Pull random keys from dictionary in Python that are not equal to one another

Question

So I'm trying to setup a multiple choice quiz via Python. I'm fairly new to Python, so my apologies up front if there is a simpler way to do this. However, I'm trying to really understand some basics before moving forward to newer techniques.

I have a dictionary. In this dictionary, I want to grab 3 random keys. I also want to make sure that these three keys are not equal (in other words, random from one another). Here is the code I have wrote so far:

import random

word_drills = {'class': 'Tell Python to make a new kind of thing.',
               'object': 'Two meanings: the most basic kind of thing, and any instance of some thing.',
               'instance': 'What you get when you tell Python to create a class.',
               'def': 'How you define a function inside a class.',
               'self': 'Inside the functions in a class, self is a variable for the instance/object being accessed.',
               'inheritance': 'The concept that one class can inherit traits from another class, much like you and your parents.',
               'composition': 'The concept that a class can be composed of other classes as parts, much like how a car has wheels.',
               'attribute': 'A property classes have that are from composition and are usually variables.',
               'is-a': 'A phrase to say that something inherits from another, as in a Salmon *** Fish',
               'has-a': 'A phrase to say that something is composed of other things or has a trait, as in a Salmon *** mouth.'}

key1 = ' '
key2 = ' '
key3 = ' '             

def nodupchoice():
    while key1 == key2 == key3: 
        key1 = random.choice(word_drills.keys())
        key2 = random.choice(word_drills.keys())
        key3 = random.choice(word_drills.keys())


nodupchoice()

print "Key #1: %s, Key #2: %s, Key #3: %s" % (key1, key2, key3)

I'm fairly sure the issue is with my while loop. I wanted to create a function that will keep running until all three keys are different from one another. Finally, it would print the result. Any ideas? Thanks in advance.

Answer 1

You can use random.sample:

>>> random.sample(word_drills, 3)
['has-a', 'attribute', 'instance']

and you don't need .keys(), iteration over a dictionary is over the keys.

Note that random.sample will return three unique values from the list you supply (i.e. it will never return 'has-a' twice):

>>> all(len(set(random.sample(word_drills, 3))) == 3 for i in range(10**5))
True

Answer 2

Use random.sample

>>> import random
>>> random.sample([1,2,3,4,5,6], 3)
[4, 5, 2]

Answer 3

Probably the easiest thing to do is to shuffle the keys:

keysShuffled = list(word_drills)
random.shuffle(keysShuffled)

Then take the first 3

threeUnique = keysShuffled[:3]

Answer 4

Maybe you would like to try Quiz Me 2.5? It is a multiple choice quizzing system with a GUI in Python 3.x.

Answer 5

As others have explained, random.sample is the best way to do what you want.

To explain why your original code didn't work:

First problem, there is a logic error with your while loop - it will end as soon as two items are different (e.g 'a' == 'a' == 'b' is True, and the loop will end), to fix this, there are a few ways:

while not (key1 != key2 != key3):

Alternatively, a set can only contain unique items, so when the length of set([key1, key2, key3]) is 3, all three values are different:

while len(set([key1, key2, key3]) != 3:

Second problem, and the reason the code will refuse to run:

key1 key2 key3 are defined as global variables (key1 = ' ' are at the top-level of your.py` file, not in a function)

You can look up a global variable in a function without issue (like where you done while key1 == key2 ...). The error occurs because you also try and assign to key1 within your loop - this confuses Python's parser, because you are looking up a global variable in the while key1 == ... part, but trying to make a new function-local variable on the next line, so it helpfully throws an error.

A rubbish way to fix this is like so:

key1 = ' '
key2 = ' '
key3 = ' '             

def nodupchoice():
    global key1
    global key2
    global key3
    while not (key1 != key2 != key3):
        key1 = random.choice(word_drills.keys())
        key2 = random.choice(word_drills.keys())
        key3 = random.choice(word_drills.keys())

Then it will work as you intended, but don't do that - global variables have their uses, but this is not such a situation..

Instead of using a global variable, you can easily return multiple values from a function:

def nodupchoice():
    key1 = ' ' # local variables, which are inaccessible outside this function
    key2 = ' '
    key3 = ' '

    while not (key1 != key2 != key3):
        key1 = random.choice(word_drills.keys())
        key2 = random.choice(word_drills.keys())
        key3 = random.choice(word_drills.keys())

    return key1, key2, key3


key1, key2, key3 = nodupchoice()

print "Key #1: %s, Key #2: %s, Key #3: %s" % (key1, key2, key3)

Oh, and even better, you can pass in the word_drills as an argument:

def nodupchoice(things):
    key1 = ' ' # local variables, which are inaccessible outside this function
    key2 = ' '
    key3 = ' '

    while not (key1 != key2 != key3):
        key1 = random.choice(things)
        key2 = random.choice(things)
        key3 = random.choice(things)

    return key1, key2, key3


key1, key2, key3 = nodupchoice(word_drills)

...and you've written a function almost identical to random.sample(word_drills, 3)!

Answer 6

I had this samme problem and wrote https://github.com/robtandy/randomdict to solve it. Maybe it will help you also.

It provides a python dict object but with the additional methods, random_key, random_value and random_item. All with O(1) run time complexity.