Problem with List argument in a Function when calling it twice

Your question is a fairly recurring topic for all of us who started in Python coming from another language. Let's go first with the solution: the correct way to do what you have in mind is like this:

def caja(cosas=None):
    if caja is None:
       caja = []
    cosas.append("lapiz") 
    cosas.append("martillo")
    cosas.append("azulejos")   
    return cosas

Now, what is the explanation for this apparently "unnatural" behavior?

To begin with, this behavior occurs only with "mutable" default values, in your case, a list is, with strings this does not happen.

In Python, a function is just another object, so when you actually define it, you're creating a new object. The default values ​​function in some way as the function object's own data. Let's see what happens from defand in subsequent invocations:

When defining the function:

• Create a new function object pointed to the name cajawithin the module's namespace
• The function is evaluated: for each default value, an object is created, in this case an empty list [] that will be a data of the created function object, for convenience we will call itlista_valor_default

The first time we run the function:

• By not passing parameters, what Python does is associate the internal variable/object cosasto the object lista_valor_default that was defined when the function was created, the first time the behavior will be as expected, we add data, but not to cosas(reference) but to the objectlista_valor_default

The following call with no parameters:

• lista_valor_defaultnow it's not an empty list anymore so it will cosaspoint to a list with the elements from the previous call, and we'll keep adding more elements to lista_valor_default.

Demonstration:

The following code serves to understand the commented behavior:

def prueba(lista=[]):
  falsanuevalista = lista
  falsanuevalista.append("Agrego el valor: {0}".format(len(falsanuevalista)))
  print("id lista: {0} id falsanuevalista: {1}".format(id(lista), id(falsanuevalista)))

  return falsanuevalista

print(prueba())
print(prueba())

id lista: 140651642166280 id falsanuevalista: 140651642166280
['Agrego el valor: 0']
id lista: 140651642166280 id falsanuevalista: 140651642166280
['Agrego el valor: 0', 'Agrego el valor: 1']

We see that at all times the idof lista(default value) is the same as the internal variable falsanuevalista, so it is shown that they appendact on the default list.

There are several alternatives, the one I mentioned above is usually the most appropriate, but eventually you could leave your function as it is defined, but the trick would be to force a copy of the default list to a new internal variable and return it, for example like this:

def caja(cosas=[]):
    nueva = cosas[:]   # El slice es una forma de crear una nueva lista      
    # nueva = list(cosas)   # o usando el constructor     
    nueva.append("lapiz") 
    nueva.append("martillo")
    nueva.append("azulejos")   
    return nueva

For more information, I recommend this article .

This is a solution:

def caja():
  cosas=[]        
  cosas.append("lapiz")
  cosas.append("martillo")
  cosas.append("azulejos")
  return cosas

As a parameter you were initializing a variable, that's why it was not emptied