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']})
富:
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,根据 description
列中的关键字匹配对每条记录进行分类。
我使用了来自 this 的帮助回答按以下方式进行:
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,但我想问一下这是否是解决此问题的最佳方法。由于我有一组更大的关键字可以指示一个类别,因此我必须创建一个巨大的字典:
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:我也想知道是否有一种方法不需要我创建如上所示的庞大字典。我可以用更小的数据结构来实现这一点吗,比如:
dict2 = {
"cab_ride" : ["LYFT", "UBER"], #....
"food" : ["STARBUCKS", "MCDONALDS", "SUBWAY", "BURGER KING"], #....
"gas" : ["HOLIDAY", "SUPERAMERICA"] #....
}
最佳答案
我认为这可以很容易地使用 df.replace
和基于正则表达式的替换来实现。然后,您可以使用 df.where
来处理“其他”情况。
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
关于python - 使用正则表达式和字典将列添加到数据框,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55752715/