在Python中random.sample和random.shuffle有什么区别

Question

我有一个包含 1500 个元素的列表a_tot，我想以随机方式将此列表分为两个列表。列表a_1将有 1300 个元素，列表a_2将有 200 个元素。我的问题是关于用 1500 个元素随机化原始列表的最佳方法。当我随机化列表时，我可以拿一个切片和 1300 个切片和另一个 200 切片。一种方法是使用 random.shuffle，另一种方法是使用 random.sample。两种方法之间的随机化质量有什么差异吗？列表 1 中的数据以及列表 2 中的数据应该是随机样本。有什么建议吗？使用随机播放：

random.shuffle(a_tot)    #get a randomized list
a_1 = a_tot[0:1300]     #pick the first 1300
a_2 = a_tot[1300:]      #pick the last 200

使用示例

new_t = random.sample(a_tot,len(a_tot))    #get a randomized list
a_1 = new_t[0:1300]     #pick the first 1300
a_2 = new_t[1300:]      #pick the last 200

Answer 1

shuffle 的来源：

def shuffle(self, x, random=None, int=int):
    """x, random=random.random -> shuffle list x in place; return None.
    Optional arg random is a 0-argument function returning a random
    float in [0.0, 1.0); by default, the standard random.random.
    """
    if random is None:
        random = self.random
    for i in reversed(xrange(1, len(x))):
        # pick an element in x[:i+1] with which to exchange x[i]
        j = int(random() * (i+1))
        x[i], x[j] = x[j], x[i]

示例的来源：

def sample(self, population, k):
    """Chooses k unique random elements from a population sequence.
    Returns a new list containing elements from the population while
    leaving the original population unchanged.  The resulting list is
    in selection order so that all sub-slices will also be valid random
    samples.  This allows raffle winners (the sample) to be partitioned
    into grand prize and second place winners (the subslices).
    Members of the population need not be hashable or unique.  If the
    population contains repeats, then each occurrence is a possible
    selection in the sample.
    To choose a sample in a range of integers, use xrange as an argument.
    This is especially fast and space efficient for sampling from a
    large population:   sample(xrange(10000000), 60)
    """
    # XXX Although the documentation says `population` is "a sequence",
    # XXX attempts are made to cater to any iterable with a __len__
    # XXX method.  This has had mixed success.  Examples from both
    # XXX sides:  sets work fine, and should become officially supported;
    # XXX dicts are much harder, and have failed in various subtle
    # XXX ways across attempts.  Support for mapping types should probably
    # XXX be dropped (and users should pass mapping.keys() or .values()
    # XXX explicitly).
    # Sampling without replacement entails tracking either potential
    # selections (the pool) in a list or previous selections in a set.
    # When the number of selections is small compared to the
    # population, then tracking selections is efficient, requiring
    # only a small set and an occasional reselection.  For
    # a larger number of selections, the pool tracking method is
    # preferred since the list takes less space than the
    # set and it doesn't suffer from frequent reselections.
    n = len(population)
    if not 0 <= k <= n:
        raise ValueError, "sample larger than population"
    random = self.random
    _int = int
    result = [None] * k
    setsize = 21        # size of a small set minus size of an empty list
    if k > 5:
        setsize += 4 ** _ceil(_log(k * 3, 4)) # table size for big sets
    if n <= setsize or hasattr(population, "keys"):
        # An n-length list is smaller than a k-length set, or this is a
        # mapping type so the other algorithm wouldn't work.
        pool = list(population)
        for i in xrange(k):         # invariant:  non-selected at [0,n-i)
            j = _int(random() * (n-i))
            result[i] = pool[j]
            pool[j] = pool[n-i-1]   # move non-selected item into vacancy
    else:
        try:
            selected = set()
            selected_add = selected.add
            for i in xrange(k):
                j = _int(random() * n)
                while j in selected:
                    j = _int(random() * n)
                selected_add(j)
                result[i] = population[j]
        except (TypeError, KeyError):   # handle (at least) sets
            if isinstance(population, list):
                raise
            return self.sample(tuple(population), k)
    return result

如您所见，在这两种情况下，随机化基本上都是由行int(random() * n)完成的。 因此，底层算法本质上是相同的。

Answer 2

shuffle（） 和 sample（） 之间有两个主要区别：

1） 随机播放将就地更改数据，因此其输入必须是可变序列。 相比之下，sample 生成一个新列表，其输入可以更加多样化（元组、字符串、xrange、字节数组、集合等）。

2）样本可以让你做更少的工作（即部分洗牌）。

通过

演示在 sample（） 方面实现 shuffle（） 是可能的，来展示两者之间的概念关系是很有趣的：

def shuffle(p):
   p[:] = sample(p, len(p))

反之亦然，根据 shuffle（） 实现 sample（）：

def sample(p, k):
   p = list(p)
   shuffle(p)
   return p[:k]

它们

在shuffle（）和sample（）的实际实现中都没有那么有效，但它确实显示了它们的概念关系。

Answer 3

随机化应该与这两个选项一样好。我会说shuffle，因为读者更直接地清楚它的作用。

Answer 4

>random.shuffle()就地打乱给定的list。它的长度保持不变。

random.sample()从给定序列中挑选n项而不进行替换（也可能是元组或其他什么，只要它有一个__len__()），并以随机顺序返回它们。

Answer 5

我认为它们完全相同，除了一个更新了原始列表，一个使用（只读）它。质量没有差异。

Answer 6

from random import shuffle
from random import sample 
x = [[i] for i in range(10)]
shuffle(x)
sample(x,10)

随机更新同一列表中的输出，但示例返回更新 列表示例在 PIC 工具中提供参数编号，但随机播放 提供相同长度输入的列表