我想将大约国家/地区名称上的2个数据框与合并时合并,但出现以下错误:
TypeError:“ NoneType”对象不可调用
请参见下面的说明代码:
cl = {'Country' : ["Brazil", "US", "Russia"], 'BL?':['No', 'No','Yes']}
clist = pd.DataFrame.from_dict(cl)
cd = {'Country' : ["Braizl", "us", "Rusia"]}
cdata = pd.DataFrame.from_dict(cd)
clist = clist.sort_values('Country')
cdata = cdata.sort_values('Country')
cdata = pd.merge_asof(cdata,clist,on='Country')
预期结果将合并两个df,而cdata df将具有'BL?'列为是/否。
提前谢谢!
答案 0 :(得分:3)
这应该使您接近,但不会100%准确。您可以使用fuzzywuzzy。 fuzzywuzzy
使用Levenshtein距离来计算两个字符串之间的差异:
from fuzzywuzzy import process
# create a choice list
choices = clist['Country'].values.tolist()
# apply fuzzywuzzy to each row using lambda expression
cdata['Close Country'] = cdata['Country'].apply(lambda x: process.extractOne(x, choices)[0])
# merge
cdata.merge(clist, left_on='Close Country', right_on='Country')
Country_x Close Country Country_y BL?
0 Braizl Brazil Brazil No
1 Rusia Russia Russia Yes
2 us US US No
您甚至可以返回百分比匹配,并且如果您只想保持匹配度大于85%,则仅保留值> n
from fuzzywuzzy import process
# create a choice list
choices = clist['Country'].values.tolist()
# apply fuzzywuzzy to each row using lambda expression
cdata['Close Country'] = cdata['Country'].apply(lambda x: process.extractOne(x, choices))
# add percent match wiht apply
cdata[['Close Country', 'Percent Match']] = cdata['Close Country'].apply(pd.Series)
# merge
cdata.merge(clist, left_on='Close Country', right_on='Country')
Country_x Close Country Percent Match Country_y BL?
0 Braizl Brazil 83 Brazil No
1 Rusia Russia 91 Russia Yes
2 us US 100 US No
您可以在合并之前执行布尔索引,以删除不匹配项然后合并:
cdata[['Close Country', 'Percent Match']] = cdata['Close Country'].apply(pd.Series)
cdata = cdata[cdata['Percent Match']>85]
或者您可以在合并后执行此操作:
merge = cdata.merge(clist, left_on='Close Country', right_on='Country')
merge[merge['Percent Match'] > 85]
fuzzywuzzy
作为process
函数的一部分返回匹配百分比。在第一个示例中,我通过调用元组的第一个元素将其删除:process.extractOne(x, choices)[0]
答案 1 :(得分:1)
鉴于您的示例,我提出了解决方案。这不是很pythonic,但是可以用! (假设你有在匹配国家名称CREATE TRIGGER Adding_Default_Date ON students
AFTER INSERT
AS Begin
UPDATE s
SET my_column = ...
FROM student AS s
JOIN inserted AS i
ON i.key = s.key
End
对于每个clist
拼写错误国家)
cdata
输出:
def get_closest(x, column):
tmp = 1000
for i2, r2 in clist.iterrows():
levenshtein = editdistance.eval(x,r2['Country'])
if levenshtein <= tmp:
tmp = levenshtein
res = r2
return res['BL?']
cdata['BL'] = cdata['Country'].apply(lambda x: get_closest(x, clist))
我正在使用editdistance库来计算levenshtein距离。 您可以使用pip安装它:
Country BL
0 Braizl No
1 us No
2 Rusia Yes