List sequential initialization

[HMS Surprise]

I thought if I could do this:

>>a = b = ''
>>a = 'a'
>>a

'a'

>>b

''

then this would behave similarly:

>>la = lb = []
>>la.append('a' )
>>la

['a']

>>lb

['a']


I thought wrong! But don't know why.

Inquiring minds want to know.....


thanx,

jh

[=?UTF-8?B?UmVuw6kgRmxlc2NoZW5iZXJn?=]

Re: List sequential initialization

HMS Surprise schrieb:

I thought if I could do this:

>a = b = ''

Bind both the names a and b to the same string object.

>a = 'a'

Bind the name a to a *new* string object with the value 'a'. This
replaces the previous binding of the name a.

>la = lb = []

Bind both the names la and lb to the same list object.

>la.append('a ')

Append the string 'a' to the *existing* list object referenced by the
name la. This modifies the object that is referenced by the name la (and
also by the name lb), it does not create a new object. The equivalent to
what you did with the strings above would have been:

la = ['a']

This would have created a new list object and bound the name la to it,
while the name lb would still reference the other list object you
created earlier.

If you want to modify the list referenced by la but not the one
referenced by lb, you need to actually *copy* the list with one of these
methods:

lb = list(la)

or

lb = la[:]

--
René

[HMS Surprise]

Re: List sequential initialization

Thanks for the explaination. It didn't seem natural and from the
tutorial I read:

A value can be assigned to several variables simultaneously:

>>x = y = z = 0 # Zero x, y and z

Maybe I infer too much....

thanks again,

jh

[Chris Mellon]

Re: List sequential initialization

On 6/12/07, HMS Surprise <john@datavoice int.comwrote:

Thanks for the explaination. It didn't seem natural and from the
tutorial I read:
>
A value can be assigned to several variables simultaneously:
>

>>x = y = z = 0 # Zero x, y and z

>
>
Maybe I infer too much....
>

And yet, your answer is right there.

"A value can be assigned to several variables simultaneously"

When you say want a value assigned to several variables, Python
doesn't assume that you actually mean you want 2 different values
assigned to them.

[Dave Baum]

Re: List sequential initialization

In article <1181670619.086 709.116730@g37g 2000prf.googleg roups.com>,
HMS Surprise <john@datavoice int.comwrote:

I thought if I could do this:

>a = b = ''

a and b refer to the same object (the empty string)

>a = 'a'

You are assigning a new value to a - it now refers to the string 'a',
while b refers to the same thing it always has (the empty string)

>a

'a'

>b

''
>
then this would behave similarly:

>la = lb = []

la and lb refer to the same object, an empty list

>la.append('a ')

You are appending 'a' to the list that la refers to.

>la

['a']

>lb

['a']

Since lb referred to the same list as la, when you modified the list via
la.append, those changes can also be seen via lb.

If instead of la.append('a'), you had done:

la = ['a']

Then it would have behaved similarly to the first example, and lb would
still refer to an empty list.

>
I thought wrong! But don't know why.

For immutable objects (such as integers, strings, and tuples), the
distinction between pointing to the same object or identical copies
isn't important since you cannot modify the objects. However, when you
use mutable objects (such as lists) and modify them, then it is
important to understand when you are dealing with the same object and
when you are copying the object.

Assignment makes a name refer to an object. Multiple names can refer to
the same object (which is what a=b=c does). If you want to make a copy
of the object, you need to do so explicitly:

>>a = [1, 2, 3]
>>b = list(a)
>>a.append(4)
>>a

[1, 2, 3, 4]

>>b

[1, 2, 3]


Dave

[Steve Holden]

Re: List sequential initialization

Chris Mellon wrote:

On 6/12/07, HMS Surprise <john@datavoice int.comwrote:

>Thanks for the explaination. It didn't seem natural and from the
>tutorial I read:
>>
> A value can be assigned to several variables simultaneously:
>>

> >>x = y = z = 0 # Zero x, y and z

>>
>>
>Maybe I infer too much....
>>

>
And yet, your answer is right there.
>
"A value can be assigned to several variables simultaneously"
>
When you say want a value assigned to several variables, Python
doesn't assume that you actually mean you want 2 different values
assigned to them.

The crucial difference between

a = b = "ab"
a = "a"

and

a = b = ['a', 'b']
a.append('c')

is that in the first case two names are bound to the immutable object
"ab". Then the first name is rebound to a different immutable object.

In the second example, both names are bound to the same mutable object,
a list. That object is then modified. The modification can be performed
using either name, and both names continue to point to the same (but now
mutated) object.

regards
Steve
--
Steve Holden +1 571 484 6266 +1 800 494 3119
Holden Web LLC/Ltd http://www.holdenweb.com
Skype: holdenweb http://del.icio.us/steve.holden
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