Question

我有这样的数据：

foo = pd.DataFrame({'id': ['A1', 'A2', 'A3', 'A4', 'A5', 'A6', 'A7', 'A8', 'A9', 'A10'], 
                    'amount': [10, 30, 40, 15, 20, 12, 55, 45, 60, 75], 
                    'description': [u'LYFT SAN FRANCISCO CA', u'XYZ STARBUCKS MINNEAPOLIS MN', u'HOLIDAY BEMIDJI MN', 
                                    u'MCDONALDS MADISON WI', u'ABC SUPERAMERICA MI', u'SUBWAY ROCHESTER MN', 
                                    u'NNT BURGER KING WI', u'UBER TRIP CA', u'superamerica CA', u'AMAZON NY']})

foo：

    id       amount description
    A1        10    LYFT SAN FRANCISCO CA
    A2        30    XYZ STARBUCKS MINNEAPOLIS MN
    A3        40    HOLIDAY BEMIDJI MN
    A4        15    MCDONALDS MADISON WI
    A5        20    ABC SUPERAMERICA MI
    A6        12    SUBWAY ROCHESTER MN
    A7        55    NNT BURGER KING WI
    A8        45    UBER TRIP CA
    A9        60    superamerica CA
    A10       75    AMAZON NY

我想创建一个新列，该列根据description列中的关键字匹配对每个记录进行分类。

我已经通过以下方式使用了this答案中的帮助：

import re    
dict1 = {
    "LYFT" : "cab_ride",
    "UBER" : "cab_ride",
    "STARBUCKS" : "Food",
    "MCDONALDS" : "Food",
    "SUBWAY" : "Food",
    "BURGER KING" : "Food",
    "HOLIDAY" : "Gas",
    "SUPERAMERICA": "Gas"
        }

def get_category_from_desc(x):
    try:
        return next(dict1[k] for k in dict1 if re.search(k, x, re.IGNORECASE))
    except:
        return "Other"

foo['category'] = foo.description.map(get_category_from_desc)

这可行，但是我想问一下这是否是解决此问题的最佳方法。由于我拥有更多可以指示类别的关键字，因此我必须创建一个庞大的字典：

dict1 = {
        "STARBUCKS" : "Food",
        "MCDONALDS" : "Food",
        "SUBWAY" : "Food",
        "BURGER KING" : "Food",
             .
             .
             .
        # ~50 more keys for "Food"

        "HOLIDAY" : "Gas",
        "SUPERAMERICA": "Gas",
             .
             .
             .
        # ~20 more keys for "Gas"

        "WALMART" : "grocery",
        "COSTCO": "grocery",
             .
             .
        # ..... ~30 more keys for "grocery"
             .
             .
        # ~ Many more categories with a large number of keys for each
}

编辑：我还想知道是否有一种方法不需要我创建一个如上所示的庞大词典。我可以使用较小的数据结构来实现这一点吗？

dict2 = {
    "cab_ride" : ["LYFT", "UBER"], #....
    "food" : ["STARBUCKS", "MCDONALDS", "SUBWAY", "BURGER KING"], #....
    "gas" : ["HOLIDAY", "SUPERAMERICA"] #....
        }

Answer 1

我认为使用df.replace和基于正则表达式的替换可以很容易地实现这一目标。然后，您可以使用df.where处理“其他”案件。

dict2 = {rf'.*{k}.*': v for k, v in dict1.items()}

cats = foo['description'].replace(dict2, regex=True)
cats.where(cats != foo['description'], 'Other')

0    cab_ride
1        Food
2         Gas
3        Food
4         Gas
5        Food
6        Food
7    cab_ride
8       Other
9       Other
Name: description, dtype: object

另一种选择是将str.extract与map一起使用：

from collections import defaultdict

dict2 = defaultdict(lambda: 'Other')
dict2.update(dict1)

foo['description'].str.extract(rf"({'|'.join(dict1)})", expand=False).map(dict2)

0    cab_ride
1        Food
2         Gas
3        Food
4         Gas
5        Food
6        Food
7    cab_ride
8       Other
9       Other
Name: description, dtype: object

Answer 2

您可以将.str访问器与extract一起使用，并在字典键上使用join来编译正则表达式。

foo = pd.DataFrame({'id': ['A1', 'A2', 'A3', 'A4', 'A5', 'A6', 'A7', 'A8', 'A9', 'A10'], 
                    'amount': [10, 30, 40, 15, 20, 12, 55, 45, 60, 75], 
                    'description': [u'LYFT SAN FRANCISCO CA', u'XYZ STARBUCKS MINNEAPOLIS MN', u'HOLIDAY BEMIDJI MN', 
                                    u'MCDONALDS MADISON WI', u'ABC SUPERAMERICA MI', u'SUBWAY ROCHESTER MN', 
                                    u'NNT BURGER KING WI', u'UBER TRIP CA', u'superamerica CA', u'AMAZON NY']})


dict1 = {
    "LYFT" : "cab_ride",
    "UBER" : "cab_ride",
    "STARBUCKS" : "Food",
    "MCDONALDS" : "Food",
    "SUBWAY" : "Food",
    "BURGER KING" : "Food",
    "HOLIDAY" : "Gas",
    "SUPERAMERICA": "Gas"
        }

regstr = '(' + '|'.join(dict1.keys()) + ')'
foo['category'] = foo['description'].str.extract(regstr).squeeze().map(dict1).fillna('Other')
print(foo)

输出：

    id  amount                   description  category
0   A1      10         LYFT SAN FRANCISCO CA  cab_ride
1   A2      30  XYZ STARBUCKS MINNEAPOLIS MN      Food
2   A3      40            HOLIDAY BEMIDJI MN       Gas
3   A4      15          MCDONALDS MADISON WI      Food
4   A5      20           ABC SUPERAMERICA MI       Gas
5   A6      12           SUBWAY ROCHESTER MN      Food
6   A7      55            NNT BURGER KING WI      Food
7   A8      45                  UBER TRIP CA  cab_ride
8   A9      60               superamerica CA     Other
9  A10      75                     AMAZON NY     Other

使用正则表达式和字典将列添加到数据框

2 个答案: