Mixing
Python list doesn't reflect variable change
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 list variables
asked Aug 22 '12 at 20:01
Flexo1515Flexo1515
5091622
add a comment |
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 list variables
asked Aug 22 '12 at 20:01
Flexo1515Flexo1515
5091622
2
related : stackoverflow.com/q/11690220/748858
– mgilson
Aug 22 '12 at 20:09
add a comment |
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 list variables
asked Aug 22 '12 at 20:01
Flexo1515Flexo1515
5091622
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 list variables
python list variables
asked Aug 22 '12 at 20:01
Flexo1515Flexo1515
5091622
asked Aug 22 '12 at 20:01
Flexo1515Flexo1515
5091622
edited Jul 27 '13 at 2:21
user212218
asked Aug 22 '12 at 20:01
Flexo1515Flexo1515
5091622
asked Aug 22 '12 at 20:01
Flexo1515Flexo1515
5091622
asked Aug 22 '12 at 20:01
Flexo1515Flexo1515
5091622
5091622
2
related : stackoverflow.com/q/11690220/748858
– mgilson
Aug 22 '12 at 20:09
add a comment |
2
related : stackoverflow.com/q/11690220/748858
– mgilson
Aug 22 '12 at 20:09
2
2
related : stackoverflow.com/q/11690220/748858
– mgilson
Aug 22 '12 at 20:09
related : stackoverflow.com/q/11690220/748858
– mgilson
Aug 22 '12 at 20:09
add a comment |
6 Answers
6
active
oldest
votes
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 "alive" balloon.
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:
>>> palin[1] = polly
>>> palin
['parrot', 'dead']
Here I simply tied the palin[1] thread to the same thing polly is tied to, whatever that might be.
Note that any collection in python, such as dict, set, 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 "dead" or "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.
answered Aug 22 '12 at 20:09
Martijn Pieters♦Martijn Pieters
710k13724782299
Thank you, I believe now that I asked the incorrect question but thank you
– Flexo1515
Aug 22 '12 at 20:13
11
I like the balloon analogy -- especially if they're helium balloons which float away when the last string is released... (+1).
– mgilson
Aug 22 '12 at 20:14
4
Does this mean garbage collection strategies correspond to weather?
– DSM
Aug 22 '12 at 20:18
8
@DSM: Exactly :-) GC is the wind that blows away any balloons not tied down. :-)
– Martijn Pieters♦
Aug 22 '12 at 20:19
1
@user1404053: then you need a mutable balloon. A user-defined class, for example, instead of a string.
– Martijn Pieters♦
Aug 22 '12 at 20:23
|
show 8 more comments
Before your second print statement, store your new values into palin:
palin = ["parrot", polly]
answered Aug 22 '12 at 20:04
girasquidgirasquid
12k23652
Anyway to do it without restoring the new value?
– Flexo1515
Aug 22 '12 at 20:07
3
@user1404053: Not the way you're asking. In Python strings are immutable, so there's no way to change the value inpollyand therefore change it everywhere else. The linepolly = "dead"is a rebinding operation rather than a mutating one
– David Robinson
Aug 22 '12 at 20:11
add a comment |
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:
polly = ["alive"]
palin = ["parrot", polly]
print(palin)
polly[0] = "dead"
print(palin)
answered Aug 22 '12 at 20:12
Mark ReedMark Reed
65.6k989124
add a comment |
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 = 'a'
>>> list = ['a',lambda: a]
>>> list[1]
<function <lambda> at 0x7feff71dc500>
>>> list[1]()
'a'
>>> a = 'b'
>>> list[1]()
'b'
answered Aug 22 '12 at 20:11
JonatanJonatan
1,73411331
Some more explanation: herelambda: ais basically a function that just returns the value stored ina. That's why you can use the commandlist[1](); it means you call the function stored inlist[1]. (Also, appropriate gravatar, @Jonatan)
– Matthew Adams
Aug 22 '12 at 20:17
Quite right, thanks for the explanation. Also I had completely forgotten my avatar, haha!
– Jonatan
Aug 22 '12 at 20:20
add a comment |
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:
>>> polly = ['alive']
>>> items = ['parrot', polly]
>>> items
['parrot', ['alive']]
>>> polly[0] = 'dead'
>>> items
['parrot', ['dead']]
answered Aug 22 '12 at 20:14
BrenBarnBrenBarn
162k20287289
add a comment |
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
answered Aug 22 '12 at 20:35
Matthew AdamsMatthew Adams
5,45031841
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 "alive" balloon.
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:
>>> palin[1] = polly
>>> palin
['parrot', 'dead']
Here I simply tied the palin[1] thread to the same thing polly is tied to, whatever that might be.
Note that any collection in python, such as dict, set, 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 "dead" or "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.
answered Aug 22 '12 at 20:09
Martijn Pieters♦Martijn Pieters
710k13724782299
Thank you, I believe now that I asked the incorrect question but thank you
– Flexo1515
Aug 22 '12 at 20:13
11
I like the balloon analogy -- especially if they're helium balloons which float away when the last string is released... (+1).
– mgilson
Aug 22 '12 at 20:14
4
Does this mean garbage collection strategies correspond to weather?
– DSM
Aug 22 '12 at 20:18
8
@DSM: Exactly :-) GC is the wind that blows away any balloons not tied down. :-)
– Martijn Pieters♦
Aug 22 '12 at 20:19
1
@user1404053: then you need a mutable balloon. A user-defined class, for example, instead of a string.
– Martijn Pieters♦
Aug 22 '12 at 20:23
|
show 8 more comments
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 "alive" balloon.
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:
>>> palin[1] = polly
>>> palin
['parrot', 'dead']
Here I simply tied the palin[1] thread to the same thing polly is tied to, whatever that might be.
Note that any collection in python, such as dict, set, 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 "dead" or "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.
answered Aug 22 '12 at 20:09
Martijn Pieters♦Martijn Pieters
710k13724782299
Thank you, I believe now that I asked the incorrect question but thank you
– Flexo1515
Aug 22 '12 at 20:13
11
I like the balloon analogy -- especially if they're helium balloons which float away when the last string is released... (+1).
– mgilson
Aug 22 '12 at 20:14
4
Does this mean garbage collection strategies correspond to weather?
– DSM
Aug 22 '12 at 20:18
8
@DSM: Exactly :-) GC is the wind that blows away any balloons not tied down. :-)
– Martijn Pieters♦
Aug 22 '12 at 20:19
1
@user1404053: then you need a mutable balloon. A user-defined class, for example, instead of a string.
– Martijn Pieters♦
Aug 22 '12 at 20:23
|
show 8 more comments
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 "alive" balloon.
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:
>>> palin[1] = polly
>>> palin
['parrot', 'dead']
Here I simply tied the palin[1] thread to the same thing polly is tied to, whatever that might be.
Note that any collection in python, such as dict, set, 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 "dead" or "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.
answered Aug 22 '12 at 20:09
Martijn Pieters♦Martijn Pieters
710k13724782299
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 "alive" balloon.
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:
>>> palin[1] = polly
>>> palin
['parrot', 'dead']
Here I simply tied the palin[1] thread to the same thing polly is tied to, whatever that might be.
Note that any collection in python, such as dict, set, 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 "dead" or "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.
answered Aug 22 '12 at 20:09
Martijn Pieters♦Martijn Pieters
710k13724782299
edited Sep 27 '16 at 11:36
answered Aug 22 '12 at 20:09
Martijn Pieters♦Martijn Pieters
710k13724782299
answered Aug 22 '12 at 20:09
Martijn Pieters♦Martijn Pieters
710k13724782299
answered Aug 22 '12 at 20:09
Martijn Pieters♦Martijn Pieters
710k13724782299
710k13724782299
Thank you, I believe now that I asked the incorrect question but thank you
– Flexo1515
Aug 22 '12 at 20:13
11
I like the balloon analogy -- especially if they're helium balloons which float away when the last string is released... (+1).
– mgilson
Aug 22 '12 at 20:14
4
Does this mean garbage collection strategies correspond to weather?
– DSM
Aug 22 '12 at 20:18
8
@DSM: Exactly :-) GC is the wind that blows away any balloons not tied down. :-)
– Martijn Pieters♦
Aug 22 '12 at 20:19
1
@user1404053: then you need a mutable balloon. A user-defined class, for example, instead of a string.
– Martijn Pieters♦
Aug 22 '12 at 20:23
|
show 8 more comments
Thank you, I believe now that I asked the incorrect question but thank you
– Flexo1515
Aug 22 '12 at 20:13
11
I like the balloon analogy -- especially if they're helium balloons which float away when the last string is released... (+1).
– mgilson
Aug 22 '12 at 20:14
4
Does this mean garbage collection strategies correspond to weather?
– DSM
Aug 22 '12 at 20:18
8
@DSM: Exactly :-) GC is the wind that blows away any balloons not tied down. :-)
– Martijn Pieters♦
Aug 22 '12 at 20:19
1
@user1404053: then you need a mutable balloon. A user-defined class, for example, instead of a string.
– Martijn Pieters♦
Aug 22 '12 at 20:23
Thank you, I believe now that I asked the incorrect question but thank you
– Flexo1515
Aug 22 '12 at 20:13
Thank you, I believe now that I asked the incorrect question but thank you
– Flexo1515
Aug 22 '12 at 20:13
11
11
I like the balloon analogy -- especially if they're helium balloons which float away when the last string is released... (+1).
– mgilson
Aug 22 '12 at 20:14
I like the balloon analogy -- especially if they're helium balloons which float away when the last string is released... (+1).
– mgilson
Aug 22 '12 at 20:14
4
4
Does this mean garbage collection strategies correspond to weather?
– DSM
Aug 22 '12 at 20:18
Does this mean garbage collection strategies correspond to weather?
– DSM
Aug 22 '12 at 20:18
8
8
@DSM: Exactly :-) GC is the wind that blows away any balloons not tied down. :-)
– Martijn Pieters♦
Aug 22 '12 at 20:19
@DSM: Exactly :-) GC is the wind that blows away any balloons not tied down. :-)
– Martijn Pieters♦
Aug 22 '12 at 20:19
1
1
@user1404053: then you need a mutable balloon. A user-defined class, for example, instead of a string.
– Martijn Pieters♦
Aug 22 '12 at 20:23
@user1404053: then you need a mutable balloon. A user-defined class, for example, instead of a string.
– Martijn Pieters♦
Aug 22 '12 at 20:23
|
show 8 more comments
Before your second print statement, store your new values into palin:
palin = ["parrot", polly]
answered Aug 22 '12 at 20:04
girasquidgirasquid
12k23652
Anyway to do it without restoring the new value?
– Flexo1515
Aug 22 '12 at 20:07
3
@user1404053: Not the way you're asking. In Python strings are immutable, so there's no way to change the value inpollyand therefore change it everywhere else. The linepolly = "dead"is a rebinding operation rather than a mutating one
– David Robinson
Aug 22 '12 at 20:11
add a comment |
Before your second print statement, store your new values into palin:
palin = ["parrot", polly]
answered Aug 22 '12 at 20:04
girasquidgirasquid
12k23652
Anyway to do it without restoring the new value?
– Flexo1515
Aug 22 '12 at 20:07
3
@user1404053: Not the way you're asking. In Python strings are immutable, so there's no way to change the value inpollyand therefore change it everywhere else. The linepolly = "dead"is a rebinding operation rather than a mutating one
– David Robinson
Aug 22 '12 at 20:11
add a comment |
Before your second print statement, store your new values into palin:
palin = ["parrot", polly]
answered Aug 22 '12 at 20:04
girasquidgirasquid
12k23652
Before your second print statement, store your new values into palin:
palin = ["parrot", polly]
answered Aug 22 '12 at 20:04
girasquidgirasquid
12k23652
answered Aug 22 '12 at 20:04
girasquidgirasquid
12k23652
answered Aug 22 '12 at 20:04
girasquidgirasquid
12k23652
answered Aug 22 '12 at 20:04
girasquidgirasquid
12k23652
12k23652
Anyway to do it without restoring the new value?
– Flexo1515
Aug 22 '12 at 20:07
3
@user1404053: Not the way you're asking. In Python strings are immutable, so there's no way to change the value inpollyand therefore change it everywhere else. The linepolly = "dead"is a rebinding operation rather than a mutating one
– David Robinson
Aug 22 '12 at 20:11
add a comment |
Anyway to do it without restoring the new value?
– Flexo1515
Aug 22 '12 at 20:07
3
@user1404053: Not the way you're asking. In Python strings are immutable, so there's no way to change the value inpollyand therefore change it everywhere else. The linepolly = "dead"is a rebinding operation rather than a mutating one
– David Robinson
Aug 22 '12 at 20:11
Anyway to do it without restoring the new value?
– Flexo1515
Aug 22 '12 at 20:07
Anyway to do it without restoring the new value?
– Flexo1515
Aug 22 '12 at 20:07
3
3
@user1404053: Not the way you're asking. In Python strings are immutable, so there's no way to change the value in
polly and therefore change it everywhere else. The line polly = "dead" is a rebinding operation rather than a mutating one– David Robinson
Aug 22 '12 at 20:11
@user1404053: Not the way you're asking. In Python strings are immutable, so there's no way to change the value in
polly and therefore change it everywhere else. The line polly = "dead" is a rebinding operation rather than a mutating one– David Robinson
Aug 22 '12 at 20:11
add a comment |
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:
polly = ["alive"]
palin = ["parrot", polly]
print(palin)
polly[0] = "dead"
print(palin)
answered Aug 22 '12 at 20:12
Mark ReedMark Reed
65.6k989124
add a comment |
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:
polly = ["alive"]
palin = ["parrot", polly]
print(palin)
polly[0] = "dead"
print(palin)
answered Aug 22 '12 at 20:12
Mark ReedMark Reed
65.6k989124
add a comment |
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:
polly = ["alive"]
palin = ["parrot", polly]
print(palin)
polly[0] = "dead"
print(palin)
answered Aug 22 '12 at 20:12
Mark ReedMark Reed
65.6k989124
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:
polly = ["alive"]
palin = ["parrot", polly]
print(palin)
polly[0] = "dead"
print(palin)
answered Aug 22 '12 at 20:12
Mark ReedMark Reed
65.6k989124
answered Aug 22 '12 at 20:12
Mark ReedMark Reed
65.6k989124
answered Aug 22 '12 at 20:12
Mark ReedMark Reed
65.6k989124
answered Aug 22 '12 at 20:12
Mark ReedMark Reed
65.6k989124
65.6k989124
add a comment |
add a comment |
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 = 'a'
>>> list = ['a',lambda: a]
>>> list[1]
<function <lambda> at 0x7feff71dc500>
>>> list[1]()
'a'
>>> a = 'b'
>>> list[1]()
'b'
answered Aug 22 '12 at 20:11
JonatanJonatan
1,73411331
Some more explanation: herelambda: ais basically a function that just returns the value stored ina. That's why you can use the commandlist[1](); it means you call the function stored inlist[1]. (Also, appropriate gravatar, @Jonatan)
– Matthew Adams
Aug 22 '12 at 20:17
Quite right, thanks for the explanation. Also I had completely forgotten my avatar, haha!
– Jonatan
Aug 22 '12 at 20:20
add a comment |
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 = 'a'
>>> list = ['a',lambda: a]
>>> list[1]
<function <lambda> at 0x7feff71dc500>
>>> list[1]()
'a'
>>> a = 'b'
>>> list[1]()
'b'
answered Aug 22 '12 at 20:11
JonatanJonatan
1,73411331
Some more explanation: herelambda: ais basically a function that just returns the value stored ina. That's why you can use the commandlist[1](); it means you call the function stored inlist[1]. (Also, appropriate gravatar, @Jonatan)
– Matthew Adams
Aug 22 '12 at 20:17
Quite right, thanks for the explanation. Also I had completely forgotten my avatar, haha!
– Jonatan
Aug 22 '12 at 20:20
add a comment |
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 = 'a'
>>> list = ['a',lambda: a]
>>> list[1]
<function <lambda> at 0x7feff71dc500>
>>> list[1]()
'a'
>>> a = 'b'
>>> list[1]()
'b'
answered Aug 22 '12 at 20:11
JonatanJonatan
1,73411331
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 = 'a'
>>> list = ['a',lambda: a]
>>> list[1]
<function <lambda> at 0x7feff71dc500>
>>> list[1]()
'a'
>>> a = 'b'
>>> list[1]()
'b'
answered Aug 22 '12 at 20:11
JonatanJonatan
1,73411331
answered Aug 22 '12 at 20:11
JonatanJonatan
1,73411331
answered Aug 22 '12 at 20:11
JonatanJonatan
1,73411331
answered Aug 22 '12 at 20:11
JonatanJonatan
1,73411331
1,73411331
Some more explanation: herelambda: ais basically a function that just returns the value stored ina. That's why you can use the commandlist[1](); it means you call the function stored inlist[1]. (Also, appropriate gravatar, @Jonatan)
– Matthew Adams
Aug 22 '12 at 20:17
Quite right, thanks for the explanation. Also I had completely forgotten my avatar, haha!
– Jonatan
Aug 22 '12 at 20:20
add a comment |
Some more explanation: herelambda: ais basically a function that just returns the value stored ina. That's why you can use the commandlist[1](); it means you call the function stored inlist[1]. (Also, appropriate gravatar, @Jonatan)
– Matthew Adams
Aug 22 '12 at 20:17
Quite right, thanks for the explanation. Also I had completely forgotten my avatar, haha!
– Jonatan
Aug 22 '12 at 20:20
Some more explanation: here
lambda: a is basically a function that just returns the value stored in a. That's why you can use the command list[1](); it means you call the function stored in list[1]. (Also, appropriate gravatar, @Jonatan)– Matthew Adams
Aug 22 '12 at 20:17
Some more explanation: here
lambda: a is basically a function that just returns the value stored in a. That's why you can use the command list[1](); it means you call the function stored in list[1]. (Also, appropriate gravatar, @Jonatan)– Matthew Adams
Aug 22 '12 at 20:17
Quite right, thanks for the explanation. Also I had completely forgotten my avatar, haha!
– Jonatan
Aug 22 '12 at 20:20
Quite right, thanks for the explanation. Also I had completely forgotten my avatar, haha!
– Jonatan
Aug 22 '12 at 20:20
add a comment |
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:
>>> polly = ['alive']
>>> items = ['parrot', polly]
>>> items
['parrot', ['alive']]
>>> polly[0] = 'dead'
>>> items
['parrot', ['dead']]
answered Aug 22 '12 at 20:14
BrenBarnBrenBarn
162k20287289
add a comment |
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:
>>> polly = ['alive']
>>> items = ['parrot', polly]
>>> items
['parrot', ['alive']]
>>> polly[0] = 'dead'
>>> items
['parrot', ['dead']]
answered Aug 22 '12 at 20:14
BrenBarnBrenBarn
162k20287289
add a comment |
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:
>>> polly = ['alive']
>>> items = ['parrot', polly]
>>> items
['parrot', ['alive']]
>>> polly[0] = 'dead'
>>> items
['parrot', ['dead']]
answered Aug 22 '12 at 20:14
BrenBarnBrenBarn
162k20287289
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:
>>> polly = ['alive']
>>> items = ['parrot', polly]
>>> items
['parrot', ['alive']]
>>> polly[0] = 'dead'
>>> items
['parrot', ['dead']]
answered Aug 22 '12 at 20:14
BrenBarnBrenBarn
162k20287289
answered Aug 22 '12 at 20:14
BrenBarnBrenBarn
162k20287289
answered Aug 22 '12 at 20:14
BrenBarnBrenBarn
162k20287289
answered Aug 22 '12 at 20:14
BrenBarnBrenBarn
162k20287289
162k20287289
add a comment |
add a comment |
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
answered Aug 22 '12 at 20:35
Matthew AdamsMatthew Adams
5,45031841
add a comment |
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
answered Aug 22 '12 at 20:35
Matthew AdamsMatthew Adams
5,45031841
add a comment |
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
answered Aug 22 '12 at 20:35
Matthew AdamsMatthew Adams
5,45031841
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
answered Aug 22 '12 at 20:35
Matthew AdamsMatthew Adams
5,45031841
answered Aug 22 '12 at 20:35
Matthew AdamsMatthew Adams
5,45031841
answered Aug 22 '12 at 20:35
Matthew AdamsMatthew Adams
5,45031841
answered Aug 22 '12 at 20:35
Matthew AdamsMatthew Adams
5,45031841
5,45031841
add a comment |
add a comment |
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related : stackoverflow.com/q/11690220/748858
– mgilson
Aug 22 '12 at 20:09