根据使用频率随机生成字母? [英] Randomly Generate Letters According to their Frequency of Use?

问题描述

如何根据常用语音中的使用频率随机生成字母?

How can I randomly generate letters according to their frequency of use in common speech?

任何伪代码都值得赞赏，但在 Java 中的实现会很棒.否则，只要朝正确的方向戳一下就会有帮助.

Any pseudo-code appreciated, but an implementation in Java would be fantastic. Otherwise just a poke in the right direction would be helpful.

注意:我不需要生成使用频率 - 我相信我可以很容易地查找它.

Note: I don't need to generate the frequencies of usage - I'm sure I can look that up easily enough.

推荐答案

我假设您将频率存储为 0 到 1 之间的浮点数，总和为 1.

I am assuming that you store the frequencies as floating point numbers between 0 and 1 that total to make 1.

首先你应该准备一个累积频率表，即那个字母和它之前所有字母的频率之和.

First you should prepare a table of cumulative frequencies, i.e. the sum of the frequency of that letter and all letters before it.

为了简化，如果你从这个频率分布开始:

To simplify, if you start with this frequency distribution:

A  0.1
B  0.3
C  0.4
D  0.2

您的累积频率表将是:

A  0.1
B  0.4 (= 0.1 + 0.3)
C  0.8 (= 0.1 + 0.3 + 0.4)
D  1.0 (= 0.1 + 0.3 + 0.4 + 0.2)

现在生成一个介于 0 和 1 之间的随机数，并查看该数字在此列表中的位置.选择累积频率最小的字母大于随机数.一些例子:

Now generate a random number between 0 and 1 and see where in this list that number lies. Choose the letter that has the smallest cumulative frequency larger than your random number. Some examples:

假设您随机选择了 0.612.这介于 0.4 和 0.8 之间，即介于 B 和 C 之间，所以你会选择 C.

Say you randomly pick 0.612. This lies between 0.4 and 0.8, i.e. between B and C, so you'd choose C.

如果你的随机数是 0.039，那么它在 0.1 之前，即在 A 之前，所以选择 A.

If your random number was 0.039, that comes before 0.1, i.e. before A, so choose A.

我希望这是有道理的，否则随时要求澄清！

I hope that makes sense, otherwise feel free to ask for clarifications!

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