGKgeOMr_0.py

No.

We[1] aren't intested in whatever Python-like language is implemented
in codeskulptor.


[1] I know I'm being a bit presumptuous writing the the plural first
person rather than the signular.  If you _are_ interested in
codeskulptor python then jump in...

-- 
Grant


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Re: I am out of trial and error again Lists

[Rustom Mody]

On Saturday, October 25, 2014 1:15:09 PM UTC+5:30, Steven D'Aprano wrote:
 Rustom Mody wrote:
 
  On Saturday, October 25, 2014 11:20:03 AM UTC+5:30, Chris Angelico wrote:
  On Sat, Oct 25, 2014 at 4:40 PM, Rustom Mody  wrote:
   Its generally accepted that side-effecting functions are not a good
   idea -- typically a function that returns something and changes global
   state.
  
  Only in certain circles. Not in Python. There are large numbers of
  functions with side effects (mutator methods like list.append,
  anything that needs lots of state like random.random, everything with
  external effect like I/O, heaps of stuff), and it is most definitely
  not frowned upon.
  
  In Python 3 (or Python 2 with the future directive), print is a
  function, print() an expression. It's not semantically a statement.
  
  Ok
  So give me a valid (ie useful) use where instead of the usual
  l=[1,2,3]
  l.append(4)
  
  we have
  
  foo(l.append(4))
 
 Your question seems to be non-sequitor. To me, it doesn't appear to have any
 relationship to Chris' comments.

I am going to leave undisturbed Seymore's thread for whom these nit-picks are 
unlikely to be helpful

Answer in another one
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Sat, 25 Oct 2014 14:23:44 -0400, Dennis Lee Bieber
wlfr...@ix.netcom.com wrote:

On Fri, 24 Oct 2014 10:38:31 -0400, Seymore4Head
Seymore4Head@Hotmail.invalid declaimed the following:


I do get the difference.  I don't actually use Python 2.  I use
CodeSkulptor.  I do have Python 3 installed.  Actually I have Python 2
installed but IDLE defaults to Python 3.  So it is a pain to actually
load Python 2.


   So don't use Idle... Open up a Windows command shell and invoke Python
without giving a script file to execute.

C:\Users\Wulfraed\Documentspython
ActivePython 2.7.5.6 (ActiveState Software Inc.) based on
Python 2.7.5 (default, Sep 16 2013, 23:11:01) [MSC v.1500 64 bit (AMD64)]
on win32
Type help, copyright, credits or license for more information.
 exit()

C:\Users\Wulfraed\Documentspython3
ActivePython 3.3.2.0 (ActiveState Software Inc.) based on
Python 3.3.2 (default, Sep 16 2013, 23:11:39) [MSC v.1600 64 bit (AMD64)]
on win32
Type help, copyright, credits or license for more information.
 exit()

C:\Users\Wulfraed\Documents

   Note that my default Python is still 2.7, but I also have Python 3.3
installed, and can access it using python3 (I can also ensure Python 2.7
by using python2)



I tried list(range(10)  I thought that would work in Python 3.  It
didn't.  I spent quite a bit of time last night trying to come up with

   Really? what did it do? SHOW US!

C:\Users\Wulfraed\Documentspython3
ActivePython 3.3.2.0 (ActiveState Software Inc.) based on
Python 3.3.2 (default, Sep 16 2013, 23:11:39) [MSC v.1600 64 bit (AMD64)]
on win32
Type help, copyright, credits or license for more information.
 list(range(10))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 print(list(range(10)))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 repr(list(range(10)))
'[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]'


   Recall that print is a function in Python 3, so the extra layer of ()
are required.

C:\Users\Wulfraed\Documentspython
ActivePython 2.7.5.6 (ActiveState Software Inc.) based on
Python 2.7.5 (default, Sep 16 2013, 23:11:01) [MSC v.1500 64 bit (AMD64)]
on win32
Type help, copyright, credits or license for more information.
 list(range(10))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 print list(range(10))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 print(list(range(10)))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 repr(list(range(10)))
'[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]'


   Print doesn't need the (), but they also don't hurt it in Python 2.
(still in Py2)

 range(10)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 xrange(10)
xrange(10)
 print range(10)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 print xrange(10)
xrange(10)
 exit
Use exit() or Ctrl-Z plus Return to exit
 exit()

   Now Py3

C:\Users\Wulfraed\Documentspython3
ActivePython 3.3.2.0 (ActiveState Software Inc.) based on
Python 3.3.2 (default, Sep 16 2013, 23:11:39) [MSC v.1600 64 bit (AMD64)]
on win32
Type help, copyright, credits or license for more information.
 range(10)
range(0, 10)
 xrange(10)
Traceback (most recent call last):
  File stdin, line 1, in module
NameError: name 'xrange' is not defined
 print(range(10))
range(0, 10)


I am not working on that any more.  I have my assignment for this week
so I am done with busy work.

Thanks for all your help
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Re: I am out of trial and error again Lists

[Denis McMahon]

On Sat, 25 Oct 2014 15:01:54 +, Grant Edwards wrote:

 On 2014-10-24, Denis McMahon denismfmcma...@gmail.com wrote:
 On Fri, 24 Oct 2014 10:38:31 -0400, Seymore4Head wrote:

 Thanks everyone for your suggestions.

 Try loading the following in codeskulptor:

 http://www.codeskulptor.org/#user38_j6kGKgeOMr_0.py

 No.

I wasn't replying to you, I was replying to S4H.

 We[1] aren't intested in whatever Python-like language is implemented in
 codeskulptor.

However S4H may be, and one thing I can be sure is that if I give him a 
cs url, he will at least be running the exact same python code I typed in 
the same python environment, and hence I know what results he should see, 
namely the same ones I saw.

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

[Denis McMahon]

On Fri, 24 Oct 2014 20:15:02 -0400, Seymore4Head wrote:

 On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
 Seymore4Head@Hotmail.invalid wrote:
 
 name=012

name is a string of 3 characters

 b=list(range(3))

b is a list of 3 numbers

 print (name[1])

name[1] is the string 1

 print (b[1])

b[1] is the number 1

 if name[1] == b[1]:
 print (Eureka!)

This didn't happen

 else:
 print (OK, I get it)

This happened

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

[Terry Reedy]

On 10/25/2014 2:23 PM, Dennis Lee Bieber wrote:

On Fri, 24 Oct 2014 10:38:31 -0400, Seymore4Head
Seymore4Head@Hotmail.invalid declaimed the following:



I do get the difference.  I don't actually use Python 2.  I use
CodeSkulptor.  I do have Python 3 installed.  Actually I have Python 2
installed but IDLE defaults to Python 3.


This is wrong.  Pythonx.y runs Idlex.y, not vice versa.
Seymore installed some version of Python 3 as default, as he should 
have, and 'Edit with Idle' opens the default version of Python, with its 
version of Idle.


 So it is a pain to actually load Python 2.

Not really, neither on linux or Windows.


So don't use Idle...


Since Seymore analysis is wrong, this advice in not the answer.
If he is running on Windows, it is dubious in that the Windows console 
interpreter is a pain to use.


 Open up a Windows command shell and invoke Python

without giving a script file to execute.

C:\Users\Wulfraed\Documentspython


python -m idlelib will open the corresponding Idle. So will the Start 
menu entry or an icon pinned to the taskbar.



C:\Users\Wulfraed\Documentspython3


You must have done something extra to make this work on Windows.

--
Terry Jan Reedy

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Re: I am out of trial and error again Lists

[Mark Lawrence]

On 25/10/2014 23:48, Terry Reedy wrote:

On 10/25/2014 2:23 PM, Dennis Lee Bieber wrote:

On Fri, 24 Oct 2014 10:38:31 -0400, Seymore4Head
Seymore4Head@Hotmail.invalid declaimed the following:



I do get the difference.  I don't actually use Python 2.  I use
CodeSkulptor.  I do have Python 3 installed.  Actually I have Python 2
installed but IDLE defaults to Python 3.


This is wrong.  Pythonx.y runs Idlex.y, not vice versa.
Seymore installed some version of Python 3 as default, as he should
have, and 'Edit with Idle' opens the default version of Python, with its
version of Idle.

  So it is a pain to actually load Python 2.

Not really, neither on linux or Windows.


So don't use Idle...


Since Seymore analysis is wrong, this advice in not the answer.
If he is running on Windows, it is dubious in that the Windows console
interpreter is a pain to use.

  Open up a Windows command shell and invoke Python

without giving a script file to execute.

C:\Users\Wulfraed\Documentspython


python -m idlelib will open the corresponding Idle. So will the Start
menu entry or an icon pinned to the taskbar.


C:\Users\Wulfraed\Documentspython3


You must have done something extra to make this work on Windows.



The Python launcher for Windows should be taken into account here 
http://legacy.python.org/dev/peps/pep-0397/


--
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

[Mark Lawrence]

On 22/10/2014 21:30, 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



Following on from the numerous responses you've had here, I've no idea 
if this helps your thought processes but there's only one way for you to 
find out http://www.greenteapress.com/thinkpython/ :)


--
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

[Mark Lawrence]

On 24/10/2014 08:05, Mark Lawrence wrote:

On 22/10/2014 21:30, 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



Following on from the numerous responses you've had here, I've no idea
if this helps your thought processes but there's only one way for you to
find out http://www.greenteapress.com/thinkpython/ :)



And another http://tinyurl.com/k26vjhr

--
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

Re: I am out of trial and error again Lists

[Seymore4Head]

On Thu, 23 Oct 2014 21:56:31 -0700 (PDT), Rustom Mody
rustompm...@gmail.com wrote:

On Thursday, October 23, 2014 10:33:57 PM UTC+5:30, Seymore4Head wrote:
 On Thu, 23 Oct 2014 15:55:35 + (UTC), Denis McMahon wrote:
 
 On Thu, 23 Oct 2014 10:04:56 -0400, Seymore4Head wrote:
 
  On Thu, 23 Oct 2014 09:15:16 + (UTC), Denis McMahon wrote:
 
 Try the following 3 commands at the console:
 
 You obviously didn't, so I'll try again. Try each of the following three 
 commands in the python console at the  prompt.
 
 1) 10
 10
 
 2) range(10)
 range(0, 10)
 
 3) str(range(10))
 'range(0, 10)'
 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 
 Show *and* describe the output in each case. Describing the output that 
 you see is actually the key here, as it will allow us to assess whether 
 you understand what you are actually seeing or not, and if you don't 
 understand the output you see in the console, then we need to fix that 
 very fundamental and basic issue before moving on to more complex stuff!
 
  Ok Thanks
 
 You were expected to answer the question in the original. I have now set 
 it as a clearer and more specific task.
 
 If you're not going to do these things that are intended to help you 
 learn some of the basic features of the language, then I and everyone 
 else here that has so far been attempting to help you are wasting our 
 time.
 
 I did try them.  I may have missed replying your to your specific
 comment, but I tried them.
 
 BTW str(range (10)) does work with Python 2 which is where I may have
 got the idea.  I happened to be using Python 3 at the time I tried to
 implement it.  It is a little confusing jumping back and forth, but
 for the moment, I am going to tough it out.
 
 I do appreciate all the help too.

Hi Seymore!

Happy to see that you are moving on from
reading much; understanding nothing; thrashing

to

reading a bit; understanding a bit
[And thanks to Denis to getting you out of your confusion-hole]

So heres a small additional question set that I promise will more than repay
you your time.

Better done in python 2. But if you use python3, below replace
range(10)
with
list(range(10))

So now in the python console, please try

a.
 range(10)

and 

b.
 print (range(10))

And then post back (without consulting google!!)¹

1. Are they same or different?

2. If same, how come different expressions are same?

3. If different whats the difference?

4. [Most important]: When would you prefer which?

=
¹ Actually its ok to consult google AFTER you try

I do get the difference.  I don't actually use Python 2.  I use
CodeSkulptor.  I do have Python 3 installed.  Actually I have Python 2
installed but IDLE defaults to Python 3.  So it is a pain to actually
load Python 2.

Range(10) stores the min max values and loads each number in between
when needed.  Ian explained that very clearly.

I tried list(range(10)  I thought that would work in Python 3.  It
didn't.  I spent quite a bit of time last night trying to come up with
the right combination of str and int commands to make range(10) work
with my simple example.   It didn't.  I am pretty frustrated.  I am
just skipping that little bit of code for the moment.

Thanks everyone for your suggestions.
-- 
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 08:05:01 +0100, Mark Lawrence
breamore...@yahoo.co.uk wrote:

On 22/10/2014 21:30, 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


Following on from the numerous responses you've had here, I've no idea 
if this helps your thought processes but there's only one way for you to 
find out http://www.greenteapress.com/thinkpython/ :)

I have at least 10 ebooks.  I will get around to reading them soon.
http://i.imgur.com/rpOcKP8.jpg

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

[Chris Angelico]

On Sat, Oct 25, 2014 at 1:38 AM, Seymore4Head
Seymore4Head@hotmail.invalid wrote:
 I tried list(range(10)  I thought that would work in Python 3.  It
 didn't.

This is your problem: You say it didn't work. That is almost *never*
the right thing to say or to think. What happened when you tried that?
Did you get a SyntaxError because of the omitted close parenthesis?
Did the interpreter prompt for more input? Did a velociraptor come out
of nowhere and try to kill you [1]?

When you come back to python-list, you should say exactly what you did
and exactly what happened, not I tried X and it didn't work. Copy
and paste from your interactive session - do NOT retype, because you
introduce new errors. It's very hard to help you when you don't
explain what you're doing, and just keep on telling us how frustrated
you are.

ChrisA

[1] http://xkcd.com/292/
-- 
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 09:12:28 +0100, Mark Lawrence
breamore...@yahoo.co.uk wrote:

On 24/10/2014 08:05, Mark Lawrence wrote:
 On 22/10/2014 21:30, 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


 Following on from the numerous responses you've had here, I've no idea
 if this helps your thought processes but there's only one way for you to
 find out http://www.greenteapress.com/thinkpython/ :)


And another http://tinyurl.com/k26vjhr

Google.   I have heard of that.  :)  
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Seymore4Head]

On Sat, 25 Oct 2014 01:51:41 +1100, Chris Angelico ros...@gmail.com
wrote:

On Sat, Oct 25, 2014 at 1:38 AM, Seymore4Head
Seymore4Head@hotmail.invalid wrote:
 I tried list(range(10)  I thought that would work in Python 3.  It
 didn't.

This is your problem: You say it didn't work. That is almost *never*
the right thing to say or to think. What happened when you tried that?
Did you get a SyntaxError because of the omitted close parenthesis?
Did the interpreter prompt for more input? Did a velociraptor come out
of nowhere and try to kill you [1]?

I understand that it makes it easier for you if I can describe better
the error I get, but by the time I ask for help here I have tried many
different things to get the error to go away.

I will try in the future to do better at this.  For now, I am just
putting that exercise behind me.  I do try to find out how to do
things before asking first.

I tried so many things last night, I had to go back to an old message
to get code that worked again.

Thanks

When you come back to python-list, you should say exactly what you did
and exactly what happened, not I tried X and it didn't work. Copy
and paste from your interactive session - do NOT retype, because you
introduce new errors. It's very hard to help you when you don't
explain what you're doing, and just keep on telling us how frustrated
you are.

ChrisA

[1] http://xkcd.com/292/
-- 
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Re: I am out of trial and error again Lists

[Chris Angelico]

On Sat, Oct 25, 2014 at 2:04 AM, Seymore4Head
Seymore4Head@hotmail.invalid wrote:
 I understand that it makes it easier for you if I can describe better
 the error I get, but by the time I ask for help here I have tried many
 different things to get the error to go away.

That's part of the problem. You let yourself get frustrated and
confused, and you still have no idea what you're doing. Ask sooner, if
you have to; or develop the discipline to keep track of what you do
and what happens. But regardless of what actually happens, it didn't
work is not a helpful thing to say.

Trust me, making it easier for us will make everything easier for you
too. Even if we were to never answer a single question of yours ever
again, learning to read error messages will benefit you more than you
can imagine.

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

[Rustom Mody]

On Friday, October 24, 2014 8:11:12 PM UTC+5:30, Seymore4Head wrote:
 On Thu, 23 Oct 2014 21:56:31 -0700 (PDT), Rustom Mody wrote:
 
 On Thursday, October 23, 2014 10:33:57 PM UTC+5:30, Seymore4Head wrote:
  On Thu, 23 Oct 2014 15:55:35 + (UTC), Denis McMahon wrote:
  
  On Thu, 23 Oct 2014 10:04:56 -0400, Seymore4Head wrote:
  
   On Thu, 23 Oct 2014 09:15:16 + (UTC), Denis McMahon wrote:
  
  Try the following 3 commands at the console:
  
  You obviously didn't, so I'll try again. Try each of the following three 
  commands in the python console at the  prompt.
  
  1) 10
  10
  
  2) range(10)
  range(0, 10)
  
  3) str(range(10))
  'range(0, 10)'
  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
  
  Show *and* describe the output in each case. Describing the output that 
  you see is actually the key here, as it will allow us to assess whether 
  you understand what you are actually seeing or not, and if you don't 
  understand the output you see in the console, then we need to fix that 
  very fundamental and basic issue before moving on to more complex stuff!
  
   Ok Thanks
  
  You were expected to answer the question in the original. I have now set 
  it as a clearer and more specific task.
  
  If you're not going to do these things that are intended to help you 
  learn some of the basic features of the language, then I and everyone 
  else here that has so far been attempting to help you are wasting our 
  time.
  
  I did try them.  I may have missed replying your to your specific
  comment, but I tried them.
  
  BTW str(range (10)) does work with Python 2 which is where I may have
  got the idea.  I happened to be using Python 3 at the time I tried to
  implement it.  It is a little confusing jumping back and forth, but
  for the moment, I am going to tough it out.
  
  I do appreciate all the help too.
 
 Hi Seymore!
 
 Happy to see that you are moving on from
 reading much; understanding nothing; thrashing
 
 to
 
 reading a bit; understanding a bit
 [And thanks to Denis to getting you out of your confusion-hole]
 
 So heres a small additional question set that I promise will more than repay
 you your time.
 
 Better done in python 2. But if you use python3, below replace
 range(10)
 with
 list(range(10))

Other details skipped

 
 I tried list(range(10)  I thought that would work in Python 3.  It
 didn't.  I spent quite a bit of time last night trying to come up with
 the right combination of str and int commands to make range(10) work
 with my simple example.   It didn't.  I am pretty frustrated.  I am
 just skipping that little bit of code for the moment.

I asked you to try
list(range(10))

Did you try EXACTLY (cut-paste) that?

You are claiming to have tried
list(range(10)

Thats one closing parenthesis less

The interaction with your version would go something like this:
[Two versions 
The KeyboardInterrupt comes from giving a control-C
Dunno what happens in codeskulptor
]

 list(range(10)
... 
... 
... 
... )
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 list(range(10)
... 
KeyboardInterrupt


 
 Thanks everyone for your suggestions.

1. You are reading too much
2. Trying to hard

Think of riding a bicycle.
Cant do it by reading many books on cycling -- thats 1.
Nor by holding the handle so hard you tremble -- thats 2.

Just relax a bit...
And take small steps

Chill... as Chris joked, no monster in the computer (or on this list!)


 Range(10) stores the min max values and loads each number in between
 when needed.

It loads?? As in 'load-up-a-van'??

When you see:

 10
10

1. Does someone (a clerk maybe) in the computer count to 10?
2. Or do you, seeing that interaction, count to 10?
   [If you do, replace the 10 by 1000]
3. Or do you, remember what it means to count to 10 without having to do it?

Now go back to your statement about 'loading' and find a better verb
-- 
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Re: I am out of trial and error again Lists

[giacomo boffi]

ERRATA CORRIGE:

 many different circumstances, by the very, very helpful folks of clp.
  many different circumstances, by the very, very helpful folks of clpy

-- 
sapete contare fino a venticinque?
Olimpia Milano Jugoplastika Split Partizan Beograd
Roberto Premier Duska Ivanovic Zarko Paspalj
-- 
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Re: I am out of trial and error again Lists

[Ian Kelly]

On Fri, Oct 24, 2014 at 9:56 AM, Rustom Mody rustompm...@gmail.com wrote:
 Range(10) stores the min max values and loads each number in between
 when needed.

 It loads?? As in 'load-up-a-van'??

As in loads into memory.

 When you see:

 10
 10

 1. Does someone (a clerk maybe) in the computer count to 10?
 2. Or do you, seeing that interaction, count to 10?
[If you do, replace the 10 by 1000]
 3. Or do you, remember what it means to count to 10 without having to do it?

I don't understand why you think any of these are implied by the word load.

 Now go back to your statement about 'loading' and find a better verb

I presume he used load because that was the word I used in my
explanatory post about the difference between range in Python 2 and
Python 3 yesterday.
-- 
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 08:56:31 -0700 (PDT), Rustom Mody
rustompm...@gmail.com wrote:

On Friday, October 24, 2014 8:11:12 PM UTC+5:30, Seymore4Head wrote:
 On Thu, 23 Oct 2014 21:56:31 -0700 (PDT), Rustom Mody wrote:
 
 On Thursday, October 23, 2014 10:33:57 PM UTC+5:30, Seymore4Head wrote:
  On Thu, 23 Oct 2014 15:55:35 + (UTC), Denis McMahon wrote:
  
  On Thu, 23 Oct 2014 10:04:56 -0400, Seymore4Head wrote:
  
   On Thu, 23 Oct 2014 09:15:16 + (UTC), Denis McMahon wrote:
  
  Try the following 3 commands at the console:
  
  You obviously didn't, so I'll try again. Try each of the following three 
  commands in the python console at the  prompt.
  
  1) 10
  10
  
  2) range(10)
  range(0, 10)
  
  3) str(range(10))
  'range(0, 10)'
  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
  
  Show *and* describe the output in each case. Describing the output that 
  you see is actually the key here, as it will allow us to assess whether 
  you understand what you are actually seeing or not, and if you don't 
  understand the output you see in the console, then we need to fix that 
  very fundamental and basic issue before moving on to more complex stuff!
  
   Ok Thanks
  
  You were expected to answer the question in the original. I have now set 
  it as a clearer and more specific task.
  
  If you're not going to do these things that are intended to help you 
  learn some of the basic features of the language, then I and everyone 
  else here that has so far been attempting to help you are wasting our 
  time.
  
  I did try them.  I may have missed replying your to your specific
  comment, but I tried them.
  
  BTW str(range (10)) does work with Python 2 which is where I may have
  got the idea.  I happened to be using Python 3 at the time I tried to
  implement it.  It is a little confusing jumping back and forth, but
  for the moment, I am going to tough it out.
  
  I do appreciate all the help too.
 
 Hi Seymore!
 
 Happy to see that you are moving on from
 reading much; understanding nothing; thrashing
 
 to
 
 reading a bit; understanding a bit
 [And thanks to Denis to getting you out of your confusion-hole]
 
 So heres a small additional question set that I promise will more than repay
 you your time.
 
 Better done in python 2. But if you use python3, below replace
 range(10)
 with
 list(range(10))

Other details skipped

 
 I tried list(range(10)  I thought that would work in Python 3.  It
 didn't.  I spent quite a bit of time last night trying to come up with
 the right combination of str and int commands to make range(10) work
 with my simple example.   It didn't.  I am pretty frustrated.  I am
 just skipping that little bit of code for the moment.

I asked you to try
list(range(10))

Did you try EXACTLY (cut-paste) that?

You are claiming to have tried
list(range(10)

Thats one closing parenthesis less

The interaction with your version would go something like this:
[Two versions 
The KeyboardInterrupt comes from giving a control-C
Dunno what happens in codeskulptor
]

 list(range(10)
... 
... 
... 
... )
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 list(range(10)
... 
KeyboardInterrupt


 
 Thanks everyone for your suggestions.

1. You are reading too much
2. Trying to hard

Think of riding a bicycle.
Cant do it by reading many books on cycling -- thats 1.
Nor by holding the handle so hard you tremble -- thats 2.

Just relax a bit...
And take small steps

Chill... as Chris joked, no monster in the computer (or on this list!)


 Range(10) stores the min max values and loads each number in between
 when needed.

It loads?? As in 'load-up-a-van'??

When you see:

 10
10

1. Does someone (a clerk maybe) in the computer count to 10?
2. Or do you, seeing that interaction, count to 10?
   [If you do, replace the 10 by 1000]
3. Or do you, remember what it means to count to 10 without having to do it?

Now go back to your statement about 'loading' and find a better verb

If I could explain to you why something doesn't work then I could fix
it myself.  I don't understand why it doesn't work.  The best I can do
is repost the code. 

When I use list(range(10)) I get:
Traceback (most recent call last):
  File C:/Functions/name to number digit.py, line 37, in module
print (nametonumber(a))#1800 438 2427 837
  File C:/Functions/name to number digit.py, line 10, in
nametonumber
if y in lst(range(1,10)):
TypeError: 'list' object is not callable

All the lines I have commented out work.  Trying to use
list(range(10)) doesn't.  (Python 3)
http://i.imgur.com/LtiCyZS.jpg

It doesn't work.
It's broke.  :)
I don't know what else to say.

import string
def nametonumber(name):
lst=[]
nx=[]
digit=[]
digit=.join(str(i) for i in range(10))
for x in name:
lst.append(x)
for y in (lst):
if y in lst(range(1,10)):
#if y in 1234567890:
#if y.isdigit():
#if y in digit:   
#if y in string.digits:
nx.append(y)
if y in  -():

Re: I am out of trial and error again Lists

[Albert-Jan Roskam]

-
On Fri, Oct 24, 2014 5:56 PM CEST Rustom Mody wrote:

On Friday, October 24, 2014 8:11:12 PM UTC+5:30, Seymore4Head wrote:
 On Thu, 23 Oct 2014 21:56:31 -0700 (PDT), Rustom Mody wrote:
 
 On Thursday, October 23, 2014 10:33:57 PM UTC+5:30, Seymore4Head wrote:
  On Thu, 23 Oct 2014 15:55:35 + (UTC), Denis McMahon wrote:
  
  On Thu, 23 Oct 2014 10:04:56 -0400, Seymore4Head wrote:
  
   On Thu, 23 Oct 2014 09:15:16 + (UTC), Denis McMahon wrote:
  
  Try the following 3 commands at the console:
  
  You obviously didn't, so I'll try again. Try each of the following three 
  commands in the python console at the  prompt.
  
  1) 10
  10
  
  2) range(10)
  range(0, 10)
  
  3) str(range(10))
  'range(0, 10)'
  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
  
  Show *and* describe the output in each case. Describing the output that 
  you see is actually the key here, as it will allow us to assess whether 
  you understand what you are actually seeing or not, and if you don't 
  understand the output you see in the console, then we need to fix that 
  very fundamental and basic issue before moving on to more complex stuff!
  
   Ok Thanks
  
  You were expected to answer the question in the original. I have now set 
  it as a clearer and more specific task.
  
  If you're not going to do these things that are intended to help you 
  learn some of the basic features of the language, then I and everyone 
  else here that has so far been attempting to help you are wasting our 
  time.
  
  I did try them.  I may have missed replying your to your specific
  comment, but I tried them.
  
  BTW str(range (10)) does work with Python 2 which is where I may have
  got the idea.  I happened to be using Python 3 at the time I tried to
  implement it.  It is a little confusing jumping back and forth, but
  for the moment, I am going to tough it out.


u0_a100@condor_umts:/ $ python
Python 3.2.2 (default, Jun 23 2014, 00:13:13)
[GCC 4.8] on linux-armv7l
Type help, copyright, credits or license for more information.
 list(range(10))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]



  I do appreciate all the help too.
 
 Hi Seymore!
 
 Happy to see that you are moving on from
 reading much; understanding nothing; thrashing
 
 to
 
 reading a bit; understanding a bit
 [And thanks to Denis to getting you out of your confusion-hole]
 
 So heres a small additional question set that I promise will more than repay
 you your time.
 
 Better done in python 2. But if you use python3, below replace
 range(10)
 with
 list(range(10))

Other details skipped


1. You are reading too much
2. Trying to hard

Think of riding a bicycle.
Cant do it by reading many books on cycling -- thats 1.
Nor by holding the handle so hard you tremble -- thats 2.

Just relax a bit...
And take small steps

Chill... as Chris joked, no monster in the computer (or on this list!)

+1 for that remark. Talking about chill: grab a couple of beers (suggest: 
sixpack) and enjoy an evening of Python!




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

Re: I am out of trial and error again Lists

[Chris Angelico]

On Sat, Oct 25, 2014 at 3:37 AM, Seymore4Head
Seymore4Head@hotmail.invalid wrote:
 When I use list(range(10)) I get:
 Traceback (most recent call last):
   File C:/Functions/name to number digit.py, line 37, in module
 print (nametonumber(a))#1800 438 2427 837
   File C:/Functions/name to number digit.py, line 10, in
 nametonumber
 if y in lst(range(1,10)):
 TypeError: 'list' object is not callable

Now, finally, you're showing us an actual line of code and an actual
traceback. And from here, we can see that you misspelled list.
That's why it isn't working.

Several people have told you to use the interactive interpreter. Please do so.

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

[Rustom Mody]

Totally befuddled myself!

Are you deliberately misspelling list to lst
and hoping the error will go away.

And Puh LEESE
dont post screen shots of good ol ASCII text
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Re: I am out of trial and error again Lists

[Rustom Mody]

On Friday, October 24, 2014 10:18:12 PM UTC+5:30, Chris Angelico wrote:
 On Sat, Oct 25, 2014 at 3:37 AM, Seymore4Head wrote:
  When I use list(range(10)) I get:
  Traceback (most recent call last):
File C:/Functions/name to number digit.py, line 37, in module
  print (nametonumber(a))#1800 438 2427 837
File C:/Functions/name to number digit.py, line 10, in
  nametonumber
  if y in lst(range(1,10)):
  TypeError: 'list' object is not callable
 
 Now, finally, you're showing us an actual line of code and an actual
 traceback. And from here, we can see that you misspelled list.
 That's why it isn't working.
 
 Several people have told you to use the interactive interpreter. Please do so.
 
 ChrisA

Right.
Good.
Sorry for being impatient Seymore

You are now making progress
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Re: I am out of trial and error again Lists

[Ian Kelly]

On Fri, Oct 24, 2014 at 10:37 AM, Seymore4Head
Seymore4Head@hotmail.invalid wrote:
 If I could explain to you why something doesn't work then I could fix
 it myself.  I don't understand why it doesn't work.  The best I can do
 is repost the code.

You don't need to be able to explain why it doesn't work. You just
need to be able to explain what you expected it to do and what it
actually did. Posting the code and the traceback that you get is a
fine start.

 if y in lst(range(1,10)):

The name of the builtin is list. It's a function* that takes an
argument and uses it to construct a list, which it returns.

lst is the name of some specific list that you're using in your
code. It's not a function, which is why the error is complaining that
it isn't callable.

*Actually it's a type object, and calling it causes an instance of the
type to be constructed, but for all intents and purposes here it works
exactly like a function.
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Sat, 25 Oct 2014 03:47:51 +1100, Chris Angelico ros...@gmail.com
wrote:

On Sat, Oct 25, 2014 at 3:37 AM, Seymore4Head
Seymore4Head@hotmail.invalid wrote:
 When I use list(range(10)) I get:
 Traceback (most recent call last):
   File C:/Functions/name to number digit.py, line 37, in module
 print (nametonumber(a))#1800 438 2427 837
   File C:/Functions/name to number digit.py, line 10, in
 nametonumber
 if y in lst(range(1,10)):
 TypeError: 'list' object is not callable

Now, finally, you're showing us an actual line of code and an actual
traceback. And from here, we can see that you misspelled list.
That's why it isn't working.

Several people have told you to use the interactive interpreter. Please do so.

ChrisA

Actually I was a little frustrated when I added that line back in as
the other lines all work.
Using list(range(10)) Doesn't throw an error but it doesn't work.

http://i.imgur.com/DTc5zoL.jpg

The interpreter.   I don't know how to use that either.

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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 09:54:23 -0700 (PDT), Rustom Mody
rustompm...@gmail.com wrote:

Totally befuddled myself!

Are you deliberately misspelling list to lst
and hoping the error will go away.

And Puh LEESE
dont post screen shots of good ol ASCII text

I didn't do that on purpose.  I make a lot of typing mistakes.
Sorry
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Re: I am out of trial and error again Lists

[Rustom Mody]

On Friday, October 24, 2014 10:06:47 PM UTC+5:30, Ian wrote:
 On Fri, Oct 24, 2014 at 9:56 AM, Rustom Mody wrote:
  Range(10) stores the min max values and loads each number in between
  when needed.
 
  It loads?? As in 'load-up-a-van'??
 
 As in loads into memory.
 
  When you see:
 
  10
  10
 
  1. Does someone (a clerk maybe) in the computer count to 10?
  2. Or do you, seeing that interaction, count to 10?
 [If you do, replace the 10 by 1000]
  3. Or do you, remember what it means to count to 10 without having to do it?
 
 I don't understand why you think any of these are implied by the word load.
 
  Now go back to your statement about 'loading' and find a better verb
 
 I presume he used load because that was the word I used in my
 explanatory post about the difference between range in Python 2 and
 Python 3 yesterday.

I would be very surprised (Ian) if we had any essential disagreement on this
subject. [JFTR I see nothing wrong with your explanation]

I would also be (pleasantly) surprised if Seymore were to benefit by these
discussions at this stage.

So is it ok if we drop it here (or start a new thread)?
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Re: I am out of trial and error again Lists

[Chris Angelico]

On Sat, Oct 25, 2014 at 4:05 AM, Ian Kelly ian.g.ke...@gmail.com wrote:
 The name of the builtin is list. It's a function* that takes an
 argument and uses it to construct a list, which it returns.

 *Actually it's a type object, and calling it causes an instance of the
 type to be constructed, but for all intents and purposes here it works
 exactly like a function.

It's callable, that's all that matters.

callable(object) - bool
Return whether the object is callable (i.e., some kind of function).

:)

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

[Albert Visser]

On Fri, 24 Oct 2014 19:03:47 +0200, Seymore4Head  
Seymore4Head@hotmail.invalid wrote:




http://i.imgur.com/DTc5zoL.jpg

The interpreter.   I don't know how to use that either.



It's what's on the left hand side of your screenshot. You can simply type  
Python statements following the  prompt and hit enter to examine the  
result, instead of pushing F5 to run your code


--
Vriendelijke groeten / Kind regards,

Albert Visser

Using Opera's mail client: http://www.opera.com/mail/
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Re: I am out of trial and error again Lists

[Rustom Mody]

On Friday, October 24, 2014 10:37:45 PM UTC+5:30, Seymore4Head wrote:
 On Fri, 24 Oct 2014 09:54:23 -0700 (PDT), Rustom Mody wrote:
 
 Totally befuddled myself!
 
 Are you deliberately misspelling list to lst
 and hoping the error will go away.
 
 And Puh LEESE
 dont post screen shots of good ol ASCII text
 
 I didn't do that on purpose.  I make a lot of typing mistakes.
 Sorry

Right
No sweat!

I was genuinely asking:
- Did you mis-spell list as lst?
- Or did you go thrashing (Steven gave a picturesque description)
just changing things until the error went?

[Believe you me, I do the same when I am in a strange place and I
am searching for something in my (suit|brief)case
]
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 19:18:12 +0200, Albert Visser
albert.vis...@gmail.com wrote:

On Fri, 24 Oct 2014 19:03:47 +0200, Seymore4Head  
Seymore4Head@hotmail.invalid wrote:


 http://i.imgur.com/DTc5zoL.jpg

 The interpreter.   I don't know how to use that either.


It's what's on the left hand side of your screenshot. You can simply type  
Python statements following the  prompt and hit enter to examine the  
result, instead of pushing F5 to run your code

I guess I am confusing the Interpreter with the debugger.  Someone
suggested I use the Interpreter to step through line by line.
I don't know how to do that.
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Re: I am out of trial and error again Lists

[Rustom Mody]

On Friday, October 24, 2014 10:55:44 PM UTC+5:30, Seymore4Head wrote:
 On Fri, 24 Oct 2014 19:18:12 +0200, Albert Visser wrote:
 
 On Fri, 24 Oct 2014 19:03:47 +0200, Seymore4Head  wrote:
 
 
  http://i.imgur.com/DTc5zoL.jpg
 
  The interpreter.   I don't know how to use that either.
 
 
 It's what's on the left hand side of your screenshot. You can simply type  
 Python statements following the  prompt and hit enter to examine the  
 result, instead of pushing F5 to run your code
 
 I guess I am confusing the Interpreter with the debugger.  Someone
 suggested I use the Interpreter to step through line by line.
 I don't know how to do that.

Dont bother with the debugger just yet.
For most python programmers, sticking a few print statements 
(expressions in python 3) in adroitly is good enough.*

For now best if you concentrate on
1. What are the features of python -- the language
2. What are the standard data types and functions -- the libraries
3. How to use and jump between the two windows of your screenshot most
   effectively. What you should and should not type in each etc

* One neat trick of using the print to debug.
Say you have a line like

nx.append(2)

and nx is not getting to be what you expect.
Change it to

nx.append(2); print(nx)

Cleaning up the print after debugging is easier than if you use a
separate line like so

nx.append(2)
print(nx)

[I think I learnt this trick from Mark Lawrence]
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Re: I am out of trial and error again Lists

[Ian Kelly]

On Fri, Oct 24, 2014 at 11:03 AM, Seymore4Head
Seymore4Head@hotmail.invalid wrote:
 Actually I was a little frustrated when I added that line back in as
 the other lines all work.
 Using list(range(10)) Doesn't throw an error but it doesn't work.

 http://i.imgur.com/DTc5zoL.jpg

 The interpreter.   I don't know how to use that either.

Try both of these in the interpreter, and observe the difference:

7 in range(10)

7 in range(10)

Do you understand what the difference between 7 and 7 is?
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 10:42:08 -0700 (PDT), Rustom Mody
rustompm...@gmail.com wrote:

On Friday, October 24, 2014 10:55:44 PM UTC+5:30, Seymore4Head wrote:
 On Fri, 24 Oct 2014 19:18:12 +0200, Albert Visser wrote:
 
 On Fri, 24 Oct 2014 19:03:47 +0200, Seymore4Head  wrote:
 
 
  http://i.imgur.com/DTc5zoL.jpg
 
  The interpreter.   I don't know how to use that either.
 
 
 It's what's on the left hand side of your screenshot. You can simply type  
 Python statements following the  prompt and hit enter to examine the  
 result, instead of pushing F5 to run your code
 
 I guess I am confusing the Interpreter with the debugger.  Someone
 suggested I use the Interpreter to step through line by line.
 I don't know how to do that.

Dont bother with the debugger just yet.
For most python programmers, sticking a few print statements 
(expressions in python 3) in adroitly is good enough.*

For now best if you concentrate on
1. What are the features of python -- the language
2. What are the standard data types and functions -- the libraries
3. How to use and jump between the two windows of your screenshot most
   effectively. What you should and should not type in each etc

* One neat trick of using the print to debug.
Say you have a line like

nx.append(2)

and nx is not getting to be what you expect.
Change it to

nx.append(2); print(nx)

Cleaning up the print after debugging is easier than if you use a
separate line like so

nx.append(2)
print(nx)

[I think I learnt this trick from Mark Lawrence]

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

[Seymore4Head]

On Fri, 24 Oct 2014 11:52:15 -0600, Ian Kelly ian.g.ke...@gmail.com
wrote:

On Fri, Oct 24, 2014 at 11:03 AM, Seymore4Head
Seymore4Head@hotmail.invalid wrote:
 Actually I was a little frustrated when I added that line back in as
 the other lines all work.
 Using list(range(10)) Doesn't throw an error but it doesn't work.

 http://i.imgur.com/DTc5zoL.jpg

 The interpreter.   I don't know how to use that either.

Try both of these in the interpreter, and observe the difference:

7 in range(10)

7 in range(10)

Do you understand what the difference between 7 and 7 is?

I do understand that.  7 is a number and 7 is a string.
What my question was...and still is...is why
Python 3 fails when I try using
y=1 800 get charter

y in range str(range(10))
should work because y is a string and str(range(10)) should be
y in str(1) fails.
It doesn't give an error it's just not True when y is a number.

These hints are just not working.  I am too thick for hints. :)
If you could use it in the code, I might understand.
The other work arounds that were posted work.
I have used them.  str(range(10)) doesn't work.

import string
def nametonumber(name):
lst=[]
nx=[]
digit=[]
digit=.join(str(i) for i in range(10))
for x in name:
lst.append(x)
for y in (lst):
if y in list(range(1,10)):
#if y in 1234567890:
#if y.isdigit():
#if y in digit:   
#if y in string.digits:
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(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 dialaho
print (nametonumber(a))

Please


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Re: I am out of trial and error again Lists

[Seymore4Head]

I meant to type:
if y in range(1,10) doesn't work.
Sigh
Sorry

On Fri, 24 Oct 2014 14:15:13 -0400, Seymore4Head
Seymore4Head@Hotmail.invalid wrote:

On Fri, 24 Oct 2014 11:52:15 -0600, Ian Kelly ian.g.ke...@gmail.com
wrote:

On Fri, Oct 24, 2014 at 11:03 AM, Seymore4Head
Seymore4Head@hotmail.invalid wrote:
 Actually I was a little frustrated when I added that line back in as
 the other lines all work.
 Using list(range(10)) Doesn't throw an error but it doesn't work.

 http://i.imgur.com/DTc5zoL.jpg

 The interpreter.   I don't know how to use that either.

Try both of these in the interpreter, and observe the difference:

7 in range(10)

7 in range(10)

Do you understand what the difference between 7 and 7 is?

I do understand that.  7 is a number and 7 is a string.
What my question was...and still is...is why
Python 3 fails when I try using
y=1 800 get charter

y in range str(range(10))
should work because y is a string and str(range(10)) should be
y in str(1) fails.
It doesn't give an error it's just not True when y is a number.

These hints are just not working.  I am too thick for hints. :)
If you could use it in the code, I might understand.
The other work arounds that were posted work.
I have used them.  str(range(10)) doesn't work.

import string
def nametonumber(name):
lst=[]
nx=[]
digit=[]
digit=.join(str(i) for i in range(10))
for x in name:
lst.append(x)
for y in (lst):
if y in list(range(1,10)):
#if y in 1234567890:
#if y.isdigit():
#if y in digit:   
#if y in string.digits:
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(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 dialaho
print (nametonumber(a))

Please

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Re: I am out of trial and error again Lists

[Mark Lawrence]

On 24/10/2014 18:03, Seymore4Head wrote:


Actually I was a little frustrated when I added that line back in as
the other lines all work.
Using list(range(10)) Doesn't throw an error but it doesn't work.

http://i.imgur.com/DTc5zoL.jpg

The interpreter.   I don't know how to use that either.



You've stated that you can use google so why not try it to find out 
about the interpreter?  Or simply navigate to docs.python.org and see 
what the contents or index tell you?  Failing that carry on charging 
around like a headless chicken and hope that the extremely patient folk 
here keep helping you out.


--
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

[Mark Lawrence]

On 24/10/2014 19:20, Seymore4Head wrote:

I meant to type:
if y in range(1,10) doesn't work.
Sigh
Sorry



How many more times, state what you expect to happen and what actually 
happens.  doesn't work is useless.  Please read this http://sscce.org/


--
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

[sohcahtoa82]

On Friday, October 24, 2014 11:17:53 AM UTC-7, Seymore4Head wrote:
 On Fri, 24 Oct 2014 11:52:15 -0600, Ian Kelly ian.g.ke...@gmail.com
 wrote:
 
 On Fri, Oct 24, 2014 at 11:03 AM, Seymore4Head
 Seymore4Head@hotmail.invalid wrote:
  Actually I was a little frustrated when I added that line back in as
  the other lines all work.
  Using list(range(10)) Doesn't throw an error but it doesn't work.
 
  http://i.imgur.com/DTc5zoL.jpg
 
  The interpreter.   I don't know how to use that either.
 
 Try both of these in the interpreter, and observe the difference:
 
 7 in range(10)
 
 7 in range(10)
 
 Do you understand what the difference between 7 and 7 is?
 
 I do understand that.  7 is a number and 7 is a string.
 What my question was...and still is...is why
 Python 3 fails when I try using
 y=1 800 get charter
 
 y in range str(range(10))
 should work because y is a string and str(range(10)) should be
 y in str(1) fails.
 It doesn't give an error it's just not True when y is a number.
 
 These hints are just not working.  I am too thick for hints. :)
 If you could use it in the code, I might understand.
 The other work arounds that were posted work.
 I have used them.  str(range(10)) doesn't work.
 
 import string
 def nametonumber(name):
 lst=[]
 nx=[]
 digit=[]
 digit=.join(str(i) for i in range(10))
 for x in name:
 lst.append(x)
 for y in (lst):
 if y in list(range(1,10)):
 #if y in 1234567890:
 #if y.isdigit():
 #if y in digit:   
 #if y in string.digits:
 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(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 dialaho
 print (nametonumber(a))
 
 Please

Your code here is actually pretty close to a correct answer.  Just a few things 
to consider...

- Why are you converting your name string to a list?  It is unnecessary.  When 
you do for y in some string, then y will still be single characters on each 
iteration of the loop.

- if y in string.digits should work fine.

- if y in list(range(1,10) won't work for two reasons: First, it creates a 
list of numbers, not strings.  Second, even if it did, it would be missing the 
0 digit.

- At the end, when you convert your list to a string, you don't need to use 
list comprehension, since nx is already a list.  number = .join(nx) should 
work fine.

Also, in general, you need to stop and slow down and think like a programmer.  
If you get an error, your instinct shouldn't be to just hack at it to make the 
error go away.  Look at the error and try to make sense of it.  Learn what the 
error means and try to fix the core problem.

And for @#$%'s sake...stop saying It isn't working and not elaborating.  
You've been told by every other post in this thread to show us what you did and 
what the error was.  You've also been told to *NOT* retype what you see and to 
copy/paste your code and the error because when you make a typo when copying, 
we might see a problem that doesn't exist and then you just get more confused.
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 19:40:39 +0100, Mark Lawrence
breamore...@yahoo.co.uk wrote:

On 24/10/2014 19:20, Seymore4Head wrote:
 I meant to type:
 if y in range(1,10) doesn't work.
 Sigh
 Sorry


How many more times, state what you expect to happen and what actually 
happens.  doesn't work is useless.  Please read this http://sscce.org/

Good suggestion.
OK  how is this?
It doesn't print what I expect.
Does it print what you expect?

name=123-xyz-abc
for x in name:
if x in range(10):
print (Range,(x))
if x in str(range(10)):
print (String range,(x))

http://i.imgur.com/EGKUpAb.jpg
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 11:57:12 -0700 (PDT), sohcahto...@gmail.com wrote:

On Friday, October 24, 2014 11:17:53 AM UTC-7, Seymore4Head wrote:
 On Fri, 24 Oct 2014 11:52:15 -0600, Ian Kelly ian.g.ke...@gmail.com
 wrote:
 
 On Fri, Oct 24, 2014 at 11:03 AM, Seymore4Head
 Seymore4Head@hotmail.invalid wrote:
  Actually I was a little frustrated when I added that line back in as
  the other lines all work.
  Using list(range(10)) Doesn't throw an error but it doesn't work.
 
  http://i.imgur.com/DTc5zoL.jpg
 
  The interpreter.   I don't know how to use that either.
 
 Try both of these in the interpreter, and observe the difference:
 
 7 in range(10)
 
 7 in range(10)
 
 Do you understand what the difference between 7 and 7 is?
 
 I do understand that.  7 is a number and 7 is a string.
 What my question was...and still is...is why
 Python 3 fails when I try using
 y=1 800 get charter
 
 y in range str(range(10))
 should work because y is a string and str(range(10)) should be
 y in str(1) fails.
 It doesn't give an error it's just not True when y is a number.
 
 These hints are just not working.  I am too thick for hints. :)
 If you could use it in the code, I might understand.
 The other work arounds that were posted work.
 I have used them.  str(range(10)) doesn't work.
 
 import string
 def nametonumber(name):
 lst=[]
 nx=[]
 digit=[]
 digit=.join(str(i) for i in range(10))
 for x in name:
 lst.append(x)
 for y in (lst):
 if y in list(range(1,10)):
 #if y in 1234567890:
 #if y.isdigit():
 #if y in digit:   
 #if y in string.digits:
 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(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 dialaho
 print (nametonumber(a))
 
 Please

Your code here is actually pretty close to a correct answer.  Just a few 
things to consider...

- Why are you converting your name string to a list?  It is unnecessary.  When 
you do for y in some string, then y will still be single characters on 
each iteration of the loop.

- if y in string.digits should work fine.

- if y in list(range(1,10) won't work for two reasons: First, it creates a 
list of numbers, not strings.  Second, even if it did, it would be missing the 
0 digit.

- At the end, when you convert your list to a string, you don't need to use 
list comprehension, since nx is already a list.  number = .join(nx) should 
work fine.

Also, in general, you need to stop and slow down and think like a programmer.  
If you get an error, your instinct shouldn't be to just hack at it to make the 
error go away.  Look at the error and try to make sense of it.  Learn what the 
error means and try to fix the core problem.

And for @#$%'s sake...stop saying It isn't working and not elaborating.  
You've been told by every other post in this thread to show us what you did 
and what the error was.  You've also been told to *NOT* retype what you see 
and to copy/paste your code and the error because when you make a typo when 
copying, we might see a problem that doesn't exist and then you just get more 
confused.

Ok  I think I may have the question you guys are looking for.
I just posted it.
See above.

But it's still broke.  :( 
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Re: I am out of trial and error again Lists

[sohcahtoa82]

On Friday, October 24, 2014 12:12:10 PM UTC-7, Seymore4Head wrote:
 On Fri, 24 Oct 2014 11:57:12 -0700 (PDT), sohcahto...@gmail.com wrote:
 
 On Friday, October 24, 2014 11:17:53 AM UTC-7, Seymore4Head wrote:
  On Fri, 24 Oct 2014 11:52:15 -0600, Ian Kelly ian.g.ke...@gmail.com
  wrote:
  
  On Fri, Oct 24, 2014 at 11:03 AM, Seymore4Head
  Seymore4Head@hotmail.invalid wrote:
   Actually I was a little frustrated when I added that line back in as
   the other lines all work.
   Using list(range(10)) Doesn't throw an error but it doesn't work.
  
   http://i.imgur.com/DTc5zoL.jpg
  
   The interpreter.   I don't know how to use that either.
  
  Try both of these in the interpreter, and observe the difference:
  
  7 in range(10)
  
  7 in range(10)
  
  Do you understand what the difference between 7 and 7 is?
  
  I do understand that.  7 is a number and 7 is a string.
  What my question was...and still is...is why
  Python 3 fails when I try using
  y=1 800 get charter
  
  y in range str(range(10))
  should work because y is a string and str(range(10)) should be
  y in str(1) fails.
  It doesn't give an error it's just not True when y is a number.
  
  These hints are just not working.  I am too thick for hints. :)
  If you could use it in the code, I might understand.
  The other work arounds that were posted work.
  I have used them.  str(range(10)) doesn't work.
  
  import string
  def nametonumber(name):
  lst=[]
  nx=[]
  digit=[]
  digit=.join(str(i) for i in range(10))
  for x in name:
  lst.append(x)
  for y in (lst):
  if y in list(range(1,10)):
  #if y in 1234567890:
  #if y.isdigit():
  #if y in digit:   
  #if y in string.digits:
  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(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 dialaho
  print (nametonumber(a))
  
  Please
 
 Your code here is actually pretty close to a correct answer.  Just a few 
 things to consider...
 
 - Why are you converting your name string to a list?  It is unnecessary.  
 When you do for y in some string, then y will still be single characters 
 on each iteration of the loop.
 
 - if y in string.digits should work fine.
 
 - if y in list(range(1,10) won't work for two reasons: First, it creates a 
 list of numbers, not strings.  Second, even if it did, it would be missing 
 the 0 digit.
 
 - At the end, when you convert your list to a string, you don't need to use 
 list comprehension, since nx is already a list.  number = .join(nx) should 
 work fine.
 
 Also, in general, you need to stop and slow down and think like a 
 programmer.  If you get an error, your instinct shouldn't be to just hack at 
 it to make the error go away.  Look at the error and try to make sense of 
 it.  Learn what the error means and try to fix the core problem.
 
 And for @#$%'s sake...stop saying It isn't working and not elaborating.  
 You've been told by every other post in this thread to show us what you did 
 and what the error was.  You've also been told to *NOT* retype what you see 
 and to copy/paste your code and the error because when you make a typo when 
 copying, we might see a problem that doesn't exist and then you just get 
 more confused.
 
 Ok  I think I may have the question you guys are looking for.
 I just posted it.
 See above.
 
 But it's still broke.  :(

str(range(10)) doesn't do what you think it does.

Run 'print(str(range(10)))' and look at what you get.
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 12:25:33 -0700 (PDT), sohcahto...@gmail.com wrote:

On Friday, October 24, 2014 12:12:10 PM UTC-7, Seymore4Head wrote:
 On Fri, 24 Oct 2014 11:57:12 -0700 (PDT), sohcahto...@gmail.com wrote:
 
 On Friday, October 24, 2014 11:17:53 AM UTC-7, Seymore4Head wrote:
  On Fri, 24 Oct 2014 11:52:15 -0600, Ian Kelly ian.g.ke...@gmail.com
  wrote:
  
  On Fri, Oct 24, 2014 at 11:03 AM, Seymore4Head
  Seymore4Head@hotmail.invalid wrote:
   Actually I was a little frustrated when I added that line back in as
   the other lines all work.
   Using list(range(10)) Doesn't throw an error but it doesn't work.
  
   http://i.imgur.com/DTc5zoL.jpg
  
   The interpreter.   I don't know how to use that either.
  
  Try both of these in the interpreter, and observe the difference:
  
  7 in range(10)
  
  7 in range(10)
  
  Do you understand what the difference between 7 and 7 is?
  
  I do understand that.  7 is a number and 7 is a string.
  What my question was...and still is...is why
  Python 3 fails when I try using
  y=1 800 get charter
  
  y in range str(range(10))
  should work because y is a string and str(range(10)) should be
  y in str(1) fails.
  It doesn't give an error it's just not True when y is a number.
  
  These hints are just not working.  I am too thick for hints. :)
  If you could use it in the code, I might understand.
  The other work arounds that were posted work.
  I have used them.  str(range(10)) doesn't work.
  
  import string
  def nametonumber(name):
  lst=[]
  nx=[]
  digit=[]
  digit=.join(str(i) for i in range(10))
  for x in name:
  lst.append(x)
  for y in (lst):
  if y in list(range(1,10)):
  #if y in 1234567890:
  #if y.isdigit():
  #if y in digit:   
  #if y in string.digits:
  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(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 dialaho
  print (nametonumber(a))
  
  Please
 
 Your code here is actually pretty close to a correct answer.  Just a few 
 things to consider...
 
 - Why are you converting your name string to a list?  It is unnecessary.  
 When you do for y in some string, then y will still be single 
 characters on each iteration of the loop.
 
 - if y in string.digits should work fine.
 
 - if y in list(range(1,10) won't work for two reasons: First, it creates 
 a list of numbers, not strings.  Second, even if it did, it would be 
 missing the 0 digit.
 
 - At the end, when you convert your list to a string, you don't need to use 
 list comprehension, since nx is already a list.  number = .join(nx) 
 should work fine.
 
 Also, in general, you need to stop and slow down and think like a 
 programmer.  If you get an error, your instinct shouldn't be to just hack 
 at it to make the error go away.  Look at the error and try to make sense 
 of it.  Learn what the error means and try to fix the core problem.
 
 And for @#$%'s sake...stop saying It isn't working and not elaborating.  
 You've been told by every other post in this thread to show us what you did 
 and what the error was.  You've also been told to *NOT* retype what you see 
 and to copy/paste your code and the error because when you make a typo when 
 copying, we might see a problem that doesn't exist and then you just get 
 more confused.
 
 Ok  I think I may have the question you guys are looking for.
 I just posted it.
 See above.
 
 But it's still broke.  :(

str(range(10)) doesn't do what you think it does.

Run 'print(str(range(10)))' and look at what you get.

Yeah, I know that.  My question is why?
The answer was that Python 3 only stores the min and max values but
you can still iterate over them.
I don't think that means what I think it means.
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[sohcahtoa82]

On Friday, October 24, 2014 12:36:23 PM UTC-7, Seymore4Head wrote:
 On Fri, 24 Oct 2014 12:25:33 -0700 (PDT), sohcahto...@gmail.com wrote:
 
 On Friday, October 24, 2014 12:12:10 PM UTC-7, Seymore4Head wrote:
  On Fri, 24 Oct 2014 11:57:12 -0700 (PDT), sohcahto...@gmail.com wrote:
  
  On Friday, October 24, 2014 11:17:53 AM UTC-7, Seymore4Head wrote:
   On Fri, 24 Oct 2014 11:52:15 -0600, Ian Kelly ian.g.ke...@gmail.com
   wrote:
   
   On Fri, Oct 24, 2014 at 11:03 AM, Seymore4Head
   Seymore4Head@hotmail.invalid wrote:
Actually I was a little frustrated when I added that line back in as
the other lines all work.
Using list(range(10)) Doesn't throw an error but it doesn't work.
   
http://i.imgur.com/DTc5zoL.jpg
   
The interpreter.   I don't know how to use that either.
   
   Try both of these in the interpreter, and observe the difference:
   
   7 in range(10)
   
   7 in range(10)
   
   Do you understand what the difference between 7 and 7 is?
   
   I do understand that.  7 is a number and 7 is a string.
   What my question was...and still is...is why
   Python 3 fails when I try using
   y=1 800 get charter
   
   y in range str(range(10))
   should work because y is a string and str(range(10)) should be
   y in str(1) fails.
   It doesn't give an error it's just not True when y is a number.
   
   These hints are just not working.  I am too thick for hints. :)
   If you could use it in the code, I might understand.
   The other work arounds that were posted work.
   I have used them.  str(range(10)) doesn't work.
   
   import string
   def nametonumber(name):
   lst=[]
   nx=[]
   digit=[]
   digit=.join(str(i) for i in range(10))
   for x in name:
   lst.append(x)
   for y in (lst):
   if y in list(range(1,10)):
   #if y in 1234567890:
   #if y.isdigit():
   #if y in digit:   
   #if y in string.digits:
   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(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 dialaho
   print (nametonumber(a))
   
   Please
  
  Your code here is actually pretty close to a correct answer.  Just a few 
  things to consider...
  
  - Why are you converting your name string to a list?  It is unnecessary.  
  When you do for y in some string, then y will still be single 
  characters on each iteration of the loop.
  
  - if y in string.digits should work fine.
  
  - if y in list(range(1,10) won't work for two reasons: First, it 
  creates a list of numbers, not strings.  Second, even if it did, it would 
  be missing the 0 digit.
  
  - At the end, when you convert your list to a string, you don't need to 
  use list comprehension, since nx is already a list.  number = .join(nx) 
  should work fine.
  
  Also, in general, you need to stop and slow down and think like a 
  programmer.  If you get an error, your instinct shouldn't be to just hack 
  at it to make the error go away.  Look at the error and try to make sense 
  of it.  Learn what the error means and try to fix the core problem.
  
  And for @#$%'s sake...stop saying It isn't working and not elaborating. 
   You've been told by every other post in this thread to show us what you 
  did and what the error was.  You've also been told to *NOT* retype what 
  you see and to copy/paste your code and the error because when you make a 
  typo when copying, we might see a problem that doesn't exist and then you 
  just get more confused.
  
  Ok  I think I may have the question you guys are looking for.
  I just posted it.
  See above.
  
  But it's still broke.  :(
 
 str(range(10)) doesn't do what you think it does.
 
 Run 'print(str(range(10)))' and look at what you get.
 
 Yeah, I know that.  My question is why?
 The answer was that Python 3 only stores the min and max values but
 you can still iterate over them.
 I don't think that means what I think it means.

You can iterate over them pretty much just means you can use them as the 
source of a 'for' loop.  But in your case, when you're calling 'for y in 
str(range(10))', you're not using 'range(10)' as the source of your loop, 
you're using the result of a str() function call, and you're calling str() on 
range(), which doesn't return a concrete value in Python 3.  If you try to 
print a range(), you're just getting a string containing your original call to 
range.

And that's why you're seeing 

Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 12:55:19 -0700 (PDT), sohcahto...@gmail.com wrote:

On Friday, October 24, 2014 12:36:23 PM UTC-7, Seymore4Head wrote:
 On Fri, 24 Oct 2014 12:25:33 -0700 (PDT), sohcahto...@gmail.com wrote:
 
 On Friday, October 24, 2014 12:12:10 PM UTC-7, Seymore4Head wrote:
  On Fri, 24 Oct 2014 11:57:12 -0700 (PDT), sohcahto...@gmail.com wrote:
  
  On Friday, October 24, 2014 11:17:53 AM UTC-7, Seymore4Head wrote:
   On Fri, 24 Oct 2014 11:52:15 -0600, Ian Kelly ian.g.ke...@gmail.com
   wrote:
   
   On Fri, Oct 24, 2014 at 11:03 AM, Seymore4Head
   Seymore4Head@hotmail.invalid wrote:
Actually I was a little frustrated when I added that line back in as
the other lines all work.
Using list(range(10)) Doesn't throw an error but it doesn't work.
   
http://i.imgur.com/DTc5zoL.jpg
   
The interpreter.   I don't know how to use that either.
   
   Try both of these in the interpreter, and observe the difference:
   
   7 in range(10)
   
   7 in range(10)
   
   Do you understand what the difference between 7 and 7 is?
   
   I do understand that.  7 is a number and 7 is a string.
   What my question was...and still is...is why
   Python 3 fails when I try using
   y=1 800 get charter
   
   y in range str(range(10))
   should work because y is a string and str(range(10)) should be
   y in str(1) fails.
   It doesn't give an error it's just not True when y is a number.
   
   These hints are just not working.  I am too thick for hints. :)
   If you could use it in the code, I might understand.
   The other work arounds that were posted work.
   I have used them.  str(range(10)) doesn't work.
   
   import string
   def nametonumber(name):
   lst=[]
   nx=[]
   digit=[]
   digit=.join(str(i) for i in range(10))
   for x in name:
   lst.append(x)
   for y in (lst):
   if y in list(range(1,10)):
   #if y in 1234567890:
   #if y.isdigit():
   #if y in digit:   
   #if y in string.digits:
   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(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 dialaho
   print (nametonumber(a))
   
   Please
  
  Your code here is actually pretty close to a correct answer.  Just a few 
  things to consider...
  
  - Why are you converting your name string to a list?  It is unnecessary. 
   When you do for y in some string, then y will still be single 
  characters on each iteration of the loop.
  
  - if y in string.digits should work fine.
  
  - if y in list(range(1,10) won't work for two reasons: First, it 
  creates a list of numbers, not strings.  Second, even if it did, it 
  would be missing the 0 digit.
  
  - At the end, when you convert your list to a string, you don't need to 
  use list comprehension, since nx is already a list.  number = 
  .join(nx) should work fine.
  
  Also, in general, you need to stop and slow down and think like a 
  programmer.  If you get an error, your instinct shouldn't be to just 
  hack at it to make the error go away.  Look at the error and try to make 
  sense of it.  Learn what the error means and try to fix the core problem.
  
  And for @#$%'s sake...stop saying It isn't working and not 
  elaborating.  You've been told by every other post in this thread to 
  show us what you did and what the error was.  You've also been told to 
  *NOT* retype what you see and to copy/paste your code and the error 
  because when you make a typo when copying, we might see a problem that 
  doesn't exist and then you just get more confused.
  
  Ok  I think I may have the question you guys are looking for.
  I just posted it.
  See above.
  
  But it's still broke.  :(
 
 str(range(10)) doesn't do what you think it does.
 
 Run 'print(str(range(10)))' and look at what you get.
 
 Yeah, I know that.  My question is why?
 The answer was that Python 3 only stores the min and max values but
 you can still iterate over them.
 I don't think that means what I think it means.

You can iterate over them pretty much just means you can use them as the 
source of a 'for' loop.  But in your case, when you're calling 'for y in 
str(range(10))', you're not using 'range(10)' as the source of your loop, 
you're using the result of a str() function call, and you're calling str() on 
range(), which doesn't return a concrete value in Python 3.  If you try to 
print a range(), you're just getting a 

Re: I am out of trial and error again Lists

[Denis McMahon]

On Fri, 24 Oct 2014 10:38:31 -0400, Seymore4Head wrote:

 I tried list(range(10)  

This is missing a )

It probably sat there waiting for you to finish the line.

list(range(10))

You have two ( in the line, you need two ) to match them. 

 I thought that would work in Python 3.  It
 didn't.

It does if you enter it properly.

also try:

str(list(range(10))) 

Note that that has three ( and three ) on the line.

-- 
Denis McMahon, denismfmcma...@gmail.com
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 20:37:31 + (UTC), Denis McMahon
denismfmcma...@gmail.com wrote:

On Fri, 24 Oct 2014 10:38:31 -0400, Seymore4Head wrote:

 I tried list(range(10)  

This is missing a )

It probably sat there waiting for you to finish the line.

list(range(10))

You have two ( in the line, you need two ) to match them. 

 I thought that would work in Python 3.  It
 didn't.

It does if you enter it properly.

also try:

str(list(range(10))) 

Note that that has three ( and three ) on the line.

I make lots of typing mistakes.  It is not that. Did you see the short
example I posted?

name=123-xyz-abc
for x in name:
if x in range(10):
print (Range,(x))
if x in str(range(10)):
print (String range,(x))

It doesn't throw an error but it doesn't print what you would expect.

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

Re: I am out of trial and error again Lists

[alister]

On Fri, 24 Oct 2014 16:58:00 -0400, Seymore4Head wrote:

 On Fri, 24 Oct 2014 20:37:31 + (UTC), Denis McMahon
 denismfmcma...@gmail.com wrote:
 
On Fri, 24 Oct 2014 10:38:31 -0400, Seymore4Head wrote:

 I tried list(range(10)

This is missing a )

It probably sat there waiting for you to finish the line.

list(range(10))

You have two ( in the line, you need two ) to match them.

 I thought that would work in Python 3.  It didn't.

It does if you enter it properly.

also try:

str(list(range(10)))

Note that that has three ( and three ) on the line.
 
 I make lots of typing mistakes.  It is not that. Did you see the short
 example I posted?
 
 name=123-xyz-abc
 for x in name:
 if x in range(10):
 print (Range,(x))
 if x in str(range(10)):
 print (String range,(x))
 
 It doesn't throw an error but it doesn't print what you would expect.

it prints what I expect, it probably does not print what you expect

you have may times been told that str(range(10)) does not do what you 
expect but you keep failing to test

you think that it crates a list of strings ['1','2','3'] but it does 
not
it creates a list  then turns the whole list into a string '[1,2,3...]'

people are suggesting you try this things in the interactive prompt 
because doing teaches far better than just reading.




-- 
Honesty's the best policy.
-- Miguel de Cervantes
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[mm0fmf]

On 24/10/2014 15:47, Seymore4Head wrote:

I have at least 10 ebooks.  I will get around to reading them soon.


Sooner would be better.

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

Re: I am out of trial and error again Lists

[Ian Kelly]

On Fri, Oct 24, 2014 at 2:58 PM, Seymore4Head
Seymore4Head@hotmail.invalid wrote:
 name=123-xyz-abc
 for x in name:
 if x in range(10):
 print (Range,(x))
 if x in str(range(10)):
 print (String range,(x))

 It doesn't throw an error but it doesn't print what you would expect.

That prints exactly what I expect it to. The first if prints nothing,
because you're testing whether a string is contained in a sequence of
ints. That will always be false. The second if prints those characters
from name that happen to be in the string range(10). That's the 1
and the a.

Apparently it doesn't print what *you* expect, which is why you need
to make your expectation clear and not assume that we will just read
your mind and immediately understand what you expect the code to do.
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Denis McMahon]

On Fri, 24 Oct 2014 14:15:13 -0400, Seymore4Head wrote:

 I do understand that.  7 is a number and 7 is a string.
 What my question was...and still is...is why Python 3 fails when I try
 using y=1 800 get charter
 
 y in range str(range(10))
 should work because y is a string and str(range(10)) should be y in
 str(1) fails.
 It doesn't give an error it's just not True when y is a number.

This is because str(range(10)) does not do what you think it does.

In python 2.x, str(range(10)) creates a string representation of the 
complete list, not a list of the string representation of the separate 
list elements. '[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]'

In python 3.x, str(range(10)) creates a string representation of the list 
object. 'range(0, 10)'

the only single digit strings in the python3 representation are 0 and 
1

To recreate the python2 behaviour in python 3, use:

str(list(range(10)))

which gives

'[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]'

howver the test:

if x.isdigit():

is much better.

But finally, with your telephone number decoder, look at:

http://www.codeskulptor.org/#user38_QnR06Upp4AH6h0Q.py

-- 
Denis McMahon, denismfmcma...@gmail.com
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 21:19:22 + (UTC), Denis McMahon
denismfmcma...@gmail.com wrote:

On Fri, 24 Oct 2014 14:15:13 -0400, Seymore4Head wrote:

 I do understand that.  7 is a number and 7 is a string.
 What my question was...and still is...is why Python 3 fails when I try
 using y=1 800 get charter
 
 y in range str(range(10))
 should work because y is a string and str(range(10)) should be y in
 str(1) fails.
 It doesn't give an error it's just not True when y is a number.

This is because str(range(10)) does not do what you think it does.

In python 2.x, str(range(10)) creates a string representation of the 
complete list, not a list of the string representation of the separate 
list elements. '[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]'

In python 3.x, str(range(10)) creates a string representation of the list 
object. 'range(0, 10)'

the only single digit strings in the python3 representation are 0 and 
1

To recreate the python2 behaviour in python 3, use:

str(list(range(10)))

which gives

'[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]'

howver the test:

if x.isdigit():

is much better.

But finally, with your telephone number decoder, look at:

http://www.codeskulptor.org/#user38_QnR06Upp4AH6h0Q.py

That is much cleaner than mine.  Nice.

I did make one more change to mine that makes it easier to read.
I changed treating all  -()  With a space.
I am still thinking about how to treat the large space if it is a
digit instead:
1 800 555 
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: I am out of trial and error again Lists

[Denis McMahon]

On Fri, 24 Oct 2014 10:38:31 -0400, Seymore4Head wrote:

 Thanks everyone for your suggestions.

Try loading the following in codeskulptor:

http://www.codeskulptor.org/#user38_j6kGKgeOMr_0.py

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

[Denis McMahon]

On Fri, 24 Oct 2014 15:07:06 -0400, Seymore4Head wrote:

 On Fri, 24 Oct 2014 19:40:39 +0100, Mark Lawrence
 breamore...@yahoo.co.uk wrote:
 
On 24/10/2014 19:20, Seymore4Head wrote:
 I meant to type:
 if y in range(1,10) doesn't work.
 Sigh Sorry


How many more times, state what you expect to happen and what actually
happens.  doesn't work is useless.  Please read this http://sscce.org/
 
 Good suggestion.
 OK  how is this?
 It doesn't print what I expect.
 Does it print what you expect?
 
 name=123-xyz-abc
 for x in name:
 if x in range(10):
 print (Range,(x))
 if x in str(range(10)):
 print (String range,(x))
 
 http://i.imgur.com/EGKUpAb.jpg

I suspect you're discovering the difference between the python2 and 
python3 range() functions, and what happens when you encapsulate them in 
string.

I've already posted about this once this evening.

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Re: I am out of trial and error again Lists

[Dave Angel]

Seymore4Head Seymore4Head@Hotmail.invalid Wrote in message:
 On Fri, 24 Oct 2014 09:54:23 -0700 (PDT), Rustom Mody
 rustompm...@gmail.com wrote:
 
Totally befuddled myself!

Are you deliberately misspelling list to lst
and hoping the error will go away.

And Puh LEESE
dont post screen shots of good ol ASCII text
 
 I didn't do that on purpose.  I make a lot of typing mistakes.
 Sorry
 

That's what copy/paste are for. Copy from your console, and paste
 into your email. Don't ever retype unless you're trying to
 frustrate us,

-- 
DaveA

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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 21:48:14 + (UTC), Denis McMahon
denismfmcma...@gmail.com wrote:

On Fri, 24 Oct 2014 10:38:31 -0400, Seymore4Head wrote:

 Thanks everyone for your suggestions.

Try loading the following in codeskulptor:

http://www.codeskulptor.org/#user38_j6kGKgeOMr_0.py

That is a useful way to test.
Thanks.
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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
Seymore4Head@Hotmail.invalid wrote:

Thanks for all the helpful replies.  I just discovered that there is
something wrong with my news feed.  Some of the messages did not make
it to me.  I can go back and read this thread in Google Groups but I
can't reply to it.

If I missed thanking or replying to anyone, that is the reason.

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

[Seymore4Head]

On Fri, 24 Oct 2014 18:09:59 -0400 (EDT), Dave Angel
da...@davea.name wrote:

Seymore4Head Seymore4Head@Hotmail.invalid Wrote in message:
 On Fri, 24 Oct 2014 09:54:23 -0700 (PDT), Rustom Mody
 rustompm...@gmail.com wrote:
 
Totally befuddled myself!

Are you deliberately misspelling list to lst
and hoping the error will go away.

And Puh LEESE
dont post screen shots of good ol ASCII text
 
 I didn't do that on purpose.  I make a lot of typing mistakes.
 Sorry
 

That's what copy/paste are for. Copy from your console, and paste
 into your email. Don't ever retype unless you're trying to
 frustrate us,

I promise I am not trying to frustrate anyone.  I know I have.
Sorry
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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
Seymore4Head@Hotmail.invalid wrote:

name=123-xyz-abc 
a=range(10)
b=list(range(10))
c=str(list(range(10)))
print (a,(a))
print (b,(b))
print (c,(c))

for x in name:
if x in a:
print (a,(x))  
if x in b:
print (b,(x))  
if x in c:
print (c,(x))

B is type list and C is type str.
I guess I am still a little too thick.  I would expect b and c to
work. 
http://i.imgur.com/dT3sEQq.jpg
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Re: I am out of trial and error again Lists

[Rustom Mody]

On Saturday, October 25, 2014 4:00:01 AM UTC+5:30, Seymore4Head wrote:
 On Fri, 24 Oct 2014 18:09:59 -0400 (EDT), Dave Angel wrote:
  Don't ever retype unless you're trying to
  frustrate us,
 
 I promise I am not trying to frustrate anyone.  I know I have.
 Sorry

No issues Seymore :-)

As far as I am concerned this has been useful and educative for me --
I found out about codeskulptor.

Though I am a bit conflicted whether it helps or confuses students.

Also many other things... What I take as minor distinctions 
between 
python 2 and 3 may not be so minor if one doesn't know whats going on.
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Re: I am out of trial and error again Lists

[Denis McMahon]

On Fri, 24 Oct 2014 17:35:34 -0400, Seymore4Head wrote:

But finally, with your telephone number decoder, look at:

http://www.codeskulptor.org/#user38_QnR06Upp4AH6h0Q.py
 
 That is much cleaner than mine.  Nice.
 
 I did make one more change to mine that makes it easier to read. I
 changed treating all  -()  With a space.
 I am still thinking about how to treat the large space if it is a digit
 instead:
 1 800 555 

Note that my decoder assumes anything other than a letter can be copied 
straight to the output string.

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

[Denis McMahon]

On Fri, 24 Oct 2014 16:58:00 -0400, Seymore4Head wrote:

 I make lots of typing mistakes.  It is not that. Did you see the short
 example I posted?
 
 name=123-xyz-abc
 for x in name:
 if x in range(10):
 print (Range,(x))
 if x in str(range(10)):
 print (String range,(x))
 
 It doesn't throw an error but it doesn't print what you would expect.

It prints exactly what I expect.

Try the following:

print(str(range(10)), type(str(range(10
print(str(list(range(10))), type(str(listr(range(10)

In python 3, str(x) just wraps x up and puts it in a string. range(x) 
generates an iterable range object.

hence str(range(10)) is a string telling you that range(10) is an iterable 
range object with certain boundaries.

However, list(iterable) expands the iterable to the full list of possible 
values, so str(list(range(10))) is a string representation of the list 
containing the values that the iterable range(10) creates.

Note that whether you're looking at a string representation of a value or 
the value itself is a lot clearer in the interpreter console where 
strings are displayed with quotes.

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

[Mark Lawrence]

On 24/10/2014 23:58, Seymore4Head wrote:

On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
Seymore4Head@Hotmail.invalid wrote:

name=123-xyz-abc
a=range(10)
b=list(range(10))
c=str(list(range(10)))
print (a,(a))
print (b,(b))
print (c,(c))

for x in name:
 if x in a:
 print (a,(x))
 if x in b:
 print (b,(x))
 if x in c:
 print (c,(x))

B is type list and C is type str.
I guess I am still a little too thick.  I would expect b and c to
work.
http://i.imgur.com/dT3sEQq.jpg



Why?

--
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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

[Rustom Mody]

On Saturday, October 25, 2014 4:30:47 AM UTC+5:30, Seymore4Head wrote:
 On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head wrote:
 
 name=123-xyz-abc 
 a=range(10)
 b=list(range(10))
 c=str(list(range(10)))
 print (a,(a))
 print (b,(b))
 print (c,(c))
 
 for x in name:
 if x in a:
 print (a,(x))  
 if x in b:
 print (b,(x))  
 if x in c:
 print (c,(x))
 
 B is type list and C is type str.
 I guess I am still a little too thick.  I would expect b and c to
 work. 

Lets simplify the problem a bit.
Do all the following in interpreter window

 name=012
 b=list(range(3))

 for x in name:  print x

 for x in b: print x

Same or different?

Now go back to Denis' nice example and put in type(x)
into each print

Same or different?
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Re: I am out of trial and error again Lists

[Denis McMahon]

On Fri, 24 Oct 2014 18:58:04 -0400, Seymore4Head wrote:

 On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
 Seymore4Head@Hotmail.invalid wrote:

OK, assuming you tried to run this in python3, not python2 or 
codeskulptor.

 name=123-xyz-abc
 a=range(10)

  a is an iterable object giving the numbers 0 through 9

 b=list(range(10))

  b is an list containing the numbers 0 through 9

 c=str(list(range(10)))

  c is an string representation of a list containing the numbers 0 
through 9

 print (a,(a))
 print (b,(b))
 print (c,(c))
 
 for x in name:

 ^ x is a string representing one character in name

 if x in a:
 print (a,(x))

  here you are looking for a string x amongst the numbers yielded by 
an iterable

 if x in b:
 print (b,(x))

  here you are comparing a string x with the elements of a list of 
numbers

 if x in c:
 print (c,(x))

  here you are comparing a string x with the characters in a string 
representation of a list of numbers

 B is type list and C is type str.
 I guess I am still a little too thick.  I would expect b and c to work.
 http://i.imgur.com/dT3sEQq.jpg

a is the range object: range(0, 9)
b is the list: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
c is the string: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

When you try and compare a character x eg 8 with the numbers yielded by 
the iterable a, none of them match because character 8 is not the same 
as number 8

When you try and compare a character x eg 8 with the elements of the 
list b, none of them match because character 8 is not the same as 
number 8

When you try and compare a character x eg 8 with the string 
representation of the list b, you get a match of x 8 to the 25th 
character of string c which is also 8.

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

[Seymore4Head]

On Fri, 24 Oct 2014 16:27:58 -0700 (PDT), Rustom Mody
rustompm...@gmail.com wrote:

On Saturday, October 25, 2014 4:30:47 AM UTC+5:30, Seymore4Head wrote:
 On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head wrote:
 
 name=123-xyz-abc 
 a=range(10)
 b=list(range(10))
 c=str(list(range(10)))
 print (a,(a))
 print (b,(b))
 print (c,(c))
 
 for x in name:
 if x in a:
 print (a,(x))  
 if x in b:
 print (b,(x))  
 if x in c:
 print (c,(x))
 
 B is type list and C is type str.
 I guess I am still a little too thick.  I would expect b and c to
 work. 

Lets simplify the problem a bit.
Do all the following in interpreter window

 name=012
 b=list(range(3))

 for x in name:  print x

 for x in b: print x

Same or different?

Now go back to Denis' nice example and put in type(x)
into each print

Same or different?

First.  The interpreter is not good for me to use even when I am using
Python 3 because I forget to add :  and I forget to put () around the
print statements.

To keep me from having to correct myself every time I use it, it is
just easier to make a short py file.

Here is mine:

name=012
b=list(range(3))
for x in name:  print (x)
print (type (x))
for x in b: print (x)
print (type (b))

I don't understand what I was supposed to learn from that.  I know
that name will be a string so x will be a string.
I would still think if you compare a 1 from a string to a 1 from a
list, it should be the same.

Obviously I am wrong, but we knew that already.
I get they are not the same, but I still think they should be.

name=012
b=list(range(3))
print (name[1])
print ([1])

1
[1]

OK  I get it.  They are not the same.  I was expecting 1


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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 23:21:43 + (UTC), Denis McMahon
denismfmcma...@gmail.com wrote:

On Fri, 24 Oct 2014 16:58:00 -0400, Seymore4Head wrote:

 I make lots of typing mistakes.  It is not that. Did you see the short
 example I posted?
 
 name=123-xyz-abc
 for x in name:
 if x in range(10):
 print (Range,(x))
 if x in str(range(10)):
 print (String range,(x))
 
 It doesn't throw an error but it doesn't print what you would expect.

It prints exactly what I expect.

Try the following:

print(str(range(10)), type(str(range(10
print(str(list(range(10))), type(str(listr(range(10)

In python 3, str(x) just wraps x up and puts it in a string. range(x) 
generates an iterable range object.

hence str(range(10)) is a string telling you that range(10) is an iterable 
range object with certain boundaries.

However, list(iterable) expands the iterable to the full list of possible 
values, so str(list(range(10))) is a string representation of the list 
containing the values that the iterable range(10) creates.

Note that whether you're looking at a string representation of a value or 
the value itself is a lot clearer in the interpreter console where 
strings are displayed with quotes.

I get it now.
Thanks
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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 19:48:16 -0400, Seymore4Head
Seymore4Head@Hotmail.invalid wrote:

On Fri, 24 Oct 2014 16:27:58 -0700 (PDT), Rustom Mody
rustompm...@gmail.com wrote:

On Saturday, October 25, 2014 4:30:47 AM UTC+5:30, Seymore4Head wrote:
 On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head wrote:
 
 name=123-xyz-abc 
 a=range(10)
 b=list(range(10))
 c=str(list(range(10)))
 print (a,(a))
 print (b,(b))
 print (c,(c))
 
 for x in name:
 if x in a:
 print (a,(x))  
 if x in b:
 print (b,(x))  
 if x in c:
 print (c,(x))
 
 B is type list and C is type str.
 I guess I am still a little too thick.  I would expect b and c to
 work. 

Lets simplify the problem a bit.
Do all the following in interpreter window

 name=012
 b=list(range(3))

 for x in name:  print x

 for x in b: print x

Same or different?

Now go back to Denis' nice example and put in type(x)
into each print

Same or different?

First.  The interpreter is not good for me to use even when I am using
Python 3 because I forget to add :  and I forget to put () around the
print statements.

To keep me from having to correct myself every time I use it, it is
just easier to make a short py file.

Here is mine:

name=012
b=list(range(3))
for x in name:  print (x)
print (type (x))
for x in b: print (x)
print (type (b))

I don't understand what I was supposed to learn from that.  I know
that name will be a string so x will be a string.
I would still think if you compare a 1 from a string to a 1 from a
list, it should be the same.

Obviously I am wrong, but we knew that already.
I get they are not the same, but I still think they should be.

name=012
b=list(range(3))
print (name[1])
print ([1])

1
[1]

OK  I get it.  They are not the same.  I was expecting 1

Wait!  I don't get it.
name=012
b=list(range(3))
print (name[1])
print (b[1])
1
1

I forgot the b

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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
Seymore4Head@Hotmail.invalid wrote:

name=012
b=list(range(3))
print (name[1])
print (b[1])
if name[1] == b[1]:
print (Eureka!)
else:
print (OK, I get it)
 
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Re: I am out of trial and error again Lists

[MRAB]

On 2014-10-25 00:57, Seymore4Head wrote:
[snip]

Wait!  I don't get it.
name=012
b=list(range(3))
print (name[1])
print (b[1])
1
1

I forgot the b


If you print the int 1, you'll see:

1

If you print the string 1, you'll see:

1

Normally you want it to print only the characters of the string. Think
how annoying it would be if every time you printed a string it appeared
in quotes:

 print(Hello world!)
'Hello world!'

How could you print just the text:

Hello world!

No, it's better that it prints the characters of the string.

One function you can use is repr:

x = 1
y = 1
print(repr(x))
print(repr(y))

This will print:

1
'1'

OK, now it's clear that x is an int and y is a string.

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Re: I am out of trial and error again Lists

[Terry Reedy]

On 10/24/2014 6:27 PM, Seymore4Head wrote:


I promise I am not trying to frustrate anyone.  I know I have.


Seymore, if you want to learn real Python, download and install 3.4.2 
and either use the Idle Shell and Editor or the interactive console 
interpreter and a decent programmer editor.


I cannot recommend CodeSkulptor to anyone.  The opening statement
# CodeSkulptor runs Python programs
is deceptive.  CodeSkulptor Python is not Python. It does not correspond 
to any x.y version of Python. It is a somewhat crippled subset of 
ancient Python (2.1) with selected additions of later features.  It does 
not have exceptions, raise, and try: except:.  These are an essential 
part of original Python and Python today  It does not have complex 
(maybe introduced in 1.5) and unicode (introduced in 2.0).  Let that 
pass.  More important, it does not have new-style classes, an essential 
new feature introduced in 2.2.  Python beginners should start with 
unified new-styled classes.  If lucky, they need never learn about the 
original old style, dis-unified type versus class system that started 
going away in 2.2, is mostly gone in 2.7. and completely gone in 3.0.


If you really want to continue with CodeSkulpter Python, you should find 
a CodeSkulpterPython list.  You cannot expect people here to know that 
legal code like

  class I(int): pass
will not run, but will fail with an exceeding cryptic message:
  Line 1: undefined: TypeError: a.$d is undefined
Nor can you expect us to know all the other limitations.

This is a list for Python.  If you want help here, get and use a real 
Python interpreter, with a proper interactive mode, as multiple people 
have suggested.


--
Terry Jan Reedy


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Re: I am out of trial and error again Lists

[Seymore4Head]

On Sat, 25 Oct 2014 01:20:53 +0100, MRAB pyt...@mrabarnett.plus.com
wrote:

On 2014-10-25 00:57, Seymore4Head wrote:
[snip]
 Wait!  I don't get it.
 name=012
 b=list(range(3))
 print (name[1])
 print (b[1])
 1
 1

 I forgot the b

If you print the int 1, you'll see:

1

If you print the string 1, you'll see:

1

Normally you want it to print only the characters of the string. Think
how annoying it would be if every time you printed a string it appeared
in quotes:

  print(Hello world!)
'Hello world!'

How could you print just the text:

Hello world!

No, it's better that it prints the characters of the string.

One function you can use is repr:

x = 1
y = 1
print(repr(x))
print(repr(y))

This will print:

1
'1'

OK, now it's clear that x is an int and y is a string.

Yes

x = 123
y = 123
z = [1,2,3]
print(repr(x))
print(repr(y))
print(repr(z))
123
'123'
[1, 2, 3]

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

[Seymore4Head]

On Fri, 24 Oct 2014 20:27:03 -0400, Terry Reedy tjre...@udel.edu
wrote:

On 10/24/2014 6:27 PM, Seymore4Head wrote:

 I promise I am not trying to frustrate anyone.  I know I have.

Seymore, if you want to learn real Python, download and install 3.4.2 
and either use the Idle Shell and Editor or the interactive console 
interpreter and a decent programmer editor.

I cannot recommend CodeSkulptor to anyone.  The opening statement
# CodeSkulptor runs Python programs
is deceptive.  CodeSkulptor Python is not Python. It does not correspond 
to any x.y version of Python. It is a somewhat crippled subset of 
ancient Python (2.1) with selected additions of later features.  It does 
not have exceptions, raise, and try: except:.  These are an essential 
part of original Python and Python today  It does not have complex 
(maybe introduced in 1.5) and unicode (introduced in 2.0).  Let that 
pass.  More important, it does not have new-style classes, an essential 
new feature introduced in 2.2.  Python beginners should start with 
unified new-styled classes.  If lucky, they need never learn about the 
original old style, dis-unified type versus class system that started 
going away in 2.2, is mostly gone in 2.7. and completely gone in 3.0.

If you really want to continue with CodeSkulpter Python, you should find 
a CodeSkulpterPython list.  You cannot expect people here to know that 
legal code like
   class I(int): pass
will not run, but will fail with an exceeding cryptic message:
   Line 1: undefined: TypeError: a.$d is undefined
Nor can you expect us to know all the other limitations.

This is a list for Python.  If you want help here, get and use a real 
Python interpreter, with a proper interactive mode, as multiple people 
have suggested.

OK.  I will.
Thanks

But the difference between Python 2 and Codeskulptor was not an issue
with this question.
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Re: I am out of trial and error again Lists

[Cameron Simpson]

On 24Oct2014 20:37, Seymore4Head Seymore4Head@Hotmail.invalid wrote:

On Sat, 25 Oct 2014 01:20:53 +0100, MRAB pyt...@mrabarnett.plus.com

One function you can use is repr:

x = 1
y = 1
print(repr(x))
print(repr(y))

This will print:

1
'1'

OK, now it's clear that x is an int and y is a string.


Yes


In particular, Python's interactive mode uses repr to print the result of any 
expression that whose value was not None:


  [/Users/cameron]fleet* python
  Python 2.7.8 (default, Oct  3 2014, 02:34:26)
  [GCC 4.2.1 Compatible Apple LLVM 5.1 (clang-503.0.40)] on darwin
  Type help, copyright, credits or license for more information.
   x=1
   y='1'
   x
  1
   y
  '1'
  

so you get this for free in that mode.

Cheers,
Cameron Simpson c...@zip.com.au

If your new theorem can be stated with great simplicity, then there
will exist a pathological exception.- Adrian Mathesis
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Re: I am out of trial and error again Lists

[Larry Hudson]

On 10/24/2014 07:38 AM, Seymore4Head wrote:
snip

I do get the difference.  I don't actually use Python 2.  I use
CodeSkulptor.  I do have Python 3 installed.  Actually I have Python 2
installed but IDLE defaults to Python 3.  So it is a pain to actually
load Python 2.



Exactly HOW are you trying to run Idle?  A default install of Py2 and Py3 in Windows should have 
also installed Idle for each version.  In my Win7 system, they are BOTH in the standard menu, 
you should be able to call up either one.


OT:  Side comment:  I rarely use Windows these days, maybe once every two or three months -- I 
MUCH prefer Linux.  Among other reasons its a far better environment for programming.  I only 
have one (active) system with Windows installed, and two others with Linux only.  Actually make 
that three, if you count my Raspberry Pi.   :-)


 -=- Larry -=-

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Re: I am out of trial and error again Lists

[Seymore4Head]

On Fri, 24 Oct 2014 19:16:21 -0700, Larry Hudson org...@yahoo.com
wrote:

On 10/24/2014 07:38 AM, Seymore4Head wrote:
snip
 I do get the difference.  I don't actually use Python 2.  I use
 CodeSkulptor.  I do have Python 3 installed.  Actually I have Python 2
 installed but IDLE defaults to Python 3.  So it is a pain to actually
 load Python 2.


Exactly HOW are you trying to run Idle?  A default install of Py2 and Py3 in 
Windows should have 
also installed Idle for each version.  In my Win7 system, they are BOTH in the 
standard menu, 
you should be able to call up either one.

OT:  Side comment:  I rarely use Windows these days, maybe once every two or 
three months -- I 
MUCH prefer Linux.  Among other reasons its a far better environment for 
programming.  I only 
have one (active) system with Windows installed, and two others with Linux 
only.  Actually make 
that three, if you count my Raspberry Pi.   :-)

  -=- Larry -=-
I have a directory of my py files.  I right click on one of the py
files and open with IDLE   Windows XP

If I try to open a py file I have for Python 2 it still opens using
IDLE 3.

I don't have many py 2 files anyway.
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Re: I am out of trial and error again Lists

[Rustom Mody]

On Saturday, October 25, 2014 5:21:01 AM UTC+5:30, Seymore4Head wrote:
 On Fri, 24 Oct 2014 16:27:58 -0700 (PDT), Rustom Mody wrote:
 
 On Saturday, October 25, 2014 4:30:47 AM UTC+5:30, Seymore4Head wrote:
  On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head wrote:
  
  name=123-xyz-abc 
  a=range(10)
  b=list(range(10))
  c=str(list(range(10)))
  print (a,(a))
  print (b,(b))
  print (c,(c))
  
  for x in name:
  if x in a:
  print (a,(x))  
  if x in b:
  print (b,(x))  
  if x in c:
  print (c,(x))
  
  B is type list and C is type str.
  I guess I am still a little too thick.  I would expect b and c to
  work. 
 
 Lets simplify the problem a bit.
 Do all the following in interpreter window
 
  name=012
  b=list(range(3))
 
  for x in name:  print x
 
  for x in b: print x
 
 Same or different?
 
 Now go back to Denis' nice example and put in type(x)
 into each print
 
 Same or different?
 
 First.  The interpreter is not good for me to use even when I am using
 Python 3 because I forget to add :  and I forget to put () around the
 print statements.
 

What would you say to a person who
- Buys a Lambhorgini
- Hitches a horse (or bullock) to it
- Moans how clumsily slow it is.

   ??

In case you dont get it:

- the interpreter is one of the factors that puts python into the hi-end class.
- the bullock/horse class is the C/Java type language where you always 
need to work through files

Here is my list of points of sliding for bullock category to Lambhorgini:
http://blog.languager.org/2012/10/functional-programming-lost-booty.html
['Interpreter' is called 'REPL' there]

 To keep me from having to correct myself every time I use it, it is
 just easier to make a short py file.

Yes the intention is right, The implementation is wrong.

Programmers consider it a virtue to be lazy.
But you have to put it some work to learn to be lazy in an effective way
And currently you are being lazy on the wrong front.

Hints:
1. A good programmer tries out things at the interpreter ONE LINE AT A TIME
2. What he tries out are usually EXPRESSIONS like eg
   str(list(range(10)))

   And not STATEMENTS like
   if x in name: 
  print x

3. IOW a good programmer rarely needs to type a colon at the interpreter

4. The least useful statement to try at the interpreter is print.

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Re: I am out of trial and error again Lists

[Rustom Mody]

On Saturday, October 25, 2014 9:17:12 AM UTC+5:30, Rustom Mody wrote:
 4. The least useful statement to try at the interpreter is print.

Yeah this is python2 thinking; in python 3, print is technically an expression.


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Re: I am out of trial and error again Lists

[Ben Finney]

Rustom Mody rustompm...@gmail.com writes:

 On Saturday, October 25, 2014 9:17:12 AM UTC+5:30, Rustom Mody wrote:
  4. The least useful statement to try at the interpreter is print.

 Yeah this is python2 thinking; in python 3, print is technically an
 expression.

This is wrong thinking. In Python 3, print is a function.

-- 
 \   “[W]hoever is able to make you absurd is able to make you |
  `\unjust.” —Voltaire |
_o__)  |
Ben Finney

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Re: I am out of trial and error again Lists

[Rustom Mody]

On Saturday, October 25, 2014 9:56:02 AM UTC+5:30, Ben Finney wrote:
 Rustom Mody writes:
 
  On Saturday, October 25, 2014 9:17:12 AM UTC+5:30, Rustom Mody wrote:
   4. The least useful statement to try at the interpreter is print.
 
  Yeah this is python2 thinking; in python 3, print is technically an
  expression.
 
 This is wrong thinking. In Python 3, print is a function.

[tl;dr at bottom]

Ok I was a bit sloppy -- should have said 'the print'
But there was no specific prior use that 'the' would refer to.
So one could say (if you like!):

|  | Python 2  | Python 3   |
| print| syntax| function   |
| print(x) | statement | expression |

I was really talking of the second row not the first

So much for being legalistic -- An approach which is ultimately not helpful.

For one thing function is one kind of expression ie its a subset not a disjoint 
relation.

More important (in this case) its bad pedagogy.
Its generally accepted that side-effecting functions are not a good idea
-- typically a function that returns something and changes global state.

In that sense its best to think of print(x) as syntactically an expression, 
semantically a statement.

Or put differently, the following error is more poorly reported in
python3 than in python2

Python 2.7.8 (default, Oct  7 2014, 17:59:21) 
[GCC 4.9.1] on linux2
Type help, copyright, credits or license for more information.
 2 + (print 2)
  File stdin, line 1
2 + (print 2)
 ^
SyntaxError: invalid syntax


Python 3.4.2 (default, Oct  8 2014, 10:45:20) 
[GCC 4.9.1] on linux
Type help, copyright, credits or license for more information.
 2 + (print (2))
2
Traceback (most recent call last):
  File stdin, line 1, in module
TypeError: unsupported operand type(s) for +: 'int' and 'NoneType'

=
tl;dr I dont think all this is very helpful to Seymore
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Re: I am out of trial and error again Lists

[Chris Angelico]

On Sat, Oct 25, 2014 at 4:40 PM, Rustom Mody rustompm...@gmail.com wrote:
 Its generally accepted that side-effecting functions are not a good idea
 -- typically a function that returns something and changes global state.

Only in certain circles. Not in Python. There are large numbers of
functions with side effects (mutator methods like list.append,
anything that needs lots of state like random.random, everything with
external effect like I/O, heaps of stuff), and it is most definitely
not frowned upon.

In Python 3 (or Python 2 with the future directive), print is a
function, print() an expression. It's not semantically a statement.

ChrisA
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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 pyshell#0, line 1, in module
.join(list(range(10)))
TypeError: sequence item 0: expected string, int found

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

--
Terry Jan Reedy

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Re: I am out of trial and error again Lists

[Larry Hudson]

On 10/22/2014 03:30 PM, Seymore4Head wrote:

On Wed, 22 Oct 2014 16:30:37 -0400, Seymore4Head
Seymore4Head@Hotmail.invalid 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))

I know you are trying to explore lists here, but I found myself somewhat intrigued with the 
problem itself, so I wrote a different version.  This version does not use lists but only 
strings.  I'm just presenting it as-is to let you try to follow the logic, but if you ask, I'll 
explain it in detail.  It turns your long sequence of if's essentially into a single line -- 
unfortunately 's' and 'z' have to be handled as special-cases, which turns that single line into 
a six-line if/elif/else set.  You might consider this line 'tricky', but I'll just say it's just 
looking at the problem from a different viewpoint.  BTW, this version accepts upper-case as well 
as lower-case.  isdigit() and isalpha() are standard string methods.


#--  Code  --
def name2number(name):
nstr = ''#  Phone-number string to return
codes = 'abcdefghijklmnopqrtuvwxy'  #  Note missing s and z

for ch in name:
if ch in  -():
nstr += ch
elif ch.isdigit():
nstr += ch
elif ch.isalpha():
ch = ch.lower()
#   S and Z are special cases
if ch == 's':
nstr += '7'
elif ch == 'z':
nstr += '9'
else:
nstr += str(codes.index(ch) // 3 + 2)
return nstr
#---  End of Code  -

A possible variation would be to make nstr a list instead of a string, and use .append() instead 
of the +=, and finally return the string by using join() on the list.  Also, the if and first 
elif in the for could be combined:  if ch in  -() or ch.isdigit():


 -=- Larry -=-

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Re: I am out of trial and error again Lists

[Mark Lawrence]

On 23/10/2014 02:57, Seymore4Head wrote:

On Wed, 22 Oct 2014 21:35:19 -0400, Seymore4Head
Seymore4Head@Hotmail.invalid wrote:


On Thu, 23 Oct 2014 02:31:57 +0100, MRAB pyt...@mrabarnett.plus.com
wrote:


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

On Thu, 23 Oct 2014 11:05:08 +1100, Steven D'Aprano
steve+comp.lang.pyt...@pearwood.info 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 stdin, line 1, in module
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 stdin, line 1, in module
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.  :(



If you were to read and digest what is written it would help.  You're 
trying to run IDLE.  We're talking the interactive interpreter.  If (at 
least on Windows) you run a command prompt and then type pythoncr you 
should see something like this.


Python 3.4.2 (v3.4.2:ab2c023a9432, Oct  6 2014, 22:16:31) [MSC v.1600 64 
bit (AMD64)] on win32

Type help, copyright, credits or license for more information.

--
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

[Ian Kelly]

On Thu, Oct 23, 2014 at 1:20 AM, Mark Lawrence breamore...@yahoo.co.uk wrote:
 If you were to read and digest what is written it would help.  You're trying
 to run IDLE.  We're talking the interactive interpreter.

IDLE includes the interactive interpreter.

  If (at least on
 Windows) you run a command prompt and then type pythoncr you should see
 something like this.

 Python 3.4.2 (v3.4.2:ab2c023a9432, Oct  6 2014, 22:16:31) [MSC v.1600 64 bit
 (AMD64)] on win32
 Type help, copyright, credits or license for more information.

Same thing comes up when I start IDLE, so what's your point?
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Re: I am out of trial and error again Lists

[Mark Lawrence]

On 23/10/2014 08:56, Ian Kelly wrote:

On Thu, Oct 23, 2014 at 1:20 AM, Mark Lawrence breamore...@yahoo.co.uk wrote:

If you were to read and digest what is written it would help.  You're trying
to run IDLE.  We're talking the interactive interpreter.


IDLE includes the interactive interpreter.


  If (at least on
Windows) you run a command prompt and then type pythoncr you should see
something like this.

Python 3.4.2 (v3.4.2:ab2c023a9432, Oct  6 2014, 22:16:31) [MSC v.1600 64 bit
(AMD64)] on win32
Type help, copyright, credits or license for more information.


Same thing comes up when I start IDLE, so what's your point?



When you run the interactive interpreter thats all you get.  The OP 
isn't the first person to try things with IDLE not realising you need to 
have a script loaded to do something.


--
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

[Denis McMahon]

On Wed, 22 Oct 2014 18:30:17 -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

In the last post I made, I suggested to you the mechanisms of using the 
python console and using code which prints out variables after every line.

Try the following 3 commands at the console:

 10
 range(10)
 str(range(10))

10 is just the number 10
range(10) is a list of 10 integers in the sequence 0 to 9
str(range(10)) is?

Please stop the stab in the dark trial and error coding followed by the 
plaintive why doesn't it work wailing, and put some effort into reading 
and understanding the manuals.

If a function doesn't do what you expect with the input you think you've 
given it, there is invariably one of three causes:

a) The input you actually gave it isn't the input you thought you gave it
b) You didn't read the description of the function, and it doesn't in 
fact do what you thought
c) Both a and b above

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

[Denis McMahon]

On Wed, 22 Oct 2014 19:11:51 -0400, Seymore4Head wrote:

 On Wed, 22 Oct 2014 22:43:14 + (UTC), Denis McMahon
 denismfmcma...@gmail.com 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.

If you don't understand the error, that is the point that you should ask 
for help, rather than make stab in the dark attempts to fix it and then 
asking why the solutions don't work.

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

[Rustom Mody]

On Thursday, October 23, 2014 1:39:32 PM UTC+5:30, Mark Lawrence wrote:
 On 23/10/2014 08:56, Ian Kelly wrote:
  On Thu, Oct 23, 2014 at 1:20 AM, Mark Lawrence  wrote:
  If you were to read and digest what is written it would help.  You're 
  trying
  to run IDLE.  We're talking the interactive interpreter.
 
  IDLE includes the interactive interpreter.
 
If (at least on
  Windows) you run a command prompt and then type pythoncr you should see
  something like this.
 
  Python 3.4.2 (v3.4.2:ab2c023a9432, Oct  6 2014, 22:16:31) [MSC v.1600 64 
  bit
  (AMD64)] on win32
  Type help, copyright, credits or license for more information.
 
  Same thing comes up when I start IDLE, so what's your point?
 
 
 When you run the interactive interpreter thats all you get.  The OP 
 isn't the first person to try things with IDLE not realising you need to 
 have a script loaded to do something.

If I do (at the shell prompt with an example)
$ python3
Python 3.4.2 (default, Oct  8 2014, 10:45:20) 
[GCC 4.9.1] on linux
Type help, copyright, credits or license for more information.
 2+3
5

And if I do
$ idle3
I get a new window containing

Python 3.4.2 (default, Oct  8 2014, 10:45:20) 
[GCC 4.9.1] on linux
Type copyright, credits or license() for more information.
 2+3
5

so...
Still not clear whats your point.
idle and python interpreter seem to be mostly the same

[As best as I can make out the OP is not using the standalone
interpreter
nor idle
nor (the many options like) python-interpreter-inside-emacs
nor ipython
nor ...
]
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