How to update a variable in multiple lists (Python 3.6)

Question

I am having a scope issue, and I know how I would solve this in Java, but Python is giving me some issues. Say I have a code as follows:

a = 1
b = 2
c = 3

list1 = [a,b,c]
list2 = [b,c]

b += 1

print list1
print list2

This code does not change the value inside the lists, and there is an easy workaround for the simple example I've given, but the code I am trying to write has 10+ lists that all have different meanings and are used in different ways, but need to communicate with each other and have all the same values for a,b,c. I need to be able to change the variable a,b,c and have all the lists update as well, is there any other way besides writing out the update for each list?

Ok, what's a better structure for this, dictionary? I could use numpy arrays, but I don't think that would solve the 'scope' problem I'm encountering — Dylan Barth
– Dylan Barth, Commented Jun 23, 2017 at 21:16

cs95 · Accepted Answer · 2017-06-23 21:19:56Z

2

You cannot store the integers by reference, such that incrementing the value in one place causes all other values to be updated. This is because ints are immutable, so changing their value causes a reference change.

>>> a = 5
>>> b = 1
>>> x = [a, b]
>>> id(b)
140508116572264
>>> id(x[1])
140508116572264
>>> b += 1
>>> id(b)
140508116572240

You can see the id of b changes after the increment, so the "b" in the loop and the "b" outside aren't the same.

What you can do, is define a wrapper class for your ints.

class MyInteger:
    def __init__(self, value):
        self.value = value

Now, build your lists with your MyIntegers:

a = MyInteger(1)
b = MyInteger(2)
c = MyInteger(3)

list1 = [a, b, c]
list2 = [b, c]

Now, when you increment the value of b, you can see the changes are reflected in the lists.

print([x.value for x in list1])
b.value += 1
print([x.value for x in list1])

This prints:

[1, 2, 3]
[1, 3, 3]

answered Jun 23, 2017 at 21:19

cs95

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5 Comments

Dylan Barth Over a year ago

Amazing! This should be a relatively simple workaround. So quick question, if I change the value of a list such as list1[1]+=1 would it change it everywhere? If not would a structure like: for i in list1: i += 1 do the trick?

cs95 Over a year ago

list1[1].value += 1, and yes. It will. i.value += 1 also works, because they reference the same custom object.

Dylan Barth Over a year ago

Thanks I'm new. Quick Q on that class you've made, I get an error "undefined name 'self'" and "undefined name 'value'", I don't usually craft independently made classes, is this an acceptable error or did I miss something?

cs95 Over a year ago

No, that is definitely not supposed to happen. You generally shouldn't use self outside the class. value can be called as soon as the object is created. Is the code in your machine as I have pasted it here?

Dylan Barth Over a year ago

Copied and pasted instead of typing what I saw and it fixed it, maybe the __ was an issue I think I may have typed one _ instead of two. Thanks for the help, error gone, answer accepted, time appreciated. Cheers!

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