如何更新python中的变量列表？

Question

This is an aggravating issue that I've come into whilst programming in python. The user can append various variables to a list however, this list when accessed later on will still display the same results as when first assigned. eg below

a=1
b=2
list = [a,b] #user defined order
print list # 1,2
a=2
print list # prints 1,2

I need list to print out 2,2. However i cannot find out a way to DYNAMICALLY update the list to accommodate ANY order of the variables (many ways are hard coding which i've seen online such as assigning list = [a,b] when needed to update however i dont know if its b,a or a,b)

Help would be much appreciated. Thank you

Edit : My question is different due to being about varaibles that need to be dynamically updated in a list rather than simply changing a list item.

Answer 1

you need to update the list and not the variable:

a = 1
b = 2
lst = [a, b]  # the list now stores the references a and b point to
              # changing a later on has no effect on this!  
lst[0] = 2

and please do not use list as variable name! this is a keyword built-in type in python and overwriting it is a very bad idea!

if you only know the value of the element in the list (and this value is unique) you could do this:

old_val = 2
new_val = 3

lst[lst.index(old_val)] = new_val

Answer 2

It's not possible if you store immutable items in a list, which Python integers are, so add a level of indirection and use mutable lists:

a,b,c,d = [1],[2],[3],[4]
L = [d,b,a,c] # user-defined order
print L
a[0] = 5
print L

Output:

[[4], [2], [1], [3]]
[[4], [2], [5], [3]]

This has the feel of an XY Problem , however. Describing the problem you are solving with this solution may elicit better solutions.

Answer 3

Why don't you use dictionary??

With dictionary:

_dict = {}
_dict["a"] = 1
_dict["b"] = 2
print _dict["a"] # prints 1

Now if you want to set and get value of variable "a":

_dict["a"] = 2
print _dict["a"] # prints 2

Answer 4

At the assignment 'a=1' an integer object is created with the value 1. That object is added to your list. At the second assignment (a=2) a new object is created, but the old one still is in the list, its value unchanged. That is what you see.

To change this behaviour, the value of the object must be changed. Ints are immutable in Python, meaning that exactly this, ie changing the value, is not possible. You'll have to define your own and then change its value. For example:

class MyInt(object):

    def __init__(self, value):
        self.setValue(value)

    def setValue(self, value):
        self.value = value

    def __repr__(self):
        return str(self.value)

a=MyInt(1)
print(dir(a))
b=2
list = [a,b] #user defined order
print (list) # 1,2
a.setValue(2)
print (list) # prints 2,2

Answer 5

It is not clean but a work around trick , you can assign the value of a is a single element list

a = [1]
b = 2

So when you assign them to a list:

lst = [a,b]

and change the value of a by calling its 0th element:

a[0] = 2

it will also change the referenced list that is a

print(lst)
#[[2],2]

likewise if you place the value in the list:

lst[0][0] = 2
print(a[0])
#2

Answer 6

You can define a function:\

a, b = 1, 2

def list_():
   return [a, b]

list_() # [1, 2]

a = 2
list_() # [2, 2]