我尝试使用random.randint(0, 100)，但一些数字相同。有没有一种方法/模块来创建唯一的随机数列表？

注意：以下代码基于答案，并在发布答案后添加。这不是问题的一部分。这是解决方案。

def getScores():
    # open files to read and write
    f1 = open("page.txt", "r");
    p1 = open("pgRes.txt", "a");

    gScores = [];
    bScores = [];
    yScores = [];

    # run 50 tests of 40 random queries to implement "bootstrapping" method 
    for i in range(50):
        # get 40 random queries from the 50
        lines = random.sample(f1.readlines(), 40);

这将返回从0到99范围内选择的10个数字的列表，没有重复。

import random
random.sample(range(100), 10)

参考您的特定代码示例，您可能希望一次从文件中读取所有行，然后从内存中的已保存列表中选择随机行。例如：

all_lines = f1.readlines()
for i in range(50):
    lines = random.sample(all_lines, 40)

这样，您只需要在循环之前实际从文件中读取一次即可。与返回文件的开头并f1.readlines()为每次循环迭代再次调用相比，这样做的效率要高得多。

您可以像下面这样使用随机模块中的随机播放功能：

import random

my_list = list(xrange(1,100)) # list of integers from 1 to 99
                              # adjust this boundaries to fit your needs
random.shuffle(my_list)
print my_list # <- List of unique random numbers

请注意，在这里shuffle方法不会像预期的那样返回任何列表，它只会对通过引用传递的列表进行随机排序。

您可以先创建一个从a到的数字列表b，其中a和b分别是列表中的最小和最大数字，然后使用Fisher-Yates算法或使用Python的random.shuffle方法对其进行混洗。

此答案中给出的解决方案有效，但如果样本量很小但总数很大（例如random.sample(insanelyLargeNumber, 10)），则可能会在内存问题上产生问题。

为了解决这个问题，我可以这样做：

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

线性同余伪随机数生成器

O（1）记忆

O（k）运算

这个问题可以用一个简单的线性同余发生器来解决。这需要恒定的内存开销（8个整数）和最多2 *（序列长度）的计算。

所有其他解决方案都使用更多的内存和更多的计算资源！如果只需要几个随机序列，则此方法将便宜得多。对于大小范围N，如果要按N唯一k序列或更大的顺序生成，我建议使用内置方法接受的解决方案，random.sample(range(N),k)因为它已在python中进行了速度优化。

码

# Return a randomized "range" using a Linear Congruential Generator
# to produce the number sequence. Parameters are the same as for 
# python builtin "range".
#   Memory  -- storage for 8 integers, regardless of parameters.
#   Compute -- at most 2*"maximum" steps required to generate sequence.
#
def random_range(start, stop=None, step=None):
    import random, math
    # Set a default values the same way "range" does.
    if (stop == None): start, stop = 0, start
    if (step == None): step = 1
    # Use a mapping to convert a standard range into the desired range.
    mapping = lambda i: (i*step) + start
    # Compute the number of numbers in this range.
    maximum = (stop - start) // step
    # Seed range with a random integer.
    value = random.randint(0,maximum)
    # 
    # Construct an offset, multiplier, and modulus for a linear
    # congruential generator. These generators are cyclic and
    # non-repeating when they maintain the properties:
    # 
    #   1) "modulus" and "offset" are relatively prime.
    #   2) ["multiplier" - 1] is divisible by all prime factors of "modulus".
    #   3) ["multiplier" - 1] is divisible by 4 if "modulus" is divisible by 4.
    # 
    offset = random.randint(0,maximum) * 2 + 1      # Pick a random odd-valued offset.
    multiplier = 4*(maximum//4) + 1                 # Pick a multiplier 1 greater than a multiple of 4.
    modulus = int(2**math.ceil(math.log2(maximum))) # Pick a modulus just big enough to generate all numbers (power of 2).
    # Track how many random numbers have been returned.
    found = 0
    while found < maximum:
        # If this is a valid value, yield it in generator fashion.
        if value < maximum:
            found += 1
            yield mapping(value)
        # Calculate the next value in the sequence.
        value = (value*multiplier + offset) % modulus

用法

此函数“ random_range”的用法与任何生成器（例如“ range”）相同。一个例子：

# Show off random range.
print()
for v in range(3,6):
    v = 2**v
    l = list(random_range(v))
    print("Need",v,"found",len(set(l)),"(min,max)",(min(l),max(l)))
    print("",l)
    print()

样品结果

Required 8 cycles to generate a sequence of 8 values.
Need 8 found 8 (min,max) (0, 7)
 [1, 0, 7, 6, 5, 4, 3, 2]

Required 16 cycles to generate a sequence of 9 values.
Need 9 found 9 (min,max) (0, 8)
 [3, 5, 8, 7, 2, 6, 0, 1, 4]

Required 16 cycles to generate a sequence of 16 values.
Need 16 found 16 (min,max) (0, 15)
 [5, 14, 11, 8, 3, 2, 13, 1, 0, 6, 9, 4, 7, 12, 10, 15]

Required 32 cycles to generate a sequence of 17 values.
Need 17 found 17 (min,max) (0, 16)
 [12, 6, 16, 15, 10, 3, 14, 5, 11, 13, 0, 1, 4, 8, 7, 2, ...]

Required 32 cycles to generate a sequence of 32 values.
Need 32 found 32 (min,max) (0, 31)
 [19, 15, 1, 6, 10, 7, 0, 28, 23, 24, 31, 17, 22, 20, 9, ...]

Required 64 cycles to generate a sequence of 33 values.
Need 33 found 33 (min,max) (0, 32)
 [11, 13, 0, 8, 2, 9, 27, 6, 29, 16, 15, 10, 3, 14, 5, 24, ...]

如果从1到N的N个数字的列表是随机生成的，则可以，某些数字可能会重复。

如果要按随机顺序排列从1到N的数字列表，请使用从1到N的整数填充数组，然后使用Fisher-Yates随机播放或Python的random.shuffle()。

如果需要采样非常大的数字，则不能使用 range

random.sample(range(10000000000000000000000000000000), 10)

因为它抛出：

OverflowError: Python int too large to convert to C ssize_t

另外，如果random.sample由于范围太小而无法生成所需数量的物品

 random.sample(range(2), 1000)

它抛出：

 ValueError: Sample larger than population

此函数解决了两个问题：

import random

def random_sample(count, start, stop, step=1):
    def gen_random():
        while True:
            yield random.randrange(start, stop, step)

    def gen_n_unique(source, n):
        seen = set()
        seenadd = seen.add
        for i in (i for i in source() if i not in seen and not seenadd(i)):
            yield i
            if len(seen) == n:
                break

    return [i for i in gen_n_unique(gen_random,
                                    min(count, int(abs(stop - start) / abs(step))))]

大量使用：

print('\n'.join(map(str, random_sample(10, 2, 10000000000000000000000000000000))))

样本结果：

7822019936001013053229712669368
6289033704329783896566642145909
2473484300603494430244265004275
5842266362922067540967510912174
6775107889200427514968714189847
9674137095837778645652621150351
9969632214348349234653730196586
1397846105816635294077965449171
3911263633583030536971422042360
9864578596169364050929858013943

范围小于所请求项目数的用法：

print(', '.join(map(str, random_sample(100000, 0, 3))))

样本结果：

2, 0, 1

它还适用于负范围和步骤：

print(', '.join(map(str, random_sample(10, 10, -10, -2))))
print(', '.join(map(str, random_sample(10, 5, -5, -2))))

样本结果：

2, -8, 6, -2, -4, 0, 4, 10, -6, 8
-3, 1, 5, -1, 3

您可以使用Numpy库进行快速解答，如下所示-

给定的代码段列出了0到5之间的6个唯一数字。您可以根据需要调整参数。

import numpy as np
import random
a = np.linspace( 0, 5, 6 )
random.shuffle(a)
print(a)

输出量

[ 2.  1.  5.  3.  4.  0.]

它不把任何约束，因为我们在random.sample看到称为这里。

希望这个对你有帮助。

此处提供的答案在时间和内存方面都非常有效，但由于它使用了诸如yield的高级python构造，因此有点复杂。在简单的答案行之有效的做法，但与回答的问题是，它可以实际构建所需的集之前产生许多虚假的整数。使用人口大小= 1000，样本大小= 999进行尝试。理论上，它有可能不会终止。

下面的答案解决了这两个问题，因为它是确定性的，虽然目前还不如其他两个效率高，但它还是有一定效率的。

def randomSample(populationSize, sampleSize):
  populationStr = str(populationSize)
  dTree, samples = {}, []
  for i in range(sampleSize):
    val, dTree = getElem(populationStr, dTree, '')
    samples.append(int(val))
  return samples, dTree

函数getElem，percolateUp的定义如下

import random

def getElem(populationStr, dTree, key):
  msd  = int(populationStr[0])
  if not key in dTree.keys():
    dTree[key] = range(msd + 1)
  idx = random.randint(0, len(dTree[key]) - 1)
  key = key +  str(dTree[key][idx])
  if len(populationStr) == 1:
    dTree[key[:-1]].pop(idx)
    return key, (percolateUp(dTree, key[:-1]))
  newPopulation = populationStr[1:]
  if int(key[-1]) != msd:
    newPopulation = str(10**(len(newPopulation)) - 1)
  return getElem(newPopulation, dTree, key)

def percolateUp(dTree, key):
  while (dTree[key] == []):
    dTree[key[:-1]].remove( int(key[-1]) )
    key = key[:-1]
  return dTree

最后，对于较大的n值，平均时间约为15毫秒，如下所示：

In [3]: n = 10000000000000000000000000000000

In [4]: %time l,t = randomSample(n, 5)
Wall time: 15 ms

In [5]: l
Out[5]:
[10000000000000000000000000000000L,
 5731058186417515132221063394952L,
 85813091721736310254927217189L,
 6349042316505875821781301073204L,
 2356846126709988590164624736328L]

为了获得确定性，高效且使用基本编程结构构建的，生成无重复值的随机值列表的程序，请考虑以下extractSamples定义的函数，

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

基本思想是跟踪intervalLst从中选择所需元素的可能值的间隔。从确定的意义上说，这是确定性的，我们可以保证在固定数量的步骤（仅取决于populationSize和sampleSize）内生成样本。

要使用上述功能生成我们所需的列表，

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

我们还可以将其与较早的解决方案进行比较（以更低的种群数量值）

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

请注意，我降低了populationSize数值，因为使用random.sample解决方案时，它会为更高的值产生“内存错误” （也在此处和此处的先前答案中提到）。对于上述值，我们还可以观察到其extractSamples性能优于该random.sample方法。

PS：尽管核心方法与我之前的回答类似，但是在实现和方法上都进行了实质性修改，同时还提高了清晰度。

一个非常简单的功能，也可以解决您的问题

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]

"""

基于集合的方法（“如果返回值中有随机值，请重试”）的问题是，由于冲突（需要另一次“重试”迭代），不确定它们的运行时间，尤其是当返回大量随机值时从范围。

以下是不倾向于这种不确定性运行时的替代方法：

import bisect
import random

def fast_sample(low, high, num):
    """ Samples :param num: integer numbers in range of
        [:param low:, :param high:) without replacement
        by maintaining a list of ranges of values that
        are permitted.

        This list of ranges is used to map a random number
        of a contiguous a range (`r_n`) to a permissible
        number `r` (from `ranges`).
    """
    ranges = [high]
    high_ = high - 1
    while len(ranges) - 1 < num:
        # generate a random number from an ever decreasing
        # contiguous range (which we'll map to the true
        # random number).
        # consider an example with low=0, high=10,
        # part way through this loop with:
        #
        # ranges = [0, 2, 3, 7, 9, 10]
        #
        # r_n :-> r
        #   0 :-> 1
        #   1 :-> 4
        #   2 :-> 5
        #   3 :-> 6
        #   4 :-> 8
        r_n = random.randint(low, high_)
        range_index = bisect.bisect_left(ranges, r_n)
        r = r_n + range_index
        for i in xrange(range_index, len(ranges)):
            if ranges[i] <= r:
                # as many "gaps" we iterate over, as much
                # is the true random value (`r`) shifted.
                r = r_n + i + 1
            elif ranges[i] > r_n:
                break
        # mark `r` as another "gap" of the original
        # [low, high) range.
        ranges.insert(i, r)
        # Fewer values possible.
        high_ -= 1
    # `ranges` happens to contain the result.
    return ranges[:-1]

import random

sourcelist=[]
resultlist=[]

for x in range(100):
    sourcelist.append(x)

for y in sourcelist:
    resultlist.insert(random.randint(0,len(resultlist)),y)

print (resultlist)

如果希望确保添加的数字唯一，则可以使用Set对象

（如果使用2.7或更高版本），或者如果未使用，则导入sets模块。

正如其他人提到的那样，这意味着数字并不是真正的随机数。

采样整数而不在minval和之间进行替换maxval：

import numpy as np

minval, maxval, n_samples = -50, 50, 10
generator = np.random.default_rng(seed=0)
samples = generator.permutation(np.arange(minval, maxval))[:n_samples]

# or, if minval is 0,
samples = generator.permutation(maxval)[:n_samples]

与贾克斯：

import jax

minval, maxval, n_samples = -50, 50, 10
key = jax.random.PRNGKey(seed=0)
samples = jax.random.shuffle(key, jax.numpy.arange(minval, maxval))[:n_samples]

从win xp中的CLI：

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

在加拿大，我们有6/49乐透。我只是将上面的代码包装在lotto.bat中，然后运行C:\home\lotto.bat或just C:\home\lotto。

因为random.randint经常重复一个数字，所以我使用set，range(7)然后将其缩短为6。

有时，如果数字重复超过2倍，则结果列表长度将小于6。

编辑：但是，这random.sample(range(6,49),6)是正确的方法。

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