Question

我正在尝试创建以Pandas数据帧形式表示的文档术语矩阵。到目前为止，这是我的代码：

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

当我运行此代码时，我收到以下错误：

'list' object has no attribute 'lower'

如何摆脱这个错误？

Answer 1

将列表对象包装在str（）中以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

创建文档术语矩阵时的属性错误

1 个答案: