About python while statement and pop()

[hito koto]

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 popã€how can i to do?

i want to change this is code:

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

 

[Dave Angel]

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 popã€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.

 

[Vincent Vande Vyvre]

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 popã€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)

 

[Chris Angelico]

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 popã€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

 

[hito koto]

2014年6月12日木曜日 12時58分27秒 UTC+9 Chris Angelico:

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 popã€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:.... 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]])

 

[Chris Angelico]

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

 

[Steven D'Aprano]

I want to use while statement,

for example:... 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 += 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.

 

[hito koto]

2014年6月12日木曜日 14時43分42秒 UTC+9 Steven D'Aprano:

I want to use while statement,

for example:

... y = []

... while x !=[]:

... y.append(x.pop())

... return y

[[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
[] 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 += 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.

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]] )

 

[hito koto]

2014年6月12日木曜日 14時43分42秒 UTC+9 Steven D'Aprano:

I want to use while statement,

for example:

... y = []

... while x !=[]:

... y.append(x.pop())

... return y

[[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
[] 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 += 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.

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]] )

 

[hito koto]

2014年6月12日木曜日 14時43分42秒 UTC+9 Steven D'Aprano:

I want to use while statement,

for example:

... y = []

... while x !=[]:

... y.append(x.pop())

... return y

[[10], [5, 6, 7, 8, 9], [1, 2, 3, 4]]
[] 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 += 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.

Hi,
Thank you!

 

[Mark H Harris]

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):
[1, 2, 3, 4, 5, 6, 7]
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)

[]

 

[Mark H Harris]

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

Consider this generator variation:
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]

 

[Marko Rauhamaa]

while done==False:

Correction:

while not done:

Better Python and not bad English, either.


Marko

 

[Chris Angelico]

Consider this generator variation:
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

 

[Mark H Harris]

while not done:

Better Python and not bad English, either.

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

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]

 

[Mark H Harris]

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?

;-)

 

[Mark H Harris]

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:
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 ===================

 

[wxjmfauth]

Le jeudi 12 juin 2014 19:16:19 UTC+2, Mark H. Harris a écrit :

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:



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