算法不放回抽样？ [英] Algorithm for sampling without replacement?

问题描述

我想测试数据的特定群集发生是偶然的可能性。鲁棒的方式来做到这一点是蒙特卡洛仿真，在其中数据和组之间的关联是随机重新分配大量的时间（如10,000），和集群的一个度量是用于实际的数据与模拟，以确定AP比较值。

I am trying to test the likelihood that a particular clustering of data has occurred by chance. A robust way to do this is Monte Carlo simulation, in which the associations between data and groups are randomly reassigned a large number of times (e.g. 10,000), and a metric of clustering is used to compare the actual data with the simulations to determine a p value.

我已经得到了大部分这方面的工作，用指针映射分组的数据元素，所以我打算随机重新分配数据指针。问题：什么是一个快速的方式，无需更换样品，让每一个指针在复制数据集的随机重新分配

I've got most of this working, with pointers mapping the grouping to the data elements, so I plan to randomly reassign pointers to data. THE QUESTION: what is a fast way to sample without replacement, so that every pointer is randomly reassigned in the replicate data sets?

例如（这些数据只是一个简单的例子）：

For example (these data are just a simplified example):

数据（N = 12个值） - A组：0.1，0.2，0.4 / B组：0.5，0.6，0.8 / C组：0.4，0.5 / D组：0.2，0.2，0.3，0.5

Data (n=12 values) - Group A: 0.1, 0.2, 0.4 / Group B: 0.5, 0.6, 0.8 / Group C: 0.4, 0.5 / Group D: 0.2, 0.2, 0.3, 0.5

对于每个复制的数据集，我有同样的簇大小（A = 3，B = 3，C = 2，D = 4）和数据值，但会重新分配值集群。

For each replicate data set, I would have the same cluster sizes (A=3, B=3, C=2, D=4) and data values, but would reassign the values to the clusters.

要做到这一点，我可以在1-12的范围产生随机数，分配A组的第一个元素，然后在1-11范围内产生随机数和A组指定的第二个元素，依此类推。指针重新分配是快，我将有pre-分配的所有数据结构，但无需更换抽样看起来可能已经解决了很多次的问题。

To do this, I could generate random numbers in the range 1-12, assign the first element of group A, then generate random numbers in the range 1-11 and assign the second element in group A, and so on. The pointer reassignment is fast, and I will have pre-allocated all data structures, but the sampling without replacement seems like a problem that might have been solved many times before.

逻辑或伪code preferred。

Logic or pseudocode preferred.

谢谢！

推荐答案

见我回答这个问题<一href="http://stackoverflow.com/questions/196017/unique-random-numbers-in-o1#196065">http://stackoverflow.com/questions/196017/unique-random-numbers-in-o1#196065.同样的逻辑也应该做到你在找什么做的。

See my answer to this question http://stackoverflow.com/questions/196017/unique-random-numbers-in-o1#196065. The same logic should accomplish what you are looking to do.

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