apriori算法不调用函数,手写python实现
时间: 2024-04-29 11:26:20 浏览: 9
好的,让我来回答您的问题。apriori算法是一种关联规则挖掘算法,用于发现数据中的频繁项集。以下是手写Python实现的代码:
```
def loadDataSet():
"""
加载数据集
"""
dataSet = [['葡萄酒', '尿布', '薯片'],
['尿布', '薯片'],
['葡萄酒', '尿布', '可乐'],
['葡萄酒', '尿布', '薯片', '可乐'],
['葡萄酒', '薯片']]
return dataSet
def createC1(dataSet):
"""
创建候选项集C1
"""
C1 = []
for transaction in dataSet:
for item in transaction:
if not [item] in C1:
C1.append([item])
C1.sort()
return list(map(frozenset, C1))
def scanD(D, Ck, minSupport):
"""
扫描数据集D,从Ck生成Lk
"""
ssCnt = {}
for tid in D:
for can in Ck:
if can.issubset(tid):
if not can in ssCnt:
ssCnt[can] = 1
else:
ssCnt[can] += 1
numItems = float(len(D))
retList = []
supportData = {}
for key in ssCnt:
support = ssCnt[key] / numItems
if support >= minSupport:
retList.append(key)
supportData[key] = support
return retList, supportData
def aprioriGen(Lk, k):
"""
根据Lk生成Ck+1
"""
retList = []
lenLk = len(Lk)
for i in range(lenLk):
for j in range(i+1, lenLk):
L1 = list(Lk[i])[:k-2]
L2 = list(Lk[j])[:k-2]
L1.sort()
L2.sort()
if L1 == L2:
retList.append(Lk[i] | Lk[j])
return retList
def apriori(dataSet, minSupport=0.5):
"""
apriori算法主函数
"""
C1 = createC1(dataSet)
D = list(map(set, dataSet))
L1, supportData = scanD(D, C1, minSupport)
L = [L1]
k = 2
while (len(L[k-2]) > 0):
Ck = aprioriGen(L[k-2], k)
Lk, supK = scanD(D, Ck, minSupport)
supportData.update(supK)
L.append(Lk)
k += 1
return L, supportData
```
以上代码实现了apriori算法的核心部分,包括候选项集的生成、频繁项集的挖掘、以及支持度计算等步骤。希望能够帮到您!