Print ONLY Dictionary Name from Inside a List in Python

Question

I have a List containing Three Dictionaries as below:

lloyd = {
"name": "Lloyd",
"homework": [90.0, 97.0, 75.0, 92.0],
"quizzes": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}

alice = {
"name": "Alice",
"homework": [100.0, 92.0, 98.0, 100.0],
"quizzes": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}

tyler = {
"name": "Tyler",
"homework": [0.0, 87.0, 75.0, 22.0],
"quizzes": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}

students = [lloyd, alice, tyler]

I want to loop through the list and print only the names of dictionaries

i.e.

lloyd
alice
tyler

I do not want to print the keys or values of the dictionaries themselves.

Any suggestions in this regard will be helpful

If the names of the dictionaries are same as the attribute `name`, why not print that instead? — Quill, Aug 22 '15 at 17:03
I don't think this is possible in Python, and if it is it's not good practice. What exactly are you trying to do? This has nothing to with dictionaries. — veggie1, Aug 22 '15 at 17:06
See here: http://stackoverflow.com/questions/1538342/how-can-i-get-the-name-of-an-object-in-python — fukanchik, Aug 22 '15 at 17:10

score 2 · Answer 1 · answered Aug 22 '15 at 17:09

Objects in Python don't "have names", they are bound to variables. Consider the following code:

lst1 = [3, [4, 5, 6], 7]
lst2 = lst1

What's the "name" of the list (it can be accessed either as lst1 or lst2). What's the name of the second element of the list (only the list stores a reference to the sub-list).

The dir() function can give you a list of the names in a namespace, but that's about the best you can do.

score 2 · Answer 2 · answered Jun 27 '20 at 20:35

I know this is old but I believe I have a solution to the question asked. This can be accomplished by nesting these dictionaries into another dictionary.

Take the original code:

lloyd = {
"name": "Lloyd",
"homework": [90.0, 97.0, 75.0, 92.0],
"quizzes": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}

alice = {
"name": "Alice",
"homework": [100.0, 92.0, 98.0, 100.0],
"quizzes": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}

tyler = {
"name": "Tyler",
"homework": [0.0, 87.0, 75.0, 22.0],
"quizzes": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}

students = [lloyd, alice, tyler]

Instead of using the list of dictionaries, create a nested dictionary called users with the dictionaries of the users as members.

users = {

    'lloyd' : {
        "name": "Lloyd",
        "homework": [90.0, 97.0, 75.0, 92.0],
        "quizzes": [88.0, 40.0, 94.0],
        "tests": [75.0, 90.0]
        },

    'alice' : {
        "name": "Alice",
        "homework": [100.0, 92.0, 98.0, 100.0],
        "quizzes": [82.0, 83.0, 91.0],
        "tests": [89.0, 97.0]
        },

    'tyler' : {
        "name": "Tyler",
        "homework": [0.0, 87.0, 75.0, 22.0],
        "quizzes": [0.0, 75.0, 78.0],
        "tests": [100.0, 100.0]
        }
    }

Then a simple for loop to print the username...

for username, user_info in users.items():
    print(f"{username.title()}")

This gives the desired output..

Lloyd
Alice
Tyler

I am just learning myself so I hope this helps someone like the stack overflow community has helped me!

score 1 · Answer 3 · answered Aug 22 '15 at 17:04

1

for i in students:print(i['name'])

answered Aug 22 '15 at 17:04

Hybrid

321
3
13

@Marcus Müller Dont Know Friend. – Hybrid Aug 23 '15 at 06:49
I've re-read the question. I think OP tries to get the "handle", not the "name" element of the dictionary. – Marcus Müller Aug 23 '15 at 06:50

score 1 · Answer 4 · answered Nov 13 '20 at 16:39

1

for x in dict.items():
    print(x[0])

answered Nov 13 '20 at 16:39

Sean

11
1

2

Welcome to Stack Overflow. It would be great if you would describe your code and state how it will solve the problem. – Patrick Klein Nov 13 '20 at 17:33

score 0 · Answer 5 · answered Aug 22 '15 at 20:40

there is not __name__ or such attr in standard dict class.(functions have func.func_name) I am sure there would be some better implementations but my solution would be like below:

create a new type of dict with some additional attributes ..

class myspecial_dict(dict):
    def __init__(self,Name,**kwargs):
        super(myspecial_dict,self).__init__(kwargs)
        self.dictname=Name

mys=myspecial_dict("lloyd",**lloyd)
print mys.dictname
print mys['quizzes']

that prints out:

lloyd
[88.0, 40.0, 94.0]

if you want to do this for all:

dicts    =[lloyd,alice,tyler]
dictnames=["lloyd","alice","tyler"]
print [myspecial_dict(x,**y).dictname for x,y in zip(dictnames,dicts) ]

that yields:

['lloyd', 'alice', 'tyler']

Print ONLY Dictionary Name from Inside a List in Python

5 Answers5