How do I create a list of random numbers without duplicates?

Question

I tried using random.randint(0, 100), but some numbers were the same. Is there a method/module to create a list unique random numbers?

Note: The fol

Answer 1

A very simple function that also solves your problem

from random import randint

data = []

def unique_rand(inicial, limit, total):

        data = []

        i = 0

        while i < total:
            number = randint(inicial, limit)
            if number not in data:
                data.append(number)
                i += 1

        return data


data = unique_rand(1, 60, 6)

print(data)


"""

prints something like 

[34, 45, 2, 36, 25, 32]

"""

Answer 2

import random
result=[]
for i in range(1,50):
    rng=random.randint(1,20)
    result.append(rng)

Answer 3

In order to obtain a program that generates a list of random values without duplicates that is deterministic, efficient and built with basic programming constructs consider the function extractSamples defined below,

def extractSamples(populationSize, sampleSize, intervalLst) :
    import random
    if (sampleSize > populationSize) :
        raise ValueError("sampleSize = "+str(sampleSize) +" > populationSize (= " + str(populationSize) + ")")
    samples = []
    while (len(samples) < sampleSize) :
        i = random.randint(0, (len(intervalLst)-1))
        (a,b) = intervalLst[i]
        sample = random.randint(a,b)
        if (a==b) :
            intervalLst.pop(i)
        elif (a == sample) : # shorten beginning of interval                                                                                                                                           
            intervalLst[i] = (sample+1, b)
        elif ( sample == b) : # shorten interval end                                                                                                                                                   
            intervalLst[i] = (a, sample - 1)
        else :
            intervalLst[i] = (a, sample - 1)
            intervalLst.append((sample+1, b))
        samples.append(sample)
    return samples

The basic idea is to keep track of intervals intervalLst for possible values from which to select our required elements from. This is deterministic in the sense that we are guaranteed to generate a sample within a fixed number of steps (solely dependent on populationSize and sampleSize).

To use the above function to generate our required list,

In [3]: populationSize, sampleSize = 10**17, 10**5

In [4]: %time lst1 = extractSamples(populationSize, sampleSize, [(0, populationSize-1)])
CPU times: user 289 ms, sys: 9.96 ms, total: 299 ms
Wall time: 293 ms

We may also compare with an earlier solution (for a lower value of populationSize)

In [5]: populationSize, sampleSize = 10**8, 10**5

In [6]: %time lst = random.sample(range(populationSize), sampleSize)
CPU times: user 1.89 s, sys: 299 ms, total: 2.19 s
Wall time: 2.18 s

In [7]: %time lst1 = extractSamples(populationSize, sampleSize, [(0, populationSize-1)])
CPU times: user 449 ms, sys: 8.92 ms, total: 458 ms
Wall time: 442 ms

Note that I reduced populationSize value as it produces Memory Error for higher values when using the random.sample solution (also mentioned in previous answers here and here). For above values, we can also observe that extractSamples outperforms the random.sample approach.

P.S. : Though the core approach is similar to my earlier answer, there are substantial modifications in implementation as well as approach alongwith improvement in clarity.

Answer 4

The solution presented in this answer works, but it could become problematic with memory if the sample size is small, but the population is huge (e.g. random.sample(insanelyLargeNumber, 10)).

To fix that, I would go with this:

answer = set()
sampleSize = 10
answerSize = 0

while answerSize < sampleSize:
    r = random.randint(0,100)
    if r not in answer:
        answerSize += 1
        answer.add(r)

# answer now contains 10 unique, random integers from 0.. 100

Answer 5

From the CLI in win xp:

python -c "import random; print(sorted(set([random.randint(6,49) for i in range(7)]))[:6])"

In Canada we have the 6/49 Lotto. I just wrap the above code in lotto.bat and run C:\home\lotto.bat or just C:\home\lotto.

Because random.randint often repeats a number, I use set with range(7) and then shorten it to a length of 6.

Occasionally if a number repeats more than 2 times the resulting list length will be less than 6.

EDIT: However, random.sample(range(6,49),6) is the correct way to go.