Java中间隔重复算法的开源实现 [英] Open Source implementation of a Spaced Repetition Algorithm in Java

问题描述

我从事的项目中，间隔重复是必不可少的，但是我不是该主题的专家，所以我害怕重新发明方形齿轮。我的研究为我指出了两个不同的系统，即Leitner系统和SM系列算法。

我还没有决定哪种系统最适合我的项目。如果我要学习SM，我想我会尝试实现类似于 Anki使用的方法。 / p>

我最好的选择是使用现有的Java库。可能很简单，我只需要计算下一次重复的时间即可。

有人听说过这样的倡议吗？

解决方案

我没有看过Anki的实现，但是您看到了吗？

基本上像这样

  public static void calcuateInterval（Card card）{
 if（card.getEFactor（）< 3）{
 card.setCount（1）; 
} 
 int count = card.getCount（）; 
 int间隔= 1; 
 if（count == 2）{
 interval = 6; 
}否则，如果（count> 2）{
 interval = Math.round（card.getInterval（）* card.getEFactor（））; 
} 
 card.setInterval（interval）; 
} 
  

如果您真的想要Anki的算法，请查看 Android的Anki可在Github中获得。虽然是GPL，所以您可能需要购买许可证。

I work on a project where Spaced Repetition is essential, however I am not a specialist on the subject and I am afraid to reinvent the square wheel. My research pointed me two different systems, namely the Leitner system and the SM family of algorithms.
I haven't decided yet which system would best fit into my project. If I was to take a SM orientation, I guess I would try to implement something similar to what Anki uses.

My best option would be to use an existing Java library. It could be quite simple, all I need is to compute the time for the next repetition.
Has anyone heard of such an initiative ?

解决方案

I haven't looked at Anki's implementation but have you seen this one? quiz-me an SRS in Java.

Basically it goes like this

public static void calcuateInterval(Card card) {
  if (card.getEFactor() < 3) {
      card.setCount(1);
  }
  int count = card.getCount();
  int interval = 1;
  if (count == 2) {
      interval = 6;
  } else if (count > 2) {
     interval =  Math.round(card.getInterval() * card.getEFactor());
  }
  card.setInterval(interval);
}

If you really want Anki's algorithm, look through the source of Anki in Android available in Github. It is GPL though so you might need to buy a license.

这篇关于Java中间隔重复算法的开源实现的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！