暗原色先验图像去雾算法研究_先验算法

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暗原色先验图像去雾算法研究

Today we are going to learn about Apriori Algorithm. Before we start with that we need to know a little bit about Data Mining.

今天，我们将学习Apriori算法。 在开始之前，我们需要了解一些有关数据挖掘的知识。

**What is Data Mining ?**

**什么是数据挖掘？**

Data Mining is a non-trivial process of identifying valid, novel, potentially useful and ultimately understandable patterns in data.

数据挖掘是识别数据中有效，新颖，潜在有用且最终易于理解的模式的重要过程。

Apriori Algorithm is concerned with Data Mining and it helps us to predict information based on previous data.

Apriori算法与数据挖掘有关，它可以帮助我们根据先前的数据预测信息。

In many e-commerce websites we see a recently bought together feature or the suggestion feature after purchasing or searching for a particular item, these suggestions are based on previous purchase of that item and Apriori Algorithm can be used to make such suggestions.

在许多电子商务网站中，我们会在购买或搜索特定项目后看到最近一起购买的功能或建议功能，这些建议基于该项目的先前购买，并且可以使用Apriori算法来提出此类建议。

Before we start with Apriori we need to understand a few simple terms :

在开始使用Apriori之前，我们需要了解一些简单的术语：

**Association Mining:** It is finding different association in our data.

**关联挖掘：**正在我们的数据中找到不同的关联。

For E.g. If you are buying butter then there is a great chance that you will buy bread too so there is an association between bread and butter here.

例如，如果您要购买黄油，那么您也很有可能也会购买面包，因此这里的面包和黄油之间存在关联。

**Support:** It specifies how many of the total transactions contain these items.

**支持：**它指定总交易中有多少包含这些项目。

Support(A->B) denotes how many transactions have all items from AUB

支持(A-> B)表示从AUB获得的所有项目有多少笔交易

Therefore

因此

*  Support(A->B) = P(AUB)
    
    支持(A-> B)= P(AUB)
 *  Support(A->B) = support(B->A)
    
    支持(A-> B)=支持(B-> A)

Therefore 10% support will mean that 10% of all the transactions contain all the items in AUB.

因此，10％的支持将意味着所有交易中的10％包含AUB中的所有项目。

**Confidence:** For a transaction A->B Confidence is the number of time B is occuring when A has occurred.

**置信度：**对于事务A-> B，置信度是发生A时B发生的时间。

Note that Confidence of A->B will be different than confidence of B->A.

请注意，A-> B的置信度与B-> A的置信度不同。

Confidence(A->B) = P(AUB)/P(A).

置信度(A-> B)= P(AUB)/ P(A)。

**Support\_Count(A):** The number of transactions in which A appears.

**Support\_Count(A)：**出现A的交易数。

An itemset having number of items greater than support count is said to be frequent itemset.

具有大于支持计数的项目数的项目集被称为频繁项目集。

Apriori algorithm is used to find frequent itemset in a database of different transactions with some minimal support count. Apriori algorithm prior knowledge to do the same, therefore the name Apriori. It states that

Apriori算法用于以最少的支持次数在不同事务的数据库中查找频繁的项目集。 Apriori算法的先验知识也是如此，因此命名为Apriori。 它指出

***All subsets of a frequent itemset must be frequent.***

***频繁项目集的所有子集都必须是频繁的。***

***If an itemset is infrequent, all its supersets will be infrequent.***

***如果某个项目集不频繁，则其所有超集都将不频繁。***

Let’s go through an example :

让我们来看一个例子：

<table> 
 <tbody> 
  <tr> 
   <td><strong>Transaction ID</strong></td> 
   <td><strong>Items</strong></td> 
  </tr> 
  <tr> 
   <td>1</td> 
   <td>I1 I3 I4</td> 
  </tr> 
  <tr> 
   <td>2</td> 
   <td>I2 I3 I5</td> 
  </tr> 
  <tr> 
   <td>3</td> 
   <td>I1 I2 I3 I5</td> 
  </tr> 
  <tr> 
   <td>4</td> 
   <td>I2 I5</td> 
  </tr> 
 </tbody> 
</table>

We will first find Candidate set (denoted by Ci) which is the count of that item in Transactions.

我们将首先找到“候选集”(用C i表示)，它是“交易”中该项目的计数。

**C1:**

**C 1 ：**

<table> 
 <tbody> 
  <tr> 
   <td><strong>Items</strong></td> 
   <td><strong>Support Count</strong></td> 
  </tr> 
  <tr> 
   <td>I1</td> 
   <td>2</td> 
  </tr> 
  <tr> 
   <td>I2</td> 
   <td>3</td> 
  </tr> 
  <tr> 
   <td>I3</td> 
   <td>3</td> 
  </tr> 
  <tr> 
   <td>I4</td> 
   <td>1</td> 
  </tr> 
  <tr> 
   <td>I5</td> 
   <td>3</td> 
  </tr> 
 </tbody> 
</table>

The items whose support count is greater than or equal to a particular min support count are included in L set

L组中包含支持计数大于或等于特定最小支持计数的项目

Let’s say support count for above problem be 2

假设上述问题的支持计数为2

**L1:**

**L 1 ：**

<table> 
 <tbody> 
  <tr> 
   <td><strong>Items</strong></td> 
   <td><strong>Support Count</strong></td> 
  </tr> 
  <tr> 
   <td>I1</td> 
   <td>2</td> 
  </tr> 
  <tr> 
   <td>I2</td> 
   <td>3</td> 
  </tr> 
  <tr> 
   <td>I3</td> 
   <td>3</td> 
  </tr> 
  <tr> 
   <td>I5</td> 
   <td>3</td> 
  </tr> 
 </tbody> 
</table>

Next is the joining step we will combine the different element in L1 in order to form C2 which is candidate of size 2 then again we will go through the database and find the count of transactions having all the items. We will continue this process till we find a L set having no elements.

接下来是加入步骤，我们将合并L 1中的不同元素以形成C 2 ，它是大小为2的候选者，然后再次遍历数据库并查找具有所有项目的交易计数。 我们将继续此过程，直到找到没有元素的L集。

**C2:**

**C 2 ：**

<table> 
 <tbody> 
  <tr> 
   <td><strong>Items</strong></td> 
   <td><strong>Support Count</strong></td> 
  </tr> 
  <tr> 
   <td>I1,I2</td> 
   <td>1</td> 
  </tr> 
  <tr> 
   <td>I1,I3</td> 
   <td>2</td> 
  </tr> 
  <tr> 
   <td>I1,I5</td> 
   <td>1</td> 
  </tr> 
  <tr> 
   <td>I2,I3</td> 
   <td>2</td> 
  </tr> 
  <tr> 
   <td>I2,I5</td> 
   <td>3</td> 
  </tr> 
  <tr> 
   <td>I3,I5</td> 
   <td>2</td> 
  </tr> 
 </tbody> 
</table>

We will remove sets which have count less than min support count and form L2

我们将删除计数少于最小支持计数的集并形成L 2

**L2:**

**L 2 ：**

<table> 
 <tbody> 
  <tr> 
   <td><strong>Items</strong></td> 
   <td><strong>Support Count</strong></td> 
  </tr> 
  <tr> 
   <td>I1,I3</td> 
   <td>2</td> 
  </tr> 
  <tr> 
   <td>I2,I3</td> 
   <td>2</td> 
  </tr> 
  <tr> 
   <td>I2,I5</td> 
   <td>3</td> 
  </tr> 
  <tr> 
   <td>I3,I5</td> 
   <td>2</td> 
  </tr> 
 </tbody> 
</table>

Now we will join L2 to form C3

现在我们将加入L 2形成C 3

Note that we cannot combine \{I1,I3\} and \{I2,I5\} because then the set will contain 4 elements. The rule here is the there should be only one element in both set which are distinct all other elements should be the same.

请注意，我们不能合并\{I1，I3\}和\{I2，I5\}，因为这样集合将包含4个元素。 这里的规则是，在两个集合中都应该只有一个元素，所有其他元素都应该相同。

**C3:**

**C 3 ：**

<table> 
 <tbody> 
  <tr> 
   <td><strong>Items</strong></td> 
   <td><strong>Support Count</strong></td> 
  </tr> 
  <tr> 
   <td>I1,I2,I3</td> 
   <td>1</td> 
  </tr> 
  <tr> 
   <td>I1,I3,I5</td> 
   <td>1</td> 
  </tr> 
  <tr> 
   <td>I2,I3,I5</td> 
   <td>2</td> 
  </tr> 
 </tbody> 
</table>

**L3:**

**L 3：**

<table> 
 <tbody> 
  <tr> 
   <td><strong>Item</strong></td> 
   <td><strong>Support Count</strong></td> 
  </tr> 
  <tr> 
   <td>I2,I3,I5</td> 
   <td>2</td> 
  </tr> 
 </tbody> 
</table>

Now we cannot form C4 therefore the algorithm will terminate here.

现在我们无法形成C 4，因此算法将在此处终止。

Now we have to calculate the strong association rules. The rules having a minimum confidence are said to be strong association rules.

现在我们必须计算强关联规则。 具有最小置信度的规则被称为强关联规则。

Suppose for this example the minimum confidence be  75%.

假设此示例的最小置信度为75％。

There can be three candidates for strong association rules.

对于强关联规则，可以有三个候选者。

I2^I3->I5 = support(I2^I3)/support(I5) = ⅔  = 66.66%

I2 ^ I3-> I5 =支撑(I2 ^ I3)/支撑(I5)=⅔= 66.66％

I3^I5->I2 = support(I3^I5)/support(I2) = ⅔ = 66.66%

I3 ^ I5-> I2 =支撑(I3 ^ I5)/支撑(I2)=⅔= 66.66％

I2^I5->I3 = support(I2^I5)/support(I3) = 3/3 = 100%

I2 ^ I5-> I3 =支持(I2 ^ I5)/支持(I3)= 3/3 = 100％

So in this example the strong association rule is

因此，在此示例中，强关联规则是

I2^I5->I3.

I2 ^ I5-> I3。

So from the above example we can draw conclusion that if someone is buying I2 and I5 then he/she is most likely to buy I3 too. This is used to make suggestions while we are purchasing online.

因此，从上面的示例中我们可以得出结论，如果某人正在购买I2和I5，那么他/她也很有可能也会购买I3。 用于在我们在线购买时提出建议。

The Algorithm to calculate the frequent itemset is as below:

计算频繁项集的算法如下：

Ck : Candidate Set of size k
    Lk : Frequent set of size k
    min_sup : Minimum support count
    T : Database
     
    For all transactions t in T :
    do
    	Go through the items in t 
    		If item already present in set C1 then increase count
    		else insert item in C1 with count 1
    end
     
    For all items I in C
    do
    	If count of I > min_sup
    		L1 ← Item in C1
    end
     
    for(k=1 ; Lk!=ɸ ; k++)
    do	
    	Ck+1 ← Candidates generated from Lk
    	
    	for all transactions t in T 
    	do
    		if Ck+1 a subset of t 
    			increase count of Ck+1
    end	
    	
    For all items I in C
    	do
    	If count of I > min_sup
    			Lk+1 ← Item in Ck
    	end
    end

Comment down below if you have any queries related to Apriori Algorithm.

如果您有任何与Apriori算法有关的查询，请在下面注释。

> 翻译自: [https://www.thecrazyprogrammer.com/2019/09/apriori-algorithm.html][https_www.thecrazyprogrammer.com_2019_09_apriori-algorithm.html]

暗原色先验图像去雾算法研究

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[https_www.thecrazyprogrammer.com_2019_09_apriori-algorithm.html]: https://www.thecrazyprogrammer.com/2019/09/apriori-algorithm.html