Python variable handling, I don't understand it

Question

I can't find concise information about what is going on in this very simple program:

print 'case 1'
# a and b stay different
a = [1,2,3]
b = a
b = [4,5,6]

print 'a =',a
print 'b =',b

print
print 'case 2'
# a and b becomes equal
a = [1,2,3]
b = a
b[0] = 4 
b[1] = 5 
b[2] = 6 

print 'a =',a
print 'b =',b
print

print 'case 3'
# a and b stay different now
a = [1,2,3]
b = a[:]
b[0] = 4 
b[1] = 5 
b[2] = 6 
print 'a =',a
print 'b =',b
print

print 'case 4'
# now the funny thing
a=[1,2,[3]]
b=a[:]
b[0]    = 4 
b[1]    = 5 
b[2][0] = 6 # this modifies b and a!!!

The output of this simple test is:

case 1
a = [1, 2, 3]
b = [4, 5, 6]

case 2
a = [4, 5, 6]
b = [4, 5, 6]

case 3
a = [1, 2, 3]
b = [4, 5, 6]

case 4
a = [1, 2, [6]]
b = [4, 5, [6]]

I clearly do not understand how python handles each case. Could any one provide a link so that I can read about it, or a short explanation of what is happening?

Thanks a lot.

Answer 1

This is a fantastic visualization tool for python code.

Run your code in it and everything should become clear in a minute.

Answer 2

There are two important things here:

• Variables are just labels pointing to objects
• List are mutable in python and integers aren't.

When you find that both a and b are modified, then that's because both of them are pointing to the same object. You can do id(a) and id(b) to confirm this.

Regarding the examples, note that a[:] will create a new object which is a copy of a. However, it will be a shallow copy, not a deep copy. This explains why in example 4, you can still modify both a and b. They are pointing to different list objects, but one element is another list that is shared by both of them.

Answer 3

Case 1: The name b is rebound.

Case 2: a and b are bound to the same object.

Case 3: A shallow copy of a is bound to b. The lists are different, but the objects within the list are the same.

Case 4: A shallow copy of a is bound to b, and then one of the objects is mutated.

Rebinding does not mutate, and mutation does not rebind.

Answer 4

print 'case 1'
# a and b stay different
a = [1,2,3]
b = a              #At this point 'b' and 'a' are the same, 
                   #just names given to the list
b = [4,5,6]        #At this point you assign the name 'b' to a different list

print 'a =',a
print 'b =',b

print
print 'case 2'
# a and b becomes equal
a = [1,2,3]        #At this point 'b' and 'a' are the same, 
                   #just names given to the list
b = a
b[0] = 4           #From here you modify the list, since both 'a' and 'b' 
                   #reference the same list, you will see the change in 'a'
b[1] = 5 
b[2] = 6 


print 'case 3'
# a and b stay different now
a = [1,2,3]
b = a[:]              #At this point you COPY the elements from 'a' into a new 
                      #list that is referenced by 'b'
b[0] = 4              #From here you modify 'b' but this has no connection to 'a'
b[1] = 5 
b[2] = 6 
print 'a =',a
print 'b =',b
print

print 'case 4'
# now the funny thing
a=[1,2,[3]]
b=a[:]           #At this point you COPY the elements from 'a' into a new 
                 #list that is referenced by 'b'
b[0]    = 4      #Same as before 
b[1]    = 5 
b[2][0] = 6 # this modifies b and a!!!    #Here what happens is that 'b[2]' holds 
                #the same list as 'a[2]'. Since you only modify the element of that 
                #list that will be visible in 'a', try to see it as cases 1/2 just  
                #'recursively'. If you do b[2] = 0, that won't change 'a'