Question

Can you please clarify how it is that self.add(x) below works the same way as self.data.append(x)?
That is, how does self.add(x) know to append to the list because we have not explicitly stated self.data.add(x)? When we state y.addtwice('cat'), 'cat' is added to 'self', not self.data.

class Bag:
    def __init__(self):
        self.data=[]
    def add(self,x):
        self.data.append(x)
        return self.data
    def addtwice(self,x):
        self.add(x)
        self.add(x)
        return self.data

>>> y = Bag()
>>> y.add('dog')
['dog']
>>> y.addtwice('cat')
['dog', 'cat', 'cat']

Answer 1

Because addtwice calls methods which are defined on self, and because self.data is a “mutable type”, addtwice’s call to add will end up appending the value of self.data. add, in turn calls self.data.append

When calling a function in a computer program, you can think of the process as being a series of substitutions like this:

# -> means (substitution for)
# <= means "return"
y = Bag()
y.add('dog') -> 
     y.data.append(x) ->
         #(machine code)
     <= y.data
# at this point, at the command propmt, python will just print what was returned.
y.addtwice('cat')->
     y.add('cat')->
         y.data.append(x) ->
             #(machine code) 
         <= y.data
     #nothing cares about this return
     y.add('cat')->
         y.data.append(x) ->
             #(machine code)
         <= y.data
     #nothing cares about this return either
     <= y.data
# at this point, at the command propmt, python will just print what was returned.

self, itself, is never really appended in any of those cases though. self.data is.