# Generate 'n' unique random numbers within a range

266

I know how to generate a random number within a range in Python.

``````random.randint(numLow, numHigh)
``````

And I know I can put this in a loop to generate n amount of these numbers

``````for x in range (0, n):
listOfNumbers.append(random.randint(numLow, numHigh))
``````

However, I need to make sure each number in that list is unique. Other than a load of conditional statements, is there a straightforward way of generating n number of unique random numbers?

The important thing is that each number in the list is different to the others..

So

[12, 5, 6, 1] = good

But

[12, 5, 5, 1] = bad, because the number 5 occurs twice.

This question is tagged with `python` `random` `unique`

~ Asked on 2014-04-03 15:29:57

### The Best Answer is

435

If you just need sampling without replacement:

``````>>> import random
>>> random.sample(range(1, 100), 3)
[77, 52, 45]
``````

random.sample takes a population and a sample size `k` and returns `k` random members of the population.

If you have to control for the case where `k` is larger than `len(population)`, you need to be prepared to catch a `ValueError`:

``````>>> try:
...   random.sample(range(1, 2), 3)
... except ValueError:
...   print('Sample size exceeded population size.')
...
Sample size exceeded population size
``````

~ Answered on 2014-04-03 15:34:45

29

Generate the range of data first and then shuffle it like this

``````import random
data = range(numLow, numHigh)
random.shuffle(data)
print data
``````

By doing this way, you will get all the numbers in the particular range but in a random order.

But you can use `random.sample` to get the number of elements you need, from a range of numbers like this

``````print random.sample(range(numLow, numHigh), 3)
``````

~ Answered on 2014-04-03 15:31:46