CODEWITHSUNDEEP
pythonpython-3.xrandom
jhon.smith
jhon.smith

Reputation: 2043

Generate 2 Unique but random Elements in a List

I need to generate a list with unique elements 

parties = ['Party A', 'Party B']

I have tried this

def party_generator(size=1, chars=string.ascii_uppercase):
    parties = []
    for y in range(2):
        party = ''.join(random.choice(chars) for x in range(size))
        parties.append(''.join(['Party ', party]))
    return parties

But I am afraid that my code might generate duplicates 

party_generator()
['Party S', 'Party S']

How do I generate a list with unique elements?

Upvotes: 3

Views: 105

Answers (1)

Martijn Pieters
Martijn Pieters

Reputation: 1121406

Use a set to track if you've seen the same generated random string before:

def party_generator(size=1, chars=string.ascii_uppercase):
    parties = []
    seen = set()
    while len(parties) < 2:
        party = ''.join(random.choice(chars) for x in range(size))
        if party in seen:
            continue
        seen.add(party)
        parties.append('Party {}'.format(party))
    return parties

This will continue to generate random strings until you have 2 unique values.

You cannot easily use random.sample() here as that requires a fixed set of choices to sample from but you are generating names of a variable length. It is possible to build an object that emulates a sequence (by giving it __len__ and __getattr__ methods) and generating a specific word of length size from all possible words you can create given the chars variable:

class CharacterRange(object):
    def __init__(self, chars, size):
        self.chars, self.size = chars, size

    def __len__(self):
        return len(self.chars) ** self.size

    def __getitem__(self, item):
        if item < 0:
            item = len(self) + item
        if not 0 <= item <= len(self):
            raise IndexError('Index out of range')

        result = []
        for i in range(self.size):
            item, index = divmod(item, len(self.chars))
            result.append(self.chars[index])
        return ''.join(result[::-1])

Demo:

>>> uppercase_len1 = CharacterRange(string.uppercase, 1)
>>> len(uppercase_len1)
26
>>> uppercase_len5[0]
'A'
>>> uppercase_len5[-1]
'Z'
>>> uppercase_len1[10]
'K'
>>> uppercase_len1[24]
'Y'
>>> uppercase_len5 = CharacterRange(string.uppercase, 5)
>>> len(uppercase_len5)
11881376
>>> uppercase_len5[0]
'AAAAA'
>>> uppercase_len5[-1]
'ZZZZZ'
>>> uppercase_len5[1024]
'AABNK'
>>> uppercase_len5[1355453]
'CZDCV'

You can pass this object to random.sample():

>>> import random
>>> random.sample(uppercase_len5, 5)
['CUSQB', 'UUUWM', 'MKOFI', 'MYROU', 'AHRWA']

producing N words of length K in linear time and constant memory.

and you could simplify your code to:

def party_generator(size=1, chars=string.ascii_uppercase):
    return ['Party {}'.format(party) for party in random.sample(CharacterRange(chars, size))]

However, I do think that that would be a little overkill here for just 2 random words.

Upvotes: 2

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