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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成两个列表。列表 a_1 将有 1300 个元素,而列表 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。当我将列表随机化后,我可以使用 1300 的切片和 200 的切片。一种方法是使用 random.shuffle,另一种方法是使用 random.sample。两种方法的随机化质量有何不同?列表 1 中的数据应该是随机样本以及列表 2 中的数据。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

使用样本

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
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6 回答 6

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洗牌的来源:

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]

样品来源:

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 line 完成的int(random() * n)。因此,底层算法本质上是相同的。

于 2012-10-09T11:01:22.447 回答
1

random.shuffle()将给定list的就地洗牌。它的长度保持不变。

random.sample()从给定序列中挑选n项目而不进行替换(也可能是元组或其他任何东西,只要它有 a __len__())并以随机顺序返回它们。

于 2012-10-09T10:58:26.287 回答
1

shuffle()sample()之间有两个主要区别:

1) Shuffle 将就地改变数据,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) Sample 让您可能做更少的工作(即部分洗牌)。

通过证明可以根据sample()实现shuffle()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]

这些在 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。

于 2017-04-26T19:43:37.907 回答
0

我认为它们完全一样,除了一个更新了原始列表,一个使用(只读)它。没有质量上的差异。

于 2012-10-09T10:57:38.063 回答
0

两个选项的随机化应该一样好。我会说 go with shuffle,因为它对读者来说更清楚它的作用。

于 2012-10-09T10:59:24.710 回答
0
from random import shuffle
from random import sample 
x = [[i] for i in range(10)]
shuffle(x)
sample(x,10)

shuffle 更新同一列表中的输出,但 sample 返回更新列表 sample 在 pic 工具中提供参数的 no,但 shuffle 提供相同长度的输入列表

于 2012-10-09T11:08:17.673 回答