带有生成器/可迭代/迭代器的 Python 随机样本

Question

你知道是否有办法让 python 的 random.sample 与生成器对象一起工作。我正在尝试从一个非常大的文本语料库中获取随机样本。问题是 random.sample() 引发以下错误。

TypeError: object of type 'generator' has no len()

我在想也许有一些方法可以使用 itertools 中的某些东西来做到这一点，但通过一些搜索找不到任何东西。

一个虚构的例子:

import random
def list_item(ls):
    for item in ls:
        yield item

random.sample( list_item(range(100)), 20 )

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更新

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根据 MartinPieters 的要求，我对当前提出的三种方法进行了一些计时。结果如下。

Sampling 1000 from 10000
Using iterSample 0.0163 s
Using sample_from_iterable 0.0098 s
Using iter_sample_fast 0.0148 s

Sampling 10000 from 100000
Using iterSample 0.1786 s
Using sample_from_iterable 0.1320 s
Using iter_sample_fast 0.1576 s

Sampling 100000 from 1000000
Using iterSample 3.2740 s
Using sample_from_iterable 1.9860 s
Using iter_sample_fast 1.4586 s

Sampling 200000 from 1000000
Using iterSample 7.6115 s
Using sample_from_iterable 3.0663 s
Using iter_sample_fast 1.4101 s

Sampling 500000 from 1000000
Using iterSample 39.2595 s
Using sample_from_iterable 4.9994 s
Using iter_sample_fast 1.2178 s

Sampling 2000000 from 5000000
Using iterSample 798.8016 s
Using sample_from_iterable 28.6618 s
Using iter_sample_fast 6.6482 s

因此，当涉及到大样本时，array.insert 有一个严重的缺陷。我用来计时方法的代码

from heapq import nlargest
import random
import timeit


def iterSample(iterable, samplesize):
    results = []
    for i, v in enumerate(iterable):
        r = random.randint(0, i)
        if r < samplesize:
            if i < samplesize:
                results.insert(r, v) # add first samplesize items in random order
            else:
                results[r] = v # at a decreasing rate, replace random items

    if len(results) < samplesize:
        raise ValueError("Sample larger than population.")

    return results

def sample_from_iterable(iterable, samplesize):
    return (x for _, x in nlargest(samplesize, ((random.random(), x) for x in iterable)))

def iter_sample_fast(iterable, samplesize):
    results = []
    iterator = iter(iterable)
    # Fill in the first samplesize elements:
    for _ in xrange(samplesize):
        results.append(iterator.next())
    random.shuffle(results)  # Randomize their positions
    for i, v in enumerate(iterator, samplesize):
        r = random.randint(0, i)
        if r < samplesize:
            results[r] = v  # at a decreasing rate, replace random items

    if len(results) < samplesize:
        raise ValueError("Sample larger than population.")
    return results

if __name__ == '__main__':
    pop_sizes = [int(10e+3),int(10e+4),int(10e+5),int(10e+5),int(10e+5),int(10e+5)*5]
    k_sizes = [int(10e+2),int(10e+3),int(10e+4),int(10e+4)*2,int(10e+4)*5,int(10e+5)*2]

    for pop_size, k_size in zip(pop_sizes, k_sizes):
        pop = xrange(pop_size)
        k = k_size
        t1 = timeit.Timer(stmt='iterSample(pop, %i)'%(k_size), setup='from __main__ import iterSample,pop')
        t2 = timeit.Timer(stmt='sample_from_iterable(pop, %i)'%(k_size), setup='from __main__ import sample_from_iterable,pop')
        t3 = timeit.Timer(stmt='iter_sample_fast(pop, %i)'%(k_size), setup='from __main__ import iter_sample_fast,pop')

        print 'Sampling', k, 'from', pop_size
        print 'Using iterSample', '%1.4f s'%(t1.timeit(number=100) / 100.0)
        print 'Using sample_from_iterable', '%1.4f s'%(t2.timeit(number=100) / 100.0)
        print 'Using iter_sample_fast', '%1.4f s'%(t3.timeit(number=100) / 100.0)
        print ''

我还进行了一项测试，以检查所有方法确实确实采用了生成器的无偏样本。因此，对于所有方法，我从 10000 100000 次中采样了 1000 元素，并计算了总体中每个项目的平均出现频率，结果为是 ~.1 ，正如人们对所有三种方法所期望的那样。

Answer 1

虽然 Martijn Pieters 的答案是正确的，但当 samplesize 变大时它确实会变慢，因为在循环中使用 list.insert 可能具有二次复杂度。

在我看来，这是一种在提高性能的同时保持一致性的替代方案:

def iter_sample_fast(iterable, samplesize):
    results = []
    iterator = iter(iterable)
    # Fill in the first samplesize elements:
    try:
        for _ in xrange(samplesize):
            results.append(iterator.next())
    except StopIteration:
        raise ValueError("Sample larger than population.")
    random.shuffle(results)  # Randomize their positions
    for i, v in enumerate(iterator, samplesize):
        r = random.randint(0, i)
        if r < samplesize:
            results[r] = v  # at a decreasing rate, replace random items
    return results

对于高于 10000 的 samplesize 值，差异慢慢开始显现。 (1000000, 100000)调用次数:

• iterSample:5.05 秒
• iter_sample_fast:2.64 秒