Catherine Moroney <catherine.m.moro...@jpl.nasa.gov> writes:
> Is there a way to create a C-style pointer in (pure) Python so the
> following code will reflect the changes to the variable "a" in the
> dictionary "x"?
No, Python doesn't do pointers. Rather, objects have references and
that's how the program accesses the objects.
> For example:
>
> >>> a = 1.0
> >>> b = 2.0
> >>> x = {"a":a, "b":b}
> >>> x
> {'a': 1.0, 'b': 2.0}
> >>> a = 100.0
> >>> x
> {'a': 1.0, 'b': 2.0} ## at this point, I would like the value
> ## associated with the "a" key to be 100.0
> ## rather than 1.0
You might like that, but it's just not how Python works.
Python doesn't have C-style pointers.
Python also doesn't have variables (even though the documentation uses
that term; IMO it's a mistake, and leads to confusion similar to this).
What Python has are references to objects. One kind of reference is a
name; another kind of reference is a value in a dictionary.
The assignment operator ‘=’ is the binding operator. It binds the
reference on the left side to the object on the right side.
> If I make "a" and "b" numpy arrays, then changes that I make to the
> values of a and b show up in the dictionary x.
Yes, because the objects are mutable; you can change them and existing
references are still referring to the same object. They don't “show up
in the dictionary”; the dictionary item is just referring to the same
object it did before you made the change.
> My understanding is that when I redefine the value of "a"
Please think of it, instead, as re-binding the name ‘a’ to a new value.
> that Python is creating a brand-new float with the value of 100.0,
Yes (or at least that's the abstraction being presented to you; it may
not be implemented exactly that way, but it's sufficient that we Python
programmers treat it that way).
> whereas when I use numpy arrays I am merely assigning a new value to
> the same object.
No, you're modifying the object.
A numpy array itself contains references. By altering one of the
elements in an array, you are re-binding one of its references to a
different number.
> Is there some way to rewrite the code above so the change of "a" from
> 1.0 to 100.0 is reflected in the dictionary. I would like to use
> simple datatypes such as floats, rather than numpy arrays or classes.
Please follow the Python tutorial <URL:http://docs.python.org/tutorial/>
from beginning to end. Not just read, but do it: work through the
exercises to understand what each one is teaching you.
That will give you a firm grounding in Python's data model, including
mutable and immutable types, references and binding.
Do bear in mind what I said above, though, about “variable” being a
misleading term, and ignore its implications from the C language.
--
\ “Good judgement comes from experience. Experience comes from |
`\ bad judgement.” —Frederick P. Brooks |
_o__) |
Ben Finney
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