When I write this code:
polly = "alive" palin = ["parrot", polly] print(palin) polly = "dead" print(palin)
I thought it would output this:
"['parrot', 'alive']" "['parrot', 'dead']"
However, it doesn’t. How do I get it to output that?
Python variables hold references to values. Thus, when you define the
palin list, you pass in the value referenced by
polly, not the variable itself.
You should imagine values as balloons, with variables being threads tied to those balloons.
"alive" is a balloon,
polly is just a thread to that balloon, and the
palin list has a different thread tied to that same balloon. In python, a list is simply a series of threads, all numbered starting at 0.
What you do next is tie the
polly string to a new balloon
"dead", but the list is still holding on to the old thread tied to the
You can replace that thread to
"alive" held by the list by reassigning the list by index to refer to each thread; in your example that’s thread
1] = polly palin ['parrot', 'dead']palin[
Here I simply tied the
palin thread to the same thing
polly is tied to, whatever that might be.
Note that any collection in python, such as
tuple, etc. are simply collections of threads too. Some of these can have their threads swapped out for different threads, such as lists and dicts, and that’s what makes something in python “mutable”.
Strings on the other hand, are not mutable. Once you define a string like
"alive", it’s one balloon. You can tie it down with a thread (a variable, a list, or whatever), but you cannot replace letters inside of it. You can only tie that thread to a completely new string.
Most things in python can act like balloons. Integers, strings, lists, functions, instances, classes, all can be tied down to a variable, or tied into a container.
You may want to read Ned Batchelder’s treatise on Python names too.
Before your second print statement, store your new values into
palin = ["parrot", polly]
When you put a string in a list, the list holds a copy of the string. It doesn’t matter whether the string was originally a variable, a literal value, the result of a function call, or something else; by the time the list sees it, it’s just a string value. Changing whatever generated the string later will never affect the list.
If you want to store a reference to a value that will notice when that value changes, the usual mechanism is to use a list containing the “referenced” value. Applying that to your example, you wind up with a nested list. Example:
polly = ["alive"] palin = ["parrot", polly] print(palin) polly = "dead" print(palin)
The list will contain values only, not references to variables as you would like. You could however store a lambda in the list, and have the lambda look up the value of your variable.
'a' list = ['a',lambda: a] list <function <lambda> at 0x7feff71dc500> list() 'a' a = 'b' list() 'b'a =
You can’t. Assignment to a bare name is Python always only rebinds the name, and you cannot customize or monitor this operation.
What you can do is make
polly a mutable object instead of a string, and mutate its value instead of rebinding the name. A simple example:
'alive'] items = ['parrot', polly] items ['parrot', ['alive']] polly = 'dead' items ['parrot', ['dead']]polly = [
The other answers have explained what’s going on well.
This is one of the (several) problems that motivate the use of objects. For example, one might do this:
class Animal: def __init__(self, aniType, name): self.aniType = aniType self.name = name self.isAlive = True def kill(self): self.isAlive = False def getName(self): return self.name def getType(self): return self.aniType def isLiving(self): return self.isAlive polly = Animal("parrot", "polly") print(polly.getName()+' the '+polly.getType()+' is alive?') print(polly.isLiving()) polly.kill() print(polly.getName()+' the '+polly.getType()+' is alive?') print(polly.isLiving())
It may look like a lot of code at first for a simple task, but objects are often the way to go for things like this, because they help keep everything organized.
Here’s the output of that program:
polly the parrot is alive? True polly the parrot is alive? False