随机生成信根据其使用频率？ [英] Randomly Generate Letters According to their Frequency of Use?

问题描述

我怎么能随意根据普通话使用它们的频率产生信吗？

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

任何伪code AP preciated，但在Java的实现将是非常美妙。否则，在正确的方向只是一捅将是有益的。

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

请注意：我并不需要生成使用的频率 - 我敢肯定，我可以看看，多达很轻松地

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

我假设你存储的频率为浮点数字0和1之间，道达尔，使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prepare累积频率表，也就是这封信的频率面前的总和，所有字母。

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

要简化，如果你开始用这种频率分布：

To simplify, if you start with this frequency distribution:

A  0.1
B  0.3
C  0.4
D  0.2

您累积频率表将是：

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0和1之间的随机数，并看到在该列表中该号码的所在。选择具有最小累积频率比你的随机数较大的信。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:

假设你随机挑选0.612。这个位于0.4和0.8，即B和C之间，所以你选择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.

如果您的随机数为0.039，自带0.1之前，即过，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，否则随时要求进行澄清！

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

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