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公司:
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个记录进行分类。你知道吗
我使用了来自thisanswer的帮助,按照以下方式进行:
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)
这是可行的,但我想问一下,这是否是解决这个问题的最佳方法。因为我有一个更大的关键字集,可以表示一个类别,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:我还想知道是否有一种方法不需要我像上面所示那样创建一个庞大的字典。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"] #....
}
可以将
.str
访问器与extract
一起使用,并在字典键上使用join
编译正则表达式。你知道吗输出:
我认为使用
df.replace
和基于regex的替换可以很容易地实现这一点。然后可以使用df.where
来处理“其他”情况。你知道吗另一个选项是将
str.extract
与map
一起使用:相关问题 更多 >
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