Apriori算法Python实现

Apriori如果数据挖掘算法的头发模式挖掘鼻祖，从60年代开始流行，该算法非常简单朴素的思维。首先挖掘长度1频繁模式，然后k=2

这些频繁模式的长度合并k频繁模式。计算它们的频繁的数目，并确保其充分k-1集长度为频繁，值是，为了避免反复。合并的时候。仅仅合并那些前k-2个字符都同样，而k-1的字符一边是少于还有一边的。

下面是算法的Python实现：

__author__ = 'linfuyuan'
min_frequency = int(raw_input('please input min_frequency:'))
file_name = raw_input('please input the transaction file:')
transactions = []


def has_infrequent_subset(candidate, Lk):
    for i in range(len(candidate)):
        subset = candidate[:-1]
        subset.sort()
        if not ''.join(subset) in Lk:
            return False
        lastitem = candidate.pop()
        candidate.insert(0, lastitem)
    return True


def countFrequency(candidate, transactions):
    count = 0
    for transaction in transactions:
        if transaction.issuperset(candidate):
            count += 1
    return count


with open(file_name) as f:
    for line in f.readlines():
        line = line.strip()
        tokens = line.split(',')
        if len(tokens) > 0:
            transaction = set(tokens)
            transactions.append(transaction)
currentFrequencySet = {}
for transaction in transactions:
    for item in transaction:
        time = currentFrequencySet.get(item, 0)
        currentFrequencySet[item] = time + 1
Lk = set()
for (itemset, count) in currentFrequencySet.items():
    if count >= min_frequency:
        Lk.add(itemset)
print ', '.join(Lk)

while len(Lk) > 0:
    newLk = set()
    for itemset1 in Lk:
        for itemset2 in Lk:
            cancombine = True
            for i in range(len(itemset1)):
                if i < len(itemset1) - 1:
                    cancombine = itemset1[i] == itemset2[i]
                    if not cancombine:
                        break
                else:
                    cancombine = itemset1[i] < itemset2[i]
                    if not cancombine:
                        break
            if cancombine:
                newitemset = []
                for char in itemset1:
                    newitemset.append(char)
                newitemset.append(itemset2[-1])
                if has_infrequent_subset(newitemset, Lk) and countFrequency(newitemset, transactions) >= min_frequency:
                    newLk.add(''.join(newitemset))
    print ', '.join(newLk)
    Lk = newLk