有效的算法随机选择频率项目 | 有效的算法随机选择频率项目

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问题描述

由于数组 N 字频对：

[ (w, f), (w, f), ..., (w, f) ]

其中，是W ，W 项是W P-1 。

Create an array of size m, and populate it so the first f entries are w, the next f entries are w, and so on, so the last f entries are w.

[ w, ..., w, w,..., w, ..., w, ..., w ]

然后使用PRNG在范围 0 ... M-1 选择 P 指数，并报告存储的话这些索引。
这需要 O（N + M + P）工作，这是不是很大，因为 M 可得多大于n。

Then use the pRNG to select p indices in the range 0...m-1, and report the words stored at those indices.
This takes O(n + m + p) work, which isn't great, since m can be much much larger than n.

通过输入数组步骤一次，计算

Step through the input array once, computing

m = ∑f = m + f

和计算后的 M 我 ，使用PRNG来生成许多 X K 的范围 0 ... M 我 1 每个 K 在 0 ... P-1 并选择是W 我 为是W Ĵ K （可能与更换的当前值w Ĵ K ）如果 x K ＆LT; ˚F我 。
这就要求 O（N + NP）的工作。

and after computing m, use the pRNG to generate a number x in the range 0...m-1 for each k in 0...p-1and select w for w (possibly replacing the current value of w) if x < f.
This requires O(n + np) work.

[ (w, f, m), (w, f, m), ..., (w, f, m) ]

然后，在 0 ... P-1 每个K，利用PRNG生成一个号码 X K 的范围 0 ... M-1 然后执行三元的阵列上的二进制搜索找到我 ST M 我 -f 我和乐; x K ＆LT;米我 ，然后选择是W 我 为是W Ĵ K 。
这就要求 O（N + P log n）的的工作。

and then, for each k in 0...p-1, use the pRNG to generate a number x in the range 0...m-1 then do binary search on the array of triples to find the i s.t. m-f ≤ x < m, and select w for w.
This requires O(n + p log n) work.

我的问题是，：有没有更有效的算法，我可以利用这一点，或者是这些爱在心里口难开

My question is: Is there a more efficient algorithm I can use for this, or are these as good as it gets?

推荐答案

好吧，我发现了另一种算法：的（也提到in这个答案）。基本上它创建的概率空间的划分，使得：

Ok, I found another algorithm: the alias method (also mentioned in this answer). Basically it creates a partition of the probability space such that:

有 N 分区，所有相同的宽度研究 ST NR = M 。
每个分区包含两个词语的一些比（存储与分区）。
每个字是W 我 ， F 我 =总和; 分区牛逼ST是W 我＆ISIN代码;牛逼 R＆倍;比（T，W 我）

There are n partitions, all of the same width r s.t. nr = m.
each partition contains two words in some ratio (which is stored with the partition).
for each word w, f = ∑ ∈ t r × ratio(t,w)

由于所有的分区都是相同的大小，选择哪一个分区可以在不断的工作来完成的（挑选指数 0，...，N-1 随机）和分区的比率然后可用于选择字用于在恒定工作（比较与两个词之间的比率一pRNGed数）。因此，这意味着 P 的选择可以在 0（P）工作完成后，给出了这样一个分区。

Since all the partitions are of the same size, selecting which partition can be done in constant work (pick an index from 0...n-1 at random), and the partition's ratio can then be used to select which word is used in constant work (compare a pRNGed number with the ratio between the two words). So this means the p selections can be done in O(p) work, given such a partition.

这样的划分存在的原因是，存在一个字是W 我 ST F 我＆LT; - [R ，当且仅当存在一个字是W 我' ST F 我'＆GT; ř，因为r是频率的平均值。

The reason that such a partitioning exists is that there exists a word w s.t. f < r, if and only if there exists a word w s.t. f > r, since r is the average of the frequencies.

由于这样一对是W 我 和是W 我' ，我们可以用一个伪字替换它们 W'我 频率 F我 = R （即重新presents 是W 我 的概率 F 我 / R 和是W 我' 的概率 1 - ˚F我 / R ）和一个新词 W'我' 的调整频率 F我' = F 我' - （R - ˚F我）分别。的所有字的平均频率仍然是r和从现有段的规则仍然适用。由于伪字具有频率r和由两个单词，频率＆ね; R，我们知道，如果我们重复这个过程中，我们永远不会让一个伪字出伪字，而这种迭代必须以正伪字这是所需的分区序列结束。

Given such a pair w and w we can replace them with a pseudo-word w' of frequency f' = r (that represents w with probability f/r and w with probability 1 - f/r) and a new word w' of adjusted frequency f' = f - (r - f) respectively. The average frequency of all the words will still be r, and the rule from the prior paragraph still applies. Since the pseudo-word has frequency r and is made of two words with frequency ≠ r, we know that if we iterate this process, we will never make a pseudo-word out of a pseudo-word, and such iteration must end with a sequence of n pseudo-words which are the desired partition.

要构建在 O（N）该分区时，

办理的词列表一次，构建两个列表：
- 在频率和乐字之一; - [R
- 在频率和GT字之一; - [R
- go through the list of the words once, constructing two lists:
  - one of words with frequency ≤ r
  - one of words with frequency > r
  - 如果其频率= r，则使之变成一个元素分区
  - 否则，拉动从其他列表中的单词，并用它来填写一个双字的分区。然后将第二个字回第一或第二列表，根据调整后的频率。
  这其实还是工作，如果分区数 Q＆GT; ñ（你必须要证明这一点不同）。如果你想确保r是不可或缺的，你不能很容易地找到一个因素① M 的ST Q＆GT; ñ，你可以垫所有的频率由 N ，所以 F我 = NF 我 ，其中更新 M'= MN 和集 R'= M 在①= N 。
  This actually still works if the number of partitions q > n (you just have to prove it differently). If you want to make sure that r is integral, and you can't easily find a factor q of m s.t. q > n, you can pad all the frequencies by a factor of n, so f' = nf, which updates m' = mn and sets r' = m when q = n.
  在任何情况下，这种算法只需要 O（N + P）的工作，我不得不认为这是最佳的。
  In any case, this algorithm only takes O(n + p) work, which I have to think is optimal.
  在红宝石：
```
def weighted_sample_with_replacement(input, p)
  n = input.size
  m = input.inject(0) { |sum,(word,freq)| sum + freq }

  # find the words with frequency lesser and greater than average
  lessers, greaters = input.map do |word,freq|
                        # pad the frequency so we can keep it integral
                        # when subdivided
                        [ word, freq*n ]
                      end.partition do |word,adj_freq|
                        adj_freq <= m
                      end

  partitions = Array.new(n) do
    word, adj_freq = lessers.shift

    other_word = if adj_freq < m
                   # use part of another word's frequency to pad
                   # out the partition
                   other_word, other_adj_freq = greaters.shift
                   other_adj_freq -= (m - adj_freq)
                   (other_adj_freq <= m ? lessers : greaters) << [ other_word, other_adj_freq ]
                   other_word
                 end

    [ word, other_word , adj_freq ]
  end

  (0...p).map do
    # pick a partition at random
    word, other_word, adj_freq = partitions[ rand(n) ]
    # select the first word in the partition with appropriate
    # probability
    if rand(m) < adj_freq
      word
    else
      other_word
    end
  end
end
```
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