indirectly addressing vars in Python
Forgive my newbieness - I want to refer to some variables and indirectly
alter them. Not sure if this is as easy in Python as it is in C.
Say I have three vars: oats, corn, barley
I add them to a list: myList[{oats}, {peas}, {barley}]
Then I want to past that list around and alter one of those values.
That is I want to increment the value of corn:
myList[1] = myList[1] + 1
Is there some means to do that?. Here's my little session trying to
figure this out:
oats = 1
peas = 6
myList=[]
myList
[]
myList.append(oats)
myList
[1]
myList.append(peas)
myList
[1, 6]
myList[1]= myList[1]+1
myList
[1, 7]
peas
6
So I don't seem to change the value of peas as I wished. I'm passing
the values of the vars into the list, not the vars themselves, as I
would like.
Your guidance appreciated...
Ross.
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Re: indirectly addressing vars in Python
Ross wrote: myList[1]= myList[1]+1 The problem is this makes myList[1] point to a new integer, and not the one that peas points to. Python 2.5.1 (r251:54863, Jul 10 2008, 17:25:56) [GCC 4.1.2 20070925 (Red Hat 4.1.2-33)] on linux2 Type help, copyright, credits or license for more information. oats=[1] peas=[6] mylist = [oats, peas] mylist[1][0] = mylist[1][0]+1 mylist [[1], [7]] peas [7] This is because integers are immutable, but lists are mutable. -- Jeremy Sanders http://www.jeremysanders.net/ -- http://mail.python.org/mailman/listinfo/python-list
Re: indirectly addressing vars in Python
On 2008-10-01, Ross [EMAIL PROTECTED] wrote:
Forgive my newbieness - I want to refer to some variables and
indirectly alter them. Not sure if this is as easy in Python
as it is in C.
Python doesn't have variables. It has names bound to objects.
When you do an assignment, that binds (or rebinds) a name to an
object. There is no such thing as a C-like variable (a named
region of memory with a fixed location into which you can write
different values).
Some objects (e.g. lists, dictionaries) are mutable (you can
change their value or contents), some objects (e.g. strings,
integers, floats) are not mutable.
Say I have three vars: oats, corn, barley
I add them to a list: myList[{oats}, {peas}, {barley}]
Then I want to past that list around and alter one of those
values. That is I want to increment the value of corn:
myList[1] = myList[1] + 1
Is there some means to do that?. Here's my little session
trying to figure this out:
oats = 1
peas = 6
Those two lines created two integer objects with values 1 and 6
and bound the names oats and peas to those two objects.
myList=[]
myList
[]
myList.append(oats)
That line finds the object to which the name oats is
currently bound and appends that object to the list.
myList
[1]
myList.append(peas)
Likewise for the object to which the name peas is currently
bound.
myList
[1, 6]
myList[1]= myList[1]+1
That line creates a new integer object (whose value happens to
be 7) and replaces the object at position 1 in the list with
the new object.
myList
[1, 7]
peas
6
So I don't seem to change the value of peas as I wished.
Correct. The name peas is still bound to the same object it
was before
I'm passing the values of the vars into the list, not the vars
themselves, as I would like.
There are no vars as the word is used in the context of C
programming. Just names and objects.
Here's an article explaining it:
http://rg03.wordpress.com/2007/04/21/semantics-of-python-variable-names-from-a-c-perspective/
A couple other good references:
http://starship.python.net/crew/mwh/hacks/objectthink.html
http://effbot.org/zone/python-objects.htm
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Re: indirectly addressing vars in Python
On Wed, Oct 1, 2008 at 7:53 AM, Ross [EMAIL PROTECTED] wrote:
Forgive my newbieness - I want to refer to some variables and indirectly
alter them. Not sure if this is as easy in Python as it is in C.
Say I have three vars: oats, corn, barley
I add them to a list: myList[{oats}, {peas}, {barley}]
You mean:
myList = [oats, peas, barley]
And you're not adding *variables* to a list, you're adding *values* to
the list. The list elements (and the Python runtime) have *no idea*
what variables they are/were bound to.
Then I want to past that list around and alter one of those values. That is
I want to increment the value of corn:
myList[1] = myList[1] + 1
This won't do what you're expecting. Integers in Python are immutable,
so instead of changing the value of 'corn', you're calculating a new
int object and overwriting the first element of the list with it.
Is there some means to do that?. Here's my little session trying to figure
this out:
oats = 1
peas = 6
myList=[]
myList
[]
myList.append(oats)
myList
[1]
myList.append(peas)
myList
[1, 6]
myList[1]= myList[1]+1
myList
[1, 7]
peas
6
So I don't seem to change the value of peas as I wished. I'm passing the
values of the vars into the list, not the vars themselves, as I would like.
Your guidance appreciated...
Python uses *call-by-object*, not call-by-value or call-by-reference
semantics, so unlike C/C++ but like Java you can't make a reference
to a variable and use that to non-locally rebind the variable to a new
value. To do what you want, you need to create a mutable value that
can be updated. You could code a MutableInt class wrapping 'int', or
you could use a dictionary to hold the values and then always refer to
the values using the dictionary. There are a few other ways to do it.
Hope that elucidates it for you somewhat. Then again I am a little
short on sleep :)
Cheers,
Chris
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http://rebertia.com
Ross.
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Re: indirectly addressing vars in Python
Chris Rebert a écrit :
On Wed, Oct 1, 2008 at 7:53 AM, Ross [EMAIL PROTECTED] wrote:
Forgive my newbieness - I want to refer to some variables and indirectly
alter them. Not sure if this is as easy in Python as it is in C.
Say I have three vars: oats, corn, barley
I add them to a list: myList[{oats}, {peas}, {barley}]
You mean:
myList = [oats, peas, barley]
And you're not adding *variables* to a list, you're adding *values*
s/values/references to objects/, actually
(snip)
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Re: indirectly addressing vars in Python
On Wed, 01 Oct 2008 10:53:08 -0400, Ross wrote:
Forgive my newbieness - I want to refer to some variables and indirectly
alter them. Not sure if this is as easy in Python as it is in C.
Say I have three vars: oats, corn, barley
I add them to a list: myList[{oats}, {peas}, {barley}]
Then I want to past that list around and alter one of those values. That
is I want to increment the value of corn:
myList[1] = myList[1] + 1
Is there some means to do that?. Here's my little session trying to
figure this out:
oats = 1
peas = 6
myList=[]
myList
[]
myList.append(oats)
myList
[1]
myList.append(peas)
myList
[1, 6]
myList[1]= myList[1]+1
myList
[1, 7]
peas
6
So I don't seem to change the value of peas as I wished. I'm passing
the values of the vars into the list, not the vars themselves, as I
would like.
Your guidance appreciated...
Ross.
Short answer: Python is not C.
Long answer: In python, integers are immutable (they do not change), i.e.
if you do:
a = 1
b = 2
c = a + b
a + b creates a new integer object that has the value 3 and assign it to c
Other examples of immutable values are: number types, string, tuple, etc.
In python, variable is usually called name. The concept of variable
and name is slightly different.
In C, a variable contains an object. In python, a name contains a pointer
to an object. (the term pointer is used a bit loosely here)
In python, a list is a list of pointers to objects.
i.e.:
|-- Hello World
myList |-- 1
|-- 6
when you do, for example, myList[2] + 1
what you're doing is:
1. fetch the value of myList[2] (i.e. 6)
2. do an arithmetic addition of 6 + 1
3. create a new integer object with value 7
4. bind myList[2] to that new integer object (_new integer object_ since
integer is immutable, you cannot change its value[1])
in short, pea is left intact since pea points to integer object 6.
beware though, that things are different if myList[2] is a mutable value
instead and the operation is an in-place operation, e.g. list.sort
a = [1, 2]
b = [3, 2]
myList.append(a)
myList.append(b)
myList[1].sort()
myList
[[1, 2], [2, 3]]
b
[2, 3]
in this case a, b is mutable object, so:
myList.append(x) binds myList to the list object pointed by x
myList[1].sort() is an in-place sort to the list object [3, 2] (i.e. it
mutates the list [3, 2] instead of creating a new list[1]).
in this case:
myList |-- [1, 2]
|-- [3, 2]
myList[1].sort(), being an in-place sort, modifies the list object [3, 2]
in-place. Since b points to the same list object, b's list is changed too.
To summarize, you don't usually use that line of thinking in python. If
you really insists, though is not pythonic (and is really absurd anyway),
you may use this:
a = [1]
b = [2]
myList.append(a)
myList.append(b)
myList[1][0] = myList[1][0] + 1
myList
[[1], [2]]
b
[3]
[1] Well, not always though, in CPython (the standard python reference
implementation), small immutable values are cached as speed optimization.
This is a safe operation, except for identity testing. But that doesn't
matter much, since identity testing of immutable is moot.
[2] If you want a sorting operation that returns a new list, use sorted
(list)
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