所以我有一个类似的数据帧
Rank State/Union territory NSDP Per Capita (Nominal)(2019–20)[1][2] state_id
0 1 Goa 466585.0 30.0
1 2 Sikkim 425656.0 11.0
2 3 Delhi 376143.0 NaN
3 4 Chandigarh NaN 4.0
4 5 Haryana 247207.0 6.0
5 6 Telangana 225756.0 0.0
6 7 Karnataka 223246.0 29.0
7 8 Kerala 221904.0 32.0
8 9 Puducherry 220949.0 34.0
9 10 Andaman and Nicobar Islands 219842.0 NaN
10 11 Tamil Nadu 218599.0 33.0
11 12 Gujarat 216329.0 24.0
12 13 Mizoram 204018.0 15.0
13 14 Uttarakhand 202895.0 5.0
14 15 Maharashtra 202130.0 27.0
15 16 Himachal Pradesh 190255.0 2.0
16 17 Andhra Pradesh 168480.0 28.0
17 18 Arunachal Pradesh 164615.0 NaN
18 19 Punjab 161083.0 3.0
20 20 Nagaland 130282.0 13.0
21 21 Tripura 125630.0 16.0
22 22 Rajasthan 115492.0 8.0
23 23 West Bengal 115348.0 19.0
24 24 Odisha 98896.0 21.0
25 25 Chhattisgarh 105281.0 22.0
26 26 Jammu and Kashmir 102882.0 NaN
27 27 Madhya Pradesh 103288.0 23.0
28 28 Meghalaya 92174.0 17.0
29 29 Assam 90758.0 18.0
30 30 Manipur 84746.0 14.0
31 31 Jharkhand 79873.0 20.0
32 32 Uttar Pradesh 65704.0 9.0
33 33 Bihar 46664.0 10.0
我的另一本字典有
{'Telangana': 0, 'Andaman & Nicobar Island': 35, 'Andhra Pradesh': 28, 'Arunanchal Pradesh': 12, 'Assam': 18, 'Bihar': 10, 'Chhattisgarh': 22, 'Daman
& Diu': 25, 'Goa': 30, 'Gujarat': 24, 'Haryana': 6, 'Himachal Pradesh': 2, 'Jammu & Kashmir': 1, 'Jharkhand': 20, 'Karnataka': 29, 'Kerala': 32, 'Lakshadweep': 31, 'Madhya Pradesh': 23, 'Maharashtra': 27, 'Manipur': 14, 'Chandigarh': 4, 'Puducherry': 34, 'Punjab': 3, 'Rajasthan': 8, 'Sikkim': 11, 'Tamil Nadu': 33, 'Tripura': 16, 'Uttar Pradesh': 9, 'Uttarakhand': 5, 'West Bengal': 19, 'Odisha': 21, 'Dadara & Nagar Havelli': 26, 'Meghalaya': 17, 'Mizoram': 15, 'Nagaland': 13, 'NCT of Delhi': 7}
所以你可能已经看到了问题，Andaman and Nicobar Islands在两者中都存在，但拼写不同，就像字典中的' Andaman & Nicobar Island'一样。这使得最后一列NaN9 10 Andaman and Nicobar Islands 219842.0 NaN
如何将其与difflib库相结合？
我试过
df_19_20['State/Union territory'] = df_19_20['State/Union territory'].apply(get_close_matches(df_19_20['State/Union territory'], id_d.keys()))
和
df_19_20['State/Union territory'] = get_close_matches(df_19_20['State/Union territory'], id_d.keys())
我有什么东西不见了吗？如何处理列以获得最佳匹配？
问题出现在df.apply的应用程序中
CCD_ 5需要被赋予一个函数，该函数接受正在迭代的每一行的值。您还需要清除返回list的get_close_matches的返回，因此您需要获取第一个元素
df_19_20['State/Union territory'].apply(lambda x: get_close_matches(x, id_d.keys())[0])
应该工作