订单最频繁的模式(Most frequent pattern with an order)

订单最频繁的模式(Most frequent pattern with an order)

我正在寻找像Apriori这样的算法，但是需要订购。

我需要一个算法来找到最频繁的模式。 例如：ABCDE ABC ACBD DECF

最常见的模式：ABC

顺序非常重要。 ACB不应该被看作是因为ACB在我的例子中只有一次，但ABC是三次。

有没有算法，或者是我自己编写代码的最佳解决方案？

感谢帮助。

i'm searching for an algorithm like Apriori but with order.

I need an algorithm to find most frequent pattern. Example: A B C D E A B C A C B D D E C F

the most frequent pattern: A B C

The order is very important. A C B should not be viewed because A C B is only one times in my example, but A B C three times.

Is there an algorithm or is the best solution to code it by my own?

Thanks for help.

最满意答案

您可以扫描序列并将其存储在map

String sequence = "ABCDEABCACBDDECF"; char[] chars = sequence.toCharArray(); int i = 0; Map<String,Integer> map = new HashMap<String,Integer>(); while(i < chars.length - 2){ String pattern = "" + chars[i] + chars[i+1] + chars[i+2]; System.out.println(pattern); Integer population = map.get(pattern); if(population == null){ population = 1; }else{ population++; } map.put(pattern, population); i++; } System.out.println(map);

输出是

{ACB=1, DEA=1, BCA=1, ABC=2, BCD=1, BDD=1, DEC=1, CAC=1, CBD=1, DDE=1, CDE=1, EAB=1, ECF=1}

您可以轻松地按值排序地图： http : //www.programcreek.com/2013/03/java-sort-map-by-value/或者只是扫描地图寻找最高人口

顺便说一句：模式ABC人口其实是2而不是3

You can scan the sequence and store populations in a map

String sequence = "ABCDEABCACBDDECF"; char[] chars = sequence.toCharArray(); int i = 0; Map<String,Integer> map = new HashMap<String,Integer>(); while(i < chars.length - 2){ String pattern = "" + chars[i] + chars[i+1] + chars[i+2]; System.out.println(pattern); Integer population = map.get(pattern); if(population == null){ population = 1; }else{ population++; } map.put(pattern, population); i++; } System.out.println(map);

The output is

{ACB=1, DEA=1, BCA=1, ABC=2, BCD=1, BDD=1, DEC=1, CAC=1, CBD=1, DDE=1, CDE=1, EAB=1, ECF=1}

You can easily sort the map by values : http://www.programcreek.com/2013/03/java-sort-map-by-value/ or just scan the map looking for the highest population

By the way: population of pattern ABC actually is 2 not 3

订单最频繁的模式(Most frequent pattern with an order)

我正在寻找像Apriori这样的算法，但是需要订购。

我需要一个算法来找到最频繁的模式。 例如：ABCDE ABC ACBD DECF

最常见的模式：ABC

顺序非常重要。 ACB不应该被看作是因为ACB在我的例子中只有一次，但ABC是三次。

有没有算法，或者是我自己编写代码的最佳解决方案？

感谢帮助。

i'm searching for an algorithm like Apriori but with order.

I need an algorithm to find most frequent pattern. Example: A B C D E A B C A C B D D E C F

the most frequent pattern: A B C

The order is very important. A C B should not be viewed because A C B is only one times in my example, but A B C three times.

Is there an algorithm or is the best solution to code it by my own?

Thanks for help.

最满意答案

您可以扫描序列并将其存储在map

String sequence = "ABCDEABCACBDDECF"; char[] chars = sequence.toCharArray(); int i = 0; Map<String,Integer> map = new HashMap<String,Integer>(); while(i < chars.length - 2){ String pattern = "" + chars[i] + chars[i+1] + chars[i+2]; System.out.println(pattern); Integer population = map.get(pattern); if(population == null){ population = 1; }else{ population++; } map.put(pattern, population); i++; } System.out.println(map);

输出是

{ACB=1, DEA=1, BCA=1, ABC=2, BCD=1, BDD=1, DEC=1, CAC=1, CBD=1, DDE=1, CDE=1, EAB=1, ECF=1}

您可以轻松地按值排序地图： http : //www.programcreek.com/2013/03/java-sort-map-by-value/或者只是扫描地图寻找最高人口

顺便说一句：模式ABC人口其实是2而不是3

You can scan the sequence and store populations in a map

String sequence = "ABCDEABCACBDDECF"; char[] chars = sequence.toCharArray(); int i = 0; Map<String,Integer> map = new HashMap<String,Integer>(); while(i < chars.length - 2){ String pattern = "" + chars[i] + chars[i+1] + chars[i+2]; System.out.println(pattern); Integer population = map.get(pattern); if(population == null){ population = 1; }else{ population++; } map.put(pattern, population); i++; } System.out.println(map);

The output is

{ACB=1, DEA=1, BCA=1, ABC=2, BCD=1, BDD=1, DEC=1, CAC=1, CBD=1, DDE=1, CDE=1, EAB=1, ECF=1}

You can easily sort the map by values : http://www.programcreek.com/2013/03/java-sort-map-by-value/ or just scan the map looking for the highest population

By the way: population of pattern ABC actually is 2 not 3