Making Phonebook in python : i want to get this screen by fixing my current code

Question

I made my code like below....

But as i input the data such as spam & number, previous data is deleted.

So i'd like to make multiple value in one key... (i think using list is kinda good method)

For example,

Key: spam - Value: 01012341111, 01099991234,

Key: spam - Value: 01032938962, 01023421232, 01023124242

enter image description here

In summary, I want to get this print (attached Picture)

this is my code:

enter code here
phonebook = dict()
ntype = ['spam', 'friend', 'family', 'etc']
trash =[]
spam = []
friend = []
family = []
etc = []
while True:
    a = input("Input your number and ntype : ")
    b = a.split()
    i = 0
    j = 0
    if a == 'exit':
        print("end program")
        break
        exit()
    elif a == "print spam":
        print("*** spam numbers: ")
        print('*', phonebook['spam'])
    elif a == "print numbers":
        print("*** numbers:")
    for t in ntype:
        try:
            print("**", t)
            print('*',phonebook[t])
        except KeyError:
            continue
    print("** trash")
    print("*", phonebook['trash'])
else:
    while True:
        try:
            if ntype[j] in b:
                for k in b:
                    if list(k)[0] == '0' and len(k) >= 10 and len(k) <= 11:
                        if k in phonebook.values():
                            print("Already Registered")
                        else:
                            phonebook[ntype[j]] = k
                            print(phonebook)
                break
            else:
                j+=1
        except IndexError:
            if list(b[i])[0] == '0' and len(b[i]) >= 10 and len(b[i]) <= 11:
                if b[i] in phonebook.values():
                    print("Already Registered")
                else:
                    phonebook['trash'] = b[i]
                    print(phonebook)
                break
            else:
                break

Answer 1

You should use list for that. The problem is that you cannot append to a value that has not yet been set.

>>> d = {}
>>> d['a'].append('value')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 'a'

And, as you saw, assigning multiple times to the same key won't do the trick.

>>> d = {}
>>> d['a'] = 'value1'
>>> d['a'] = 'value2'
>>> d
{'a': 'value2'}

So, in order to make it work you could initialize all your possible keys with an empty list:

>>> d = {}
>>> possible_keys = ['a', 'b']
>>> for k in possible_keys:
...     d[k] = []
... 
>>> d['a'].append('value1')
>>> d['a'].append('value2')
>>> d['b'].append('value3')
>>> d['b'].append('value4')
>>> d
{'b': ['value3', 'value4'], 'a': ['value1', 'value2']}

This works but it's just tiring. Initializing dicts is a very common use case, so a method was added to dict to add a default value if it has not yet been set:

>>> d = {}
>>> d.setdefault('a', []).append('value1')
>>> d.setdefault('a', []).append('value2')
>>> d.setdefault('b', []).append('value3')
>>> d.setdefault('b', []).append('value4')
>>> d
{'b': ['value3', 'value4'], 'a': ['value1', 'value2']}

But then again you would have to remember to call setdefault every time. To solve this the default library offers defaultdict.

>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> d['a'].append('value1')
>>> d['a'].append('value2')
>>> d['b'].append('value3')
>>> d['b'].append('value4')
>>> d['a']
['value1', 'value2']
>>> d['b']
['value3', 'value4']

Which may just be what you need.

Hope I could help. ;)