所以我有一个这样的数据框
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
我的另一本字典也有
{'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}
因此,您可能已经看到了问题,Andaman and Nicobar Islands
在这两种语言中都存在,但拼写不同,就像词典中的' Andaman & Nicobar Island'
。
这使得最后一列成为NaN
9 10 Andaman and Nicobar Islands 219842.0 NaN
如何将其与difflib库相结合
我试过了
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()))
及
df_19_20['State/Union territory'] = get_close_matches(df_19_20['State/Union territory'], id_d.keys())
有什么我遗漏的吗?如何处理列以获得最佳匹配
问题在于
df.apply
的应用df.apply
需要给定一个函数,该函数从它所迭代的每一行中获取值。您还需要清除返回get_close_matches
的返回,它返回一个list
,因此需要获取第一个元素应该有用
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