I know how to generate a random number within a range in Python.
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..
[12, 5, 6, 1] = good
[12, 5, 5, 1] = bad, because the number 5 occurs twice.
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~ Asked on 2014-04-03 15:29:57
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
>>> 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
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