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About python while statement and pop()

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  About python while statement and pop() hito koto <hitokoto2014@gmail.com> - 2014-06-11 20:12 -0700
    Re:About python while statement and pop() Dave Angel <davea@davea.name> - 2014-06-11 22:34 -0500
    Re: About python while statement and pop() Vincent Vande Vyvre <vincent.vandevyvre@swing.be> - 2014-06-12 05:40 +0200
    Re: About python while statement and pop() Chris Angelico <rosuav@gmail.com> - 2014-06-12 13:58 +1000
      Re: About python while statement and pop() hito koto <hitokoto2014@gmail.com> - 2014-06-11 21:56 -0700
        Re: About python while statement and pop() Chris Angelico <rosuav@gmail.com> - 2014-06-12 15:08 +1000
        Re: About python while statement and pop() Steven D'Aprano <steve@pearwood.info> - 2014-06-12 05:43 +0000
          Re: About python while statement and pop() hito koto <hitokoto2014@gmail.com> - 2014-06-11 23:06 -0700
          Re: About python while statement and pop() hito koto <hitokoto2014@gmail.com> - 2014-06-12 00:13 -0700
          Re: About python while statement and pop() hito koto <hitokoto2014@gmail.com> - 2014-06-12 02:51 -0700
    Re: About python while statement and pop() Mark H Harris <harrismh777@gmail.com> - 2014-06-12 11:41 -0500
    Re: About python while statement and pop() Mark H Harris <harrismh777@gmail.com> - 2014-06-12 11:49 -0500
      Re: About python while statement and pop() Marko Rauhamaa <marko@pacujo.net> - 2014-06-12 19:55 +0300
        Re: About python while statement and pop() Mark H Harris <harrismh777@gmail.com> - 2014-06-12 12:07 -0500
      Re: About python while statement and pop() Chris Angelico <rosuav@gmail.com> - 2014-06-13 02:57 +1000
        Re: About python while statement and pop() Mark H Harris <harrismh777@gmail.com> - 2014-06-12 12:10 -0500
        Re: About python while statement and pop() Mark H Harris <harrismh777@gmail.com> - 2014-06-12 12:16 -0500
          Re: About python while statement and pop() wxjmfauth@gmail.com - 2014-06-12 11:37 -0700

#73196 — About python while statement and pop()

Hello,all
I'm first time,

I want to make a while statement which can function the same x.pop () and without the use of pop、how can i to do?

i want to change this is code:

def foo(x):
    y = []
    while x !=[]:
        y.append(x.pop())
    return y

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#73197

hito koto <hitokoto2014@gmail.com> Wrote in message:
> Hello,all
> I'm first time,
> 
> I want to make a while statement which can function the same x.pop () and without the use of pop、how can i to do?

No idea what the question means. Are you just trying to rewrite
 the loop in a python implementation where pop is broken?
 


> 
> i want to change this is code:
> 
> def foo(x):
>     y = []
>     while x !=[]:
>         y.append(x.pop())
>     return y

Perhaps you're looking for the extend method. 

-- 
DaveA

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#73198

Le 12/06/2014 05:12, hito koto a écrit :
> Hello,all
> I'm first time,
>
> I want to make a while statement which can function the same x.pop () and without the use of pop、how can i to do?
>
> i want to change this is code:
>
> def foo(x):
>      y = []
>      while x !=[]:
>          y.append(x.pop())
>      return y
Something like that :

def foo(x):
     return reversed(x)

-- 
Vincent V.V.
Oqapy <https://launchpad.net/oqapy> . Qarte 
<https://launchpad.net/qarte> . PaQager <https://launchpad.net/paqager>

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#73199

On Thu, Jun 12, 2014 at 1:40 PM, Vincent Vande Vyvre
<vincent.vandevyvre@swing.be> wrote:
> Le 12/06/2014 05:12, hito koto a écrit :
>
>> Hello,all
>> I'm first time,
>>
>> I want to make a while statement which can function the same x.pop () and
>> without the use of pop、how can i to do?
>>
>> i want to change this is code:
>>
>> def foo(x):
>>      y = []
>>      while x !=[]:
>>          y.append(x.pop())
>>      return y
>
> Something like that :
>
> def foo(x):
>     return reversed(x)

That doesn't do the same thing, though. Given a list x, the original
function will empty that list and return a new list in reverse order,
but yours will return a reversed iterator over the original list
without changing it. This is more accurate, but still not identical,
and probably not what the OP's teacher is looking for:

def foo(x):
    y = x[::-1]
    x[:] = []
    return y

If the mutation of x is unimportant, it can simply be:

def foo(x):
    return x[::-1]

ChrisA

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#73203

2014年6月12日木曜日 12時58分27秒 UTC+9 Chris Angelico:
> On Thu, Jun 12, 2014 at 1:40 PM, Vincent Vande Vyvre
> 
> <vincent.vandevyvre@swing.be> wrote:
> 
> > Le 12/06/2014 05:12, hito koto a écrit :
> 
> >
> 
> >> Hello,all
> 
> >> I'm first time,
> 
> >>
> 
> >> I want to make a while statement which can function the same x.pop () and
> 
> >> without the use of pop、how can i to do?
> 
> >>
> 
> >> i want to change this is code:
> 
> >>
> 
> >> def foo(x):
> 
> >>      y = []
> 
> >>      while x !=[]:
> 
> >>          y.append(x.pop())
> 
> >>      return y
> 
> >
> 
> > Something like that :
> 
> >
> 
> > def foo(x):
> 
> >     return reversed(x)
> 
> 
> 
> That doesn't do the same thing, though. Given a list x, the original
> 
> function will empty that list and return a new list in reverse order,
> 
> but yours will return a reversed iterator over the original list
> 
> without changing it. This is more accurate, but still not identical,
> 
> and probably not what the OP's teacher is looking for:
> 
> 
> 
> def foo(x):
> 
>     y = x[::-1]
> 
>     x[:] = []
> 
>     return y
> 
> 
> 
> If the mutation of x is unimportant, it can simply be:
> 
> 
> 
> def foo(x):
> 
>     return x[::-1]
> 
> 
> 
> ChrisA


I want to use while statement,

for example:
>>> def foo(x):
...     y = []
...     while x !=[]:
...         y.append(x.pop())
...     return y
...
>>> print foo(a)
[[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
>>> a
[]   but this is empty　
>>> so,I want to leave a number of previous (a = [[1, 2, 3, 4],[5, 6, 7, 8, 9],[10]])

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#73204

On Thu, Jun 12, 2014 at 2:56 PM, hito koto <hitokoto2014@gmail.com> wrote:
> I want to use while statement,

This sounds like homework. Go back to your teacher/tutor for
assistance, rather than asking us to do the work for you; or at very
least, word your question in such a way that we can help you to learn,
rather than just give you the answer.

Second problem: You're using Google Groups. This makes your posts
messy, especially when you quote someone else's text. Please either
fix your posts before sending them, or read and post by some other
means, such as the mailing list:

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

ChrisA

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#73205

On Wed, 11 Jun 2014 21:56:06 -0700, hito koto wrote:

> I want to use while statement,
> 
> for example:
>>>> def foo(x):
> ...     y = []
> ...     while x !=[]:
> ...         y.append(x.pop())
> ...     return y
> ...
>>>> print foo(a)
> [[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
>>>> a
> []   but this is empty
>>>> so,I want to leave a number of previous (a = [[1, 2, 3, 4],[5, 6, 7,
>>>> 8, 9],[10]])


I wouldn't use a while statement. The easy way is:

py> a = [[1, 2, 3, 4],[5, 6, 7, 8, 9],[10]]
py> y = a[::-1]
py> print y
[[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
py> print a
[[1, 2, 3, 4], [5, 6, 7, 8, 9], [10]]

If you MUST use a while loop, then you need something like this:


def foo(x):
    y = []
    index = 0
    while index < len(x):
        y.append(x[i])
        i += 1
    return y


This does not copy in reverse order. To make it copy in reverse order, 
index should start at len(x) - 1 and end at 0.



-- 
Steven

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#73207

2014年6月12日木曜日 14時43分42秒 UTC+9 Steven D'Aprano:
> On Wed, 11 Jun 2014 21:56:06 -0700, hito koto wrote:
> 
> 
> 
> > I want to use while statement,
> 
> > 
> 
> > for example:
> 
> >>>> def foo(x):
> 
> > ...     y = []
> 
> > ...     while x !=[]:
> 
> > ...         y.append(x.pop())
> 
> > ...     return y
> 
> > ...
> 
> >>>> print foo(a)
> 
> > [[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
> 
> >>>> a
> 
> > []   but this is empty
> 
> >>>> so,I want to leave a number of previous (a = [[1, 2, 3, 4],[5, 6, 7,
> 
> >>>> 8, 9],[10]])
> 
> 
> 
> 
> 
> I wouldn't use a while statement. The easy way is:
> 
> 
> 
> py> a = [[1, 2, 3, 4],[5, 6, 7, 8, 9],[10]]
> 
> py> y = a[::-1]
> 
> py> print y
> 
> [[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
> 
> py> print a
> 
> [[1, 2, 3, 4], [5, 6, 7, 8, 9], [10]]
> 
> 
> 
> If you MUST use a while loop, then you need something like this:
> 
> 
> 
> 
> 
> def foo(x):
> 
>     y = []
> 
>     index = 0
> 
>     while index < len(x):
> 
>         y.append(x[i])
> 
>         i += 1
> 
>     return y
> 
> 
> 
> 
> 
> This does not copy in reverse order. To make it copy in reverse order, 
> 
> index should start at len(x) - 1 and end at 0.
> 
> 
> 
> 
> 
> 
> 
> -- 
> 
> Steven

Hi, Steven:
Thanks,

My goal is to be able to in many ways python

Sorry, I was mistake,
I want to leave a number of previous (a = [[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]] )

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#73209

2014年6月12日木曜日 14時43分42秒 UTC+9 Steven D'Aprano:
> On Wed, 11 Jun 2014 21:56:06 -0700, hito koto wrote:
> 
> 
> 
> > I want to use while statement,
> 
> > 
> 
> > for example:
> 
> >>>> def foo(x):
> 
> > ...     y = []
> 
> > ...     while x !=[]:
> 
> > ...         y.append(x.pop())
> 
> > ...     return y
> 
> > ...
> 
> >>>> print foo(a)
> 
> > [[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
> 
> >>>> a
> 
> > []   but this is empty
> 
> >>>> so,I want to leave a number of previous (a = [[1, 2, 3, 4],[5, 6, 7,
> 
> >>>> 8, 9],[10]])
> 
> 
> 
> 
> 
> I wouldn't use a while statement. The easy way is:
> 
> 
> 
> py> a = [[1, 2, 3, 4],[5, 6, 7, 8, 9],[10]]
> 
> py> y = a[::-1]
> 
> py> print y
> 
> [[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
> 
> py> print a
> 
> [[1, 2, 3, 4], [5, 6, 7, 8, 9], [10]]
> 
> 
> 
> If you MUST use a while loop, then you need something like this:
> 
> 
> 
> 
> 
> def foo(x):
> 
>     y = []
> 
>     index = 0
> 
>     while index < len(x):
> 
>         y.append(x[i])
> 
>         i += 1
> 
>     return y
> 
> 
> 
> 
> 
> This does not copy in reverse order. To make it copy in reverse order, 
> 
> index should start at len(x) - 1 and end at 0.
> 
> 
> 
> 
> 
> 
> 
> -- 
> 
> Steven

Hi, Steven:
Thanks,

My goal is to be able to in many ways python

Sorry, I was mistake,
I want to leave a number of previous (a = [[10], [9, 8, 7, 6, 5], [4, 3, 2, 1]] )

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#73214

2014年6月12日木曜日 14時43分42秒 UTC+9 Steven D'Aprano:
> On Wed, 11 Jun 2014 21:56:06 -0700, hito koto wrote:
> 
> 
> 
> > I want to use while statement,
> 
> > 
> 
> > for example:
> 
> >>>> def foo(x):
> 
> > ...     y = []
> 
> > ...     while x !=[]:
> 
> > ...         y.append(x.pop())
> 
> > ...     return y
> 
> > ...
> 
> >>>> print foo(a)
> 
> > [[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
> 
> >>>> a
> 
> > []   but this is empty
> 
> >>>> so,I want to leave a number of previous (a = [[1, 2, 3, 4],[5, 6, 7,
> 
> >>>> 8, 9],[10]])
> 
> 
> 
> 
> 
> I wouldn't use a while statement. The easy way is:
> 
> 
> 
> py> a = [[1, 2, 3, 4],[5, 6, 7, 8, 9],[10]]
> 
> py> y = a[::-1]
> 
> py> print y
> 
> [[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
> 
> py> print a
> 
> [[1, 2, 3, 4], [5, 6, 7, 8, 9], [10]]
> 
> 
> 
> If you MUST use a while loop, then you need something like this:
> 
> 
> 
> 
> 
> def foo(x):
> 
>     y = []
> 
>     index = 0
> 
>     while index < len(x):
> 
>         y.append(x[i])
> 
>         i += 1
> 
>     return y
> 
> 
> 
> 
> 
> This does not copy in reverse order. To make it copy in reverse order, 
> 
> index should start at len(x) - 1 and end at 0.
> 
> 
> 
> 
> 
> 
> 
> -- 
> 
> Steven

Hi,
Thank you!

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#73220

On 6/11/14 10:12 PM, hito koto wrote:
> i want to change this is code:
>
> def foo(x):
>      y = []
>      while x !=[]:
>          y.append(x.pop())
>      return y
>

Consider this generator (all kinds of permutations on the idea):

 >>> L1
[1, 2, 3, 4, 5, 6, 7]

 >>> def poplist(L):
	while True:
		yield L[::-1][:1:]
		L = L[::-1][1::][::-1]

		
 >>> pop = poplist(L1)

 >>> next(pop)
[7]
 >>> next(pop)
[6]
 >>> next(pop)
[5]
 >>> next(pop)
[4]
 >>> next(pop)
[3]
 >>> next(pop)
[2]
 >>> next(pop)
[1]
 >>> next(pop)
[]
 >>> next(pop)
[]

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#73221

On 6/11/14 10:12 PM, hito koto wrote:

> def foo(x):
>      y = []
>      while x !=[]:
>          y.append(x.pop())
>      return y
>

Consider this generator variation:

 >>> def poplist(L):
	done = False
	while done==False:
		yield L[::-1][:1:]
		L = L[::-1][1::][::-1]
		if len(L)==0: done=True

		
 >>> L1=[1, 2, 3, 4, 5, 6, 7]

 >>> for n in poplist(L1):
	print(n)
	
[7]
[6]
[5]
[4]
[3]
[2]
[1]
 >>>

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#73222

> 	while done==False:

Correction:

   while not done:

Better Python and not bad English, either.


Marko

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#73226

On 6/12/14 11:55 AM, Marko Rauhamaa wrote:
>     while not done:
>
> Better Python and not bad English, either.

... and taking Marko's good advice, what I think you really wanted:


 >>> def poplist(L):
	done = False
	while not done:
		yield L[::-1][:1:]
		L = L[::-1][1::][::-1]
		if len(L)==0: done=True

		
 >>> L=[1, 2, 3, 4, 5, 6, 7]

 >>> m=[]

 >>> pop = poplist(L)

 >>> for n in poplist(L):
	m.append(n[0])

 >>> m
[7, 6, 5, 4, 3, 2, 1]
 >>>

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#73223

On Fri, Jun 13, 2014 at 2:49 AM, Mark H Harris <harrismh777@gmail.com> wrote:
> Consider this generator variation:
>
>>>> def poplist(L):
>         done = False
>         while done==False:
>
>                 yield L[::-1][:1:]
>                 L = L[::-1][1::][::-1]
>                 if len(L)==0: done=True

Why not just "while L"? Or are you deliberately trying to ensure that
cheating will be detectable?

ChrisA

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#73227

On 6/12/14 11:57 AM, Chris Angelico wrote:
> On Fri, Jun 13, 2014 at 2:49 AM, Mark H Harris <harrismh777@gmail.com> wrote:
>> Consider this generator variation:
>>
>>>>> def poplist(L):
>>          done = False
>>          while done==False:
>>
>>                  yield L[::-1][:1:]
>>                  L = L[::-1][1::][::-1]
>>                  if len(L)==0: done=True
>
> Why not just "while L"? Or are you deliberately trying to ensure that
> cheating will be detectable?

;-)

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#73228

On 6/12/14 11:57 AM, Chris Angelico wrote:
>>>>> def poplist(L):
>>          done = False
>>          while done==False:
>>
>>                  yield L[::-1][:1:]
>>                  L = L[::-1][1::][::-1]
>>                  if len(L)==0: done=True
>
> Why not just "while L"?

OK,  here it is with Chris' excellent advice:

 >>> def poplist(L):
	while L:
		yield L[::-1][:1:]
		L = L[::-1][1::][::-1]

		
 >>> L=[1, 2, 3, 4, 5, 6, 7]
 >>> m=[]
 >>> pop = poplist(L)
 >>> for n in poplist(L):
	m.append(n[0])

	
 >>> m
[7, 6, 5, 4, 3, 2, 1]
 >>>


==========  aaaaah  ===================

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#73233

Le jeudi 12 juin 2014 19:16:19 UTC+2, Mark H. Harris a écrit :
> On 6/12/14 11:57 AM, Chris Angelico wrote:
> 
> >>>>> def poplist(L):
> 
> >>          done = False
> 
> >>          while done==False:
> 
> >>
> 
> >>                  yield L[::-1][:1:]
> 
> >>                  L = L[::-1][1::][::-1]
> 
> >>                  if len(L)==0: done=True
> 
> >
> 
> > Why not just "while L"?
> 
> 
> 
> OK,  here it is with Chris' excellent advice:
> 
> 
> 
>  >>> def poplist(L):
> 
> 	while L:
> 
> 		yield L[::-1][:1:]
> 
> 		L = L[::-1][1::][::-1]
> 
> 
> 
> 		
> 
>  >>> L=[1, 2, 3, 4, 5, 6, 7]
> 
>  >>> m=[]
> 
>  >>> pop = poplist(L)
> 
>  >>> for n in poplist(L):
> 
> 	m.append(n[0])
> 
> 
> 
> 	
> 
>  >>> m
> 
> [7, 6, 5, 4, 3, 2, 1]
> 
>  >>>
> 
> 
> 
> 
> 
> ==========  aaaaah  ===================

>>> a = [1, 2, 3, 4, 5, 6, 7]
>>> b = a[:]  # si nécessaire, wenn nötig
>>> b.reverse()
>>> 
>>> a
[1, 2, 3, 4, 5, 6, 7]
>>> b
[7, 6, 5, 4, 3, 2, 1]

jmf

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