Python实现数据挖掘十大算法之Apriori

最近刚学Python，就用Python写了一个算法，由于不熟练的原因，实现得并不好，但结果是对的，代码如下：

# -*- coding: UTF8 -*-
import sys
import copy

def init_pass(T):
    C = {}
    for t in T:
        for i in t:
            if i in C.keys():
                C[i] += 1
            else:
                C[i] = 1
    return C

def candidate_gen(F):
    C = []
    k = len(F[0]) + 1
    for f1 in F:
        for f2 in F:
            if f1[k-2] < f2[k-2]:
                c = copy.copy(f1)
                c.append(f2[k-2])
                flag = True
                for i in range(0,k-1):
                    s = copy.copy(c)
                    s.pop(i)
                    if s not in F:
                        flag = False
                        break
                if flag and c not in C:
                    C.append(c)
    return C

def compare_list(A,B):
    if len(A) <= len(B):
        for a in A:
            if a not in B:
                return False
    else:
        for b in B:
            if b not in A:
                return False
    return True
    
def apriori(T, minsup):
    C = []
    init = init_pass(T)
    keys = init.keys()
    keys.sort()
    C.append(keys)
    n = len(T)
    F = [[]]
    for f in C[0]:
        if init[f]*1.0/n >= minsup:
            F[0].append([f])
    k = 1
    while F[k-1] != []:
        C.append(candidate_gen(F[k-1]))
        F.append([])
        for c in C[k]:
            count = 0;
            for t in T:
                if compare_list(c,t):
                    count += 1
            if count*1.0/n >= minsup:
                F[k].append(c)
        k += 1
    U = []
    for f in F:
        for x in f:
            U.append(x)
    return U
T = [['A','B','C','D'],['B','C','E'],['A','B','C','E'],['B','D','E'],['A','B','C','D']]
F = apriori(T, 0.9)
print F