在python pandas中实现Apriori的最佳方法

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在熊猫中实施Apriori算法的最佳方法是什么?到目前为止,我一直坚持转换使用for循环提取模式.从for循环开始的一切都不起作用.在熊猫中有没有矢量化的方法呢?

import pandas as pd
import numpy as np

trans=pd.read_table('output.txt',header=None,index_col=0)

def apriori(trans,support=4):
    ts=pd.get_dummies(trans.unstack().dropna()).groupby(level=1).sum()
    #user input

    collen,rowlen  =ts.shape

    #max length of items
    tssum=ts.sum(axis=1)
    maxlen=tssum.loc[tssum.idxmax()]

    items=list(ts.columns)

    results=[]
    #loop through items
    for c in  range(1,maxlen):
        #generate patterns
        pattern=[]
        for n in  len(pattern):
            #calculate support
            pattern=['supp']=pattern.sum/rowlen
            #filter by support level
            Condit=pattern['supp']> support
            pattern=pattern[Condit]
            results.append(pattern)
   return results

results =apriori(trans)
print results

当我插入支持3时

        a  b  c  d  e
0                    
11      1  1  1  0  0
666     1  0  0  1  1
10101   0  1  1  1  0
1010    1  1  1  1  0
414147  0  1  1  0  0
10101   1  1  0  1  0
1242    0  0  0  1  1
101     1  1  1  1  0
411     0  0  1  1  1
444     1  1  1  0  0

它应该输出类似的东西

   Pattern   support
    a         6
    b         7
    c         7
    d         7
    e         3
    a,b       5
    a,c       4
    a,d       4
最佳答案
假设我明白你所追求的是什么,也许吧

from itertools import combinations
def get_support(df):
    pp = []
    for cnum in range(1,len(df.columns)+1):
        for cols in combinations(df,cnum):
            s = df[list(cols)].all(axis=1).sum()
            pp.append([",".join(cols),s])
    sdf = pd.DataFrame(pp,columns=["Pattern","Support"])
    return sdf

会让你开始:

>>> s = get_support(df)
>>> s[s.Support >= 3]
   Pattern  Support
0        a        6
1        b        7
2        c        7
3        d        7
4        e        3
5      a,b        5
6      a,c        4
7      a,d        4
9      b,c        6
10     b,d        4
12     c,d        4
14     d,e        3
15   a,b,c        4
16   a,d        3
21   b,c,d        3

[15 rows x 2 columns]
原文链接:https://www.f2er.com/python/439686.html

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