Python 在一个范围内生成'n'个唯一的随机数
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Generate 'n' unique random numbers within a range
提问by Chris Headleand
I know how to generate a random number within a range in Python.
我知道如何在 Python 中生成一个范围内的随机数。
random.randint(numLow, numHigh)
And I know I can put this in a loop to generate n amount of these numbers
我知道我可以把它放在一个循环中来生成 n 个这些数字
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?
但是，我需要确保该列表中的每个数字都是唯一的。除了大量的条件语句之外，是否有一种直接的方法可以生成 n 个唯一的随机数？
The important thing is that each number in the list is different to the others..
重要的是列表中的每个数字都与其他数字不同。
So
所以
[12, 5, 6, 1] = good
[12, 5, 6, 1] = 好
But
但
[12, 5, 5, 1] = bad, because the number 5 occurs twice.
[12, 5, 5, 1] = bad，因为数字 5 出现了两次。
采纳答案by Two-Bit Alchemist
If you just need sampling without replacement:
如果您只需要取样而不更换：
>>> import random
>>> random.sample(range(1, 100), 3)
[77, 52, 45]
random.sampletakes a population and a sample size kand returns krandom members of the population.
random.sample采用总体和样本大小，k并返回k总体中的随机成员。
If you have to control for the case where kis larger than len(population), you need to be prepared to catch a ValueError:
如果您必须控制k大于的情况，则len(population)需要准备捕获ValueError：
>>> try:
... random.sample(range(1, 2), 3)
... except ValueError:
... print('Sample size exceeded population size.')
...
Sample size exceeded population size
回答by thefourtheye
Generate the range of data first and then shuffle it like this
先生成数据范围，然后像这样shuffle
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.
通过这样做，您将获得特定范围内的所有数字，但顺序是随机的。
But you can use random.sampleto get the number of elements you need, from a range of numbers like this
但是您可以使用random.sample从这样的数字范围中获取所需元素的数量
print random.sample(range(numLow, numHigh), 3)
回答by mhlester
You could add to a setuntil you reach n:
您可以添加到 aset直到达到n：
setOfNumbers = set()
while len(setOfNumbers) < n:
setOfNumbers.add(random.randint(numLow, numHigh))
Be careful of having a smaller range than will fit in n. It will loop forever, unable to find new numbers to insert up to n
小心使用比适合的范围更小的范围n。它将永远循环，无法找到要插入的新数字n
回答by maxbublis
You could use the random.samplefunction from the standard libraryto select kelements from a population:
您可以使用标准库中的random.sample函数从种群中选择k 个元素：
import random
random.sample(range(low, high), n)
In case of a rather large range of possible numbers, you could use itertools.islicewith an infinite random generator:
如果可能的数字范围相当大，您可以使用itertools.islice无限随机生成器：
import itertools
import random
def random_gen(low, high):
while True:
yield random.randrange(low, high)
gen = random_gen(1, 100)
items = list(itertools.islice(gen, 10)) # Take first 10 random elements
After the question update it is now clear that you need ndistinct (unique) numbers.
问题更新后，现在很明显您需要n 个不同（唯一）的数字。
import itertools
import random
def random_gen(low, high):
while True:
yield random.randrange(low, high)
gen = random_gen(1, 100)
items = set()
# Try to add elem to set until set length is less than 10
for x in itertools.takewhile(lambda x: len(items) < 10, gen):
items.add(x)
