Python 3.1 - 大型列表采样期间的内存错误

Question

输入列表可以超过100万个数字.当我使用较小的'重复'运行以下代码时,它很好;

def sample(x):
    length = 1000000 
    new_array = random.sample((list(x)),length)
    return (new_array)

def repeat_sample(x):    
    i = 0
    repeats = 100
    list_of_samples = []
    for i in range(repeats):
       list_of_samples.append(sample(x))
    return(list_of_samples)

repeat_sample(large_array)

然而,使用诸如上述100的高重复,导致MemoryError.回溯如下;

Traceback (most recent call last):
  File "C:\Python31\rnd.py", line 221, in <module>
    STORED_REPEAT_SAMPLE = repeat_sample(STORED_ARRAY)
  File "C:\Python31\rnd.py", line 129, in repeat_sample
    list_of_samples.append(sample(x))
  File "C:\Python31\rnd.py", line 121, in sample
    new_array = random.sample((list(x)),length)
  File "C:\Python31\lib\random.py", line 309, in sample
    result = [None] * k
MemoryError

我假设我的内存不足.我不知道如何解决这个问题.

感谢您的时间!

Answer 1

扩展我的评论:

假设您对每个样本所做的处理是计算其均值.

def mean(samplelists):
    means = []
    n = float(len(samplelists[0]))
    for sample in samplelists:
        mean = sum(sample)/n
        means.append(mean)
    return means

calc_means(repeat_sample(large_array))

这会让你在内存中保留所有这些列表.你可以像这样轻松得到它:

def mean(sample, n):
    n = float(n)
    mean = sum(sample)/n
    return mean

def sample(x):
    length = 1000000 
    new_array = random.sample(x, length)
    return new_array

def repeat_means(x):    
    repeats = 100
    list_of_means = []
    for i in range(repeats):
        list_of_means.append(mean(sample(x)))
    return list_of_means    

repeat_means(large_array)

但这仍然不够好......你只需要构建你的结果列表就可以做到这一切:

import random

def sampling_mean(population, k, times):
    # Part of this is lifted straight from random.py
    _int = int
    _random = random.random

    n = len(population)
    kf = float(k)
    result = []

    if not 0 <= k <= n:
        raise ValueError, "sample larger than population"

    for t in range(times):
        selected = set()
        sum_ = 0
        selected_add = selected.add

        for i in xrange(k):
            j = _int(_random() * n)
            while j in selected:
                j = _int(_random() * n)
            selected_add(j)
            sum_ += population[j]

        mean = sum_/kf
        result.append(mean)
    return result

sampling_mean(x, 1000000, 100)

现在,您的算法可以像这样简化吗？