从python中具有固定数量元素的集合中进行非常快速的采样 [英] Very fast sampling from a set with fixed number of elements in python

问题描述

我需要从固定大小的集合中随机抽样一个数字，进行一些计算，然后将新数字放回集合中. (需要的样本数量非常大)

I need to sample uniformly at random a number from a set with fixed size, do some calculation, and put the new number back into the set. (The number samples needed is very large)

我试图将数字存储在列表中，并使用random.choice()选择一个元素，将其删除，然后附加新元素.但这太慢了！

I've tried to store the numbers in a list and use random.choice() to pick an element, remove it, and then append the new element. But that's way too slow!

我正在考虑将数字存储在numpy数组中，对索引列表进行采样，并对每个索引执行计算.

I'm thinking to store the numbers in a numpy array, sample a list of indices, and for each index perform the calculation.

• 有没有更快的方法来完成此过程?

推荐答案

Python列表在内部以数组形式实现(例如Java ArrayList s，C ++ std::vector s等)，因此从中间删除元素是相对较慢:所有后续元素都必须重新编制索引. (请参见 http://www.laurentluce.com/posts/python-list-implementation/以获得更多信息.)由于元素的顺序似乎与您无关，因此建议您只使用random.randint(0, len(L) - 1)选择索引i，然后使用L[i] = calculation(L[i])进行更新第i个元素.

Python lists are implemented internally as arrays (like Java ArrayLists, C++ std::vectors, etc.), so removing an element from the middle is relatively slow: all subsequent elements have to be reindexed. (See http://www.laurentluce.com/posts/python-list-implementation/ for more on this.) Since the order of elements doesn't seem to be relevant to you, I'd recommend you just use random.randint(0, len(L) - 1) to choose an index i, then use L[i] = calculation(L[i]) to update the ith element.

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