python - 如何从 Python 中具有两个或多个字符串值的冻结集中获取字符串值？

Question

import pandas as pd
from apyori import apriori
from collections import defaultdict
ds=pd.read_csv('event.csv',header=None)
num_records=len(ds)
print(num_records)
records=[]
for i in range(0,num_records):
    records.append([str(ds.values[i,j])for j in range(0,6)])
association_rule=apriori(records,min_support=0.01,min_confidence=0.50,min_lift=8,min_length=2)
association_Results=list(association_rule)
print(association_Results)
results=[]
for item in association_Results:
        pair = item[0]
       items=[x for x in pair]
       value0=str(item[0])
       value1=str(item[1])
       value2=str(item[1])[:7]
       value3=str(item[2][0][2])[:7]
       rows=[value0,value1,value2,value3]
       results.append(rows)
       #print(value0)
       #print(results)
       label=['Tracks','Support','confidence','Lift']
       Event_suggestion =pd.DataFrame.from_records(results,columns=label)
       print(Event_suggestion)
k=Event_suggestion.iloc[0]['Tracks']
print(next(value for value in k))

结果：1001 Tracks ... Lift 0 freezeset({'ART', 'Digital Marketing'}) ... 1.0

[1 rows x 4 columns]
                                    Tracks  ...     Lift
0  frozenset({'ART', 'Digital Marketing'})  ...      1.0
1    frozenset({'IOT', 'Public Speaking'})  ...  0.83333

[2 rows x 4 columns]
                                             Tracks  ...     Lift
0           frozenset({'ART', 'Digital Marketing'})  ...      1.0
1             frozenset({'IOT', 'Public Speaking'})  ...  0.83333
2  frozenset({'IOT', 'Public Speaking', 'Finance'})  ...   0.6875

[3 rows x 4 columns]
f
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** 此代码是使用 Apriori 算法建议的事件。它从事件中获取轨道，看看什么是频繁项或项并建议 **

标签： pythonstringfrozenset