Python: Define own function called "randomcall"

Question

so i am stuck on a certain problem I have to solve and was wondering if you guys could help me in my situation (i am really new to python and programming by itself). So the task i got is, to define a own new function to pick out of a Dictionary called "phonebook" random people and print out to pretend calling them. Something like "randomcall(phonebook,10) and then it prints: Calling Peter at 1800650, and 9 others.

def randomcall(phonebook, number):
    import random
    
    for name in range(phonebook):
        name = random.choice(list(phonebook))
        phonebook = phonebook[name]
        print(f"Let's call {name} with {phonebook}")

phonebook = {"martin": 12345, "anna": 65478, "klaus": 5468764, "hans": 748463, "peter": 84698416, "ulrich": 3416846546, "frank": 4789749, "lukas": 798469, "ferdinand": 68465131}

randomcall(phonebook, 3)

randomcall("", 5)

I think you mean `for i in range(number)` instead of `for name in range(phonebook)`, because `name` is used later on and there is no need to make it the iterating variable, and `phonebook` is the dictionary, not the # of people you are calling. — David, Dec 18 '20 at 13:03
Also, what does `randomcall("", 5)` mean? Are you trying to call nobody there? — David, Dec 18 '20 at 13:05
You are right, the randomcall("",5) is not necessary, it was a try to call a random person, but thats not how this works. — Sammy Mousa, Dec 18 '20 at 13:11
would the example `5` in `randomcall("",5)` mean on how many people to call? — Ice Bear, Dec 18 '20 at 13:17

score 1 · Answer 1 · answered Dec 18 '20 at 13:19

You could this, number will be the range on how many people to call.

import random

def randomcall(phonebook,number):

    for _ in range(number):
        name = random.choice(list(phonebook.keys()))
        print(f'Lets call {name} with {phonebook[name]}')



phonebook = {"martin":12345,"anna":65478,"klaus":5468764,"hans":748463,"peter":84698416,"ulrich":3416846546,"frank":4789749,"lukas":798469,"ferdinand":68465131}

randomcall(phonebook,3)

Output:

Lets call lukas with 798469
Lets call peter with 84698416
Lets call klaus with 5468764

score 1 · Accepted Answer · answered Dec 18 '20 at 13:20

Your code is correct, except for several small mistakes:

As said in my comment above, you cannot use for name in range(phonebook) because phonebook is not the integral number of times that you need to iterate. number is. Also the variable name is not referenced to in the iteration; it was assigned a new value. So it can be changed into another variable, like i, which is used more commonly in iteration.
The number of the person to call is, in your code, stored under the variable phonebook, which is the variable of the phonebook dictionary. It makes it impossible to access the actual phonebook again. Another variable, like phone, can be used there.

So the completed code should look like:

def randomcall(phonebook, number):
    import random

    for name in range(number):
        name = random.choice(list(phonebook))
        phonebook = phonebook[name]
        print(f"Let's call {name} with {phonebook}")


phonebook = {"martin": 12345, "anna": 65478, "klaus": 5468764, "hans": 748463, "peter": 84698416, "ulrich": 3416846546, "frank": 4789749, "lukas": 798469, "ferdinand": 68465131}

randomcall(phonebook, 3)

Output:

Let's call anna with 65478
Let's call klaus with 5468764
Let's call hans with 748463

Also, as stated in this comment, the line randomcall("", 5) is used to call a random person. Actually, the phonebook must be passed to the function as the phonebook parameter, and if 1 is passed to the number parameter, it generates 1 random call.

Python: Define own function called "randomcall"

2 Answers2