Python: creating a dictionary: {int, class object} in a loop

Question

I'd like to create a dictonary: {int, Class} in a loop, however the class object is being overriden. I am pretty sure that this is a basic problem, but I am stuck

class Simple:
    simpleDic = {
        'name': {""},
        'age': {1}}

    def __init__(self, name, age):
        self.simpleDic['name'] = name
        self.simpleDic['age'] = age
    
    def __str__(self):
        return "{} {}\n".format(self.simpleDic['name'], self.simpleDic['age'])

def foo():
    myDict = {}
    for x in range(3):
        myDict[x] = Simple('Name' + str(x), x)
        print(('{}: {}\n'.format("Creating", myDict[x])))
    
    for key in myDict:
        print(('{}: {}\n'.format("Printing" + str(key), myDict[key])))


#### Main program here ####

foo()

The output is as follows:

Meny Issakov · Accepted Answer · 2020-10-15 17:34:38Z

1

You're problem is on the last print You're printing myDict[x] when you should print myDict[key]

x contains the last value from the first iteration, so you're practically printing the same key all over

Following your question in this comments:

class Simple:

    def __init__(self, name, age):
        self.simpleDic = {"name": name, "age": age}
    
    def __str__(self):
        return "{} {}\n".format(self.simpleDic['name'], self.simpleDic['age'])
    
    def __repr__(self):
        return "{} {}\n".format(self.simpleDic['name'], self.simpleDic['age'])    

def foo():
    myDict = {}
    for x in range(3):
        myDict[x] = Simple('Name' + str(x), x)
        print(('{}: {}\n'.format("Creating", myDict[x])))
        
    print(myDict)

edited Oct 15, 2020 at 17:34

answered Oct 15, 2020 at 16:33

Meny Issakov

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7 Comments

rublow Over a year ago

My goodness, what a stupid mistake. I am sorry about that.

Meny Issakov Over a year ago

Happens to the best of us :)

rublow Over a year ago

And what if i changed the code as in question now (replaced simple variable with a dictionary):

Meny Issakov Over a year ago

Then it would fail, as myDict belongs to the scope of the foo function, and you would get a NameError: name 'myDict' is not defined

Meny Issakov Over a year ago

You can return myDict in the foo function, and assign it to a variable on the global scope (before your loop)

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