[python] How to get a random value from dictionary?

How can I get a random pair from a dict? I'm making a game where you need to guess a capital of a country and I need questions to appear randomly.

The dict looks like {'VENEZUELA':'CARACAS'}

How can I do this?

If you don't want to use the random module, you can also try popitem():

>> d = {'a': 1, 'b': 5, 'c': 7}
>>> d.popitem()
('a', 1)
>>> d
{'c': 7, 'b': 5}
>>> d.popitem()
('c', 7)

Since the dict doesn't preserve order, by using popitem you get items in an arbitrary (but not strictly random) order from it.

Also keep in mind that popitem removes the key-value pair from dictionary, as stated in the docs.

popitem() is useful to destructively iterate over a dictionary

This works in Python 2 and Python 3:

A random key:

random.choice(list(d.keys()))

A random value

random.choice(list(d.values()))

A random key and value

random.choice(list(d.items()))

Since this is homework:

Check out random.sample() which will select and return a random element from an list. You can get a list of dictionary keys with dict.keys() and a list of dictionary values with dict.values().

>>> import random
>>> d = dict(Venezuela = 1, Spain = 2, USA = 3, Italy = 4)
>>> random.choice(d.keys())
'Venezuela'
>>> random.choice(d.keys())
'USA'

By calling random.choice on the keys of the dictionary (the countries).

To select 50 random key values from a dictionary set dict_data:

sample = random.sample(set(dict_data.keys()), 50)

b = { 'video':0, 'music':23,"picture":12 } 
random.choice(tuple(b.items())) ('music', 23) 
random.choice(tuple(b.items())) ('music', 23) 
random.choice(tuple(b.items())) ('picture', 12) 
random.choice(tuple(b.items())) ('video', 0) 

Here is a little Python code for a dictionary class that can return random keys in O(1) time. (I included MyPy types in this code for readability):

from typing import TypeVar, Generic, Dict, List
import random

K = TypeVar('K')
V = TypeVar('V')
class IndexableDict(Generic[K, V]):
    def __init__(self) -> None:
        self.keys: List[K] = []
        self.vals: List[V] = []
        self.dict: Dict[K, int] = {}

    def __getitem__(self, key: K) -> V:
        return self.vals[self.dict[key]]

    def __setitem__(self, key: K, val: V) -> None:
        if key in self.dict:
            index = self.dict[key]
            self.vals[index] = val
        else:
            self.dict[key] = len(self.keys)
            self.keys.append(key)
            self.vals.append(val)

    def __contains__(self, key: K) -> bool:
        return key in self.dict

    def __len__(self) -> int:
        return len(self.keys)

    def random_key(self) -> K:
        return self.keys[random.randrange(len(self.keys))]

With modern versions of Python(since 3), the objects returned by methods dict.keys(), dict.values() and dict.items() are view objects*. And hey can be iterated, so using directly random.choice is not possible as now they are not a list or set.

One option is to use list comprehension to do the job with random.choice:

import random

colors = {
    'purple': '#7A4198',
    'turquoise':'#9ACBC9',
    'orange': '#EF5C35',
    'blue': '#19457D',
    'green': '#5AF9B5',
    'red': ' #E04160',
    'yellow': '#F9F985'
}

color=random.choice([hex_color for color_value in colors.values()]

print(f'The new color is: {color}')

References:

• *Python 3.8: Standard Library Documentation - Built-in types: Dictionary view objects
• Python 3.8: Data Structures - List Comprehensions:

Since the original post wanted the pair:

import random
d = {'VENEZUELA':'CARACAS', 'CANADA':'TORONTO'}
country, capital = random.choice(list(d.items()))

(python 3 style)

I found this post by looking for a rather comparable solution. For picking multiple elements out of a dict, this can be used:

idx_picks = np.random.choice(len(d), num_of_picks, replace=False) #(Don't pick the same element twice)
result = dict ()
c_keys = [d.keys()] #not so efficient - unfortunately .keys() returns a non-indexable object because dicts are unordered
for i in idx_picks:
    result[c_keys[i]] = d[i]

If you don't want to use random.choice() you can try this way:

>>> list(myDictionary)[i]
'VENEZUELA'
>>> myDictionary = {'VENEZUELA':'CARACAS', 'IRAN' : 'TEHRAN'}
>>> import random
>>> i = random.randint(0, len(myDictionary) - 1)
>>> myDictionary[list(myDictionary)[i]]
'TEHRAN'
>>> list(myDictionary)[i]
'IRAN'

Try this:

import random
a = dict(....) # a is some dictionary
random_key = random.sample(a, 1)[0]

This definitely works.

I wrote this trying to solve the same problem:

https://github.com/robtandy/randomdict

It has O(1) random access to keys, values, and items.

I am assuming that you are making a quiz kind of application. For this kind of application I have written a function which is as follows:

def shuffle(q):
"""
The input of the function will 
be the dictionary of the question
and answers. The output will
be a random question with answer
"""
selected_keys = []
i = 0
while i < len(q):
    current_selection = random.choice(q.keys())
    if current_selection not in selected_keys:
        selected_keys.append(current_selection)
        i = i+1
        print(current_selection+'? '+str(q[current_selection]))

If I will give the input of questions = {'VENEZUELA':'CARACAS', 'CANADA':'TORONTO'} and call the function shuffle(questions) Then the output will be as follows:

VENEZUELA? CARACAS
CANADA? TORONTO

You can extend this further more by shuffling the options also

Try this (using random.choice from items)

import random

a={ "str" : "sda" , "number" : 123, 55 : "num"}
random.choice(list(a.items()))
#  ('str', 'sda')
random.choice(list(a.items()))[1] # getting a value
#  'num'