Question

数据集：

df['bigram'] = df['Clean_Data'].apply(lambda row: list(ngrams(word_tokenize(row), 2)))
df[:,0:1]

Id       bigram
1952043  [(Swimming,Pool),(Pool,in),(in,the),(the,roof),(roof,top),
1918916  [(Luxury,Apartments),(Apartments,consisting),(consisting,11),
1645751  [(Flat,available),(available,sale),(sale,Medavakkam),
1270503  [(Toddler,Pool),(Pool,with),(with,Jogging),(Jogging,Tracks),
1495638  [(near,medavakkam),(medavakkam,junction),(junction,calm),

我有一个python文件（Categories.py），其中包含属性/ Land功能的无监督分类。

category = [('Luxury Apartments', 'IN', 'Recreation_Ammenities'),
        ('Swimming Pool', 'IN','Recreation_Ammenities'),
        ('Toddler Pool', 'IN', 'Recreation_Ammenities'),
        ('Jogging Tracks', 'IN', 'Recreation_Ammenities')]
Recreation = [e1 for (e1, rel, e2) in category if e2=='Recreation_Ammenities']

从bigram列和类别列表中找到匹配的单词：

tokens=pd.Series(df["bigram"])
Lid=pd.Series(df["Id"])
matches = tokens.apply(lambda x: pd.Series(x).str.extractall("|".join(["({})".format(cat) for cat in Categories.Recreation])))

运行上面的代码时，我收到此错误：

AttributeError: Can only use .str accessor with string values, which use np.object_ dtype in pandas

需要帮助。

我想要的输出是：

 Id       bigram                                  Recreation_Amenities
1952043  [(Swimming,Pool),(Pool,in),(in,the),..   Swimming Pool
1918916  [(Luxury,Apartments),(Apartments,..      Luxury Apartments
1645751  [(Flat,available),(available,sale)..     
1270503  [(Toddler,Pool),(Jogging,Tracks)..      Toddler Pool,Jogging Tracks
1495638  [(near,medavakkam),..

Answer 1

这些方面的某些内容对您有用：

<div class="row">
    <div class="col.col-md-10 .col-md-offset-1">
        <h1 class="headline">Pardon Our Dust! We Are Building Something Great!</h1>
    </div>
</div>

每个二元组都由一个空格连接，以便可以测试该二元组是否包含在您的类别列表中（即<div class="row"> <div class="col"> </div> <div class=""> <h1 class="headline">Pardon Our Dust! We Are Building Something Great!</h1> </div> <div class="col"> </div> </div>）。

Answer 2

你可以按空格加入元组，然后使用双列表理解找到娱乐中出现的单词并应用即

df['Recreation_Amenities'] = df['bigram'].apply(lambda x : [j for j in Recreation if j in  [' '.join(i) for i in x]])

让我们考虑您有一个数据框

    Id      bigram
0   1270503 [(Toddler, Pool), (Pool, with), (with, Jogging), (Jogging, Tracks)]
1   1952043 [(Swimming, Pool), (Pool, in), (in, the), (the, roof), (roof, top)]
2   1918916 [(Luxury, Apartments), (Apartments, consisting), (consisting, 11)]
3   1495638 [(near, medavakkam), (medavakkam, junction), (junction, calm)]
4   1645751 [(Flat, available), (available, sale), (sale, Medavakkam)]

你有名单娱乐，即

Recreation = ['Luxury Apartments', 'Swimming Pool', 'Toddler Pool', 'Jogging Tracks']

然后

df['Recreation_Amenities'] = df['bigram'].apply(lambda x : [j for j in Recreation if j in  [' '.join(i) for i in x]])

输出：df['Recreation_Amenities']


0    [Toddler Pool, Jogging Tracks]
1    [Swimming Pool]               
2    [Luxury Apartments]           
3    []                            
4    []                            
Name: Recreation_Amenities, dtype: object

找到与ngrams匹配的单词

2 个答案: