after del list , when I use it again, prompt 'not defined'.how could i delete its element,but not itself?

[python]

after del list , when I use it again, prompt 'not defined'.how could i
delete its element,but not itself?
except list.remove(val) ,are there other ways to delete list 's
elements?

and , I wrote:

list1=[]
def method1():
global list1
list1=['a','b','c','d']
def method2():
list2=list1
list2.remove['a']
main():
global list1
method1()
method2()
print list1[0]

it prints 'b'.
How could I keep the list1 not to change when remove list2's elements?

[bearophileHUGS]

python wrote:

after del list , when I use it again, prompt 'not defined'.how could i
delete its element,but not itself?

This is a way:

a = range(10)
del a[:]
a [] a.append(20)
a

[20]

Bye,
bearophile

[Duncan Smith]

python wrote:

[snip]

How could I keep the list1 not to change when remove list2's elements?

You can't when the names list1 and list2 refer to the same list. Try
making list2 a copy of list1,

list2 = list(list1)

Duncan

[SuperHik]

be************@lycos.com wrote:

python wrote:

after del list , when I use it again, prompt 'not defined'.how could i
delete its element,but not itself?

This is a way:

a = range(10)
del a[:] or simply
a = [] a []

a.append(20)
a

[20]

Bye,
bearophile

[Bruno Desthuilliers]

python a écrit :

after del list , when I use it again, prompt 'not defined'.how could i
delete its element,but not itself?
except list.remove(val) ,are there other ways to delete list 's
elements?

and , I wrote:

list1=[]
def method1():
Why naming a function "method" ?
global list1
globals are evil.
list1=['a','b','c','d']
def method2():
list2=list1
this makes list2 point to the same object as list1.
list2.remove['a']
main():
global list1
method1()
method2()
print list1[0]

it prints 'b'.
Of course. Python's "variable" are really just names referencing
objects. Making two names pointing to the same object does'nt make two
different objects, just two references to one object.

How could I keep the list1 not to change when remove list2's elements?

Copy the list.

[Bruno Desthuilliers]

python a écrit :

after del list , when I use it again, prompt 'not defined'.how could i
delete its element,but not itself?

FWIW, del don't delete an object (not directly at least), it just delete
the name<->object association. If (and only if) it was the only name
referencing that object, then the object will be disposed too.

list1 = [1, 2]
list2 = list1
id(list1) 1078043852 id(list2) 1078043852 del list2
list1 [1, 2] list2 Traceback (most recent call last):
File "<stdin>", line 1, in ?
NameError: name 'list2' is not defined

[Piet van Oostrum]

>>>>> SuperHik <ju************@gmail.com> (S) escribió:

S> be************@lycos.com wrote:

python wrote:
after del list , when I use it again, prompt 'not defined'.how could i
delete its element,but not itself?

This is a way:
>> a = range(10)
>> del a[:]S> or simply
S> a = []>> a
[]
Then you *have* deleted the list and created a new one, which is different
from keeping the list and deleting the elements:
a = range(10)
b = a
del a[:]
a [] b [] a.append(100)
a [100] b [100] a=[]
b [100]

--
Piet van Oostrum <pi**@cs.uu.nl>
URL: http://www.cs.uu.nl/~piet [PGP 8DAE142BE17999C4]
Private email: pi**@vanoostrum.org

[Bruno Desthuilliers]

Piet van Oostrum a écrit :

>>SuperHik <ju************@gmail.com> (S) escribió:

S> be************@lycos.com wrote:

python wrote:

>after del list , when I use it again, prompt 'not defined'.how could i
>delete its element,but not itself?

This is a way:

>>>a = range(10)
>>>del a[:]
S> or simply
S> a = []

>>>a

[]

Then you *have* deleted the list

Not necessarily:

a = range(10)
b = a
id(a) 1078034316 id(b) 1078034316 a is b True a = []
b [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] id(a) 1078043724 id(b) 1078034316

What as been deleted is the association ('binding') between the name 'a'
and the list object at 1078034316. This object won't be suppressed until
there are no more names bound to it.
and created a new one, which is different
from keeping the list and deleting the elements:

indeed !-)