创建文档术语矩阵时出现属性错误

Question

我正在尝试创建以熊猫数据框的形式表示的文档术语矩阵。这是我到目前为止的代码：

df_profession['Athlete_Clean'] = df_profession['Athlete Biographies'].str.lower() 
df_profession['Athlete_Clean'] = df_profession['Athlete_Clean'].apply(lambda x: ''.join([i for i in x if not i.isdigit()])) 
df_profession['Athlete_Clean'] = df_profession['Athlete_Clean'].str.split() 
df_profession['Athlete_Clean'] = [word for word in df_profession['Athlete_Clean'] if word not in punctuation] 
df_profession['Athlete_Clean'] = [word for word in df_profession['Athlete_Clean'] if word not in stopwords.words('english')] 

profession_dtm_athlete = pandas.DataFrame(countvec.fit_transform(df_profession['Athlete_Clean']).toarray(), columns=countvec.get_feature_names(), index = df.index) 
profession_dtm_athlete 

当我运行这段代码，我得到了以下错误：

'list' object has no attribute 'lower' 

我怎样才能摆脱这种错误的？

Answer 1

包裹列表STR（）对象将它们转换为字符串：

df_profession['Athlete_Clean'] = str(df_profession['Athlete Biographies']).lower() 
df_profession['Athlete_Clean'] = df_profession['Athlete_Clean'].apply(lambda x: ''.join([i for i in x if not i.isdigit()])) 
df_profession['Athlete_Clean'] = str(df_profession['Athlete_Clean']).split() 
df_profession['Athlete_Clean'] = [word for word in df_profession['Athlete_Clean'] if word not in punctuation] 
df_profession['Athlete_Clean'] = [word for word in df_profession['Athlete_Clean'] if word not in stopwords.words('english')] 

profession_dtm_athlete = pandas.DataFrame(countvec.fit_transform(df_profession['Athlete_Clean']).toarray(), columns=countvec.get_feature_names(), index = df.index) 
profession_dtm_athlete