数据挖掘 寻找频繁项集 二重循环改进方法1 python

这是在二重循环方法的基础上的改进一
实际就相当于Apriori算法去掉剪枝步骤

import math

def read():
    f = open("data.txt", 'r')
    list1 = f.readlines()
    f.close()
    l1 = []
    for line in list1:
        x = []
        row = line.split(" ")
        for char in row:
            if char != "n" and char != "":
                num = int(char)
                x.append(num)
        if len(x) > 0:
            l1.append(x)
    return l1

def generate(lk,k):
    ckk=[]
    if k==1:
        for i in range(len(lk)):
            for j in range( i+1, len(lk) ):
                l=[lk[i],lk[j]]
                ckk.append(l)
    else:
        for i in range(len(lk)):
            for j in range( i+1, len(lk) ):
                list_union = list(set(lk[i]).union(set(lk[j])))
                if ( len(list_union)==k+1 and list_union not in ckk):
                    ckk.append(list_union)
    ckk.sort()
    return ckk

def compute_support(origin_data,ckk):
    ckk_count = []
    for i in range(len(ckk)) :
            ckk_count.append(0)
    for row in origin_data :
        for miniset in ckk :
            if ( set(miniset).issubset(row) ) :
                ckk_count[ckk.index(miniset)]+=1
                continue
    return ckk_count

def func(origin_data, support_num):
    # 先生成l1
    c1={}
    for row in origin_data:
        for num in row:
            if ( num not in c1 ):
                c1[num]=1
            else:
                c1[num]+=1
    l1 = []
    for i in c1:
        if ( c1[i]>=support_num ):
            l1.append(i)
    l1.sort()
    # 开始循环
    lk = l1
    k=1
    count = 0   # 统计该项集有多少项
    while len(lk)>0:
        # 输出频繁K项集
        print("{}项频繁集：".format(k))
        col_num = 0
        for i in lk:
            # 每输出10列就输出一个换行
            if (col_num%11==0 and col_num!=0):
                print()
                col_num=0
            if (k==1):
                print("{:>2}".format(i), end=' ')
            else:
                print(i, end=' ')
            count+=1
            col_num+=1
        print("n")
        # 获得Ck+1频繁项集的候选集
        ckk = generate(lk,k)
        # 另起一个表记录对应位置ckk每个元素出现的次数
        ckk_count = compute_support(origin_data,ckk) 
        lk.clear()
        for miniset in ckk:
            if ( ckk_count[ckk.index(miniset)]>=support_num ):
                lk.append(miniset)
        k+=1
    print("共{}项".format(count))

if __name__ == "__main__":
    support = float(input("请输入最小支持度"))
    origin_data = read()
    support_num = math.ceil( support * len(origin_data))
    print("数据行数：", len(origin_data),
        "t最小支持度为：", support, "t最小支持数：", support_num)
    func(origin_data, support_num)