将 dfs 与最近的 Lon,Lat (Python, Pandas) 进行比较

Question

我有一个大的 df1 列（Lon，Lat，V1，V2，V3）和一个大的 df2（V4，V5，Lat，Lon，V6）。 dfs 坐标不完全匹配。 df2 有不同的行号。我想要： 1) 根据 (abs(df1.Lon-df2.Lon

df1:

Lon,Lat,V1,V2,V3
-94.9324,34.9099,5.0,66.9,46.6
-103.524,34.457,6.0,186.7,3.8
-92.5145,38.7823,4.0,188.7,273.5
-92.5143,37.3182,2.0,78.8,218.4
-92.5142,36.6965,5.0,98.5,27.7
-89.2187,36.4448,7.3,79.8,35.8

df2:

V4,V5,Lat,Lon,V6
20190329,10,35.0,-94.9,105.9
20180329,11,34.5,-103.5,305.9
20170329,15,38.7,-92.5,206.0
20160329,14,36.5,-89.22,402.1
20150329,13,36.7,-92.6,316.1
20140329,05,37.4,-92.5,290.0
20130329,05,33.8,-89.2,250.0

df3:

Lon,Lat,V1,V6
-94.9324,34.9099,5.0,105.9
-103.524,34.457,6.0,305.9
-92.5145,38.7823,4.0,206.0
-92.5143,37.3182,2.0,290.0
-92.5142,36.6965,5.0,316.1
-89.2187,36.4448,7.3,402.1

不同的代码不起作用：

df3 = df1.loc[~((abs(df2.Lat - df1.Lat) <= 0.11) & (abs(df2.Lon - df1.Lon) <= 0.11))]
df3 = df1.where((abs(df1[df1.Lon] - df2[df2.Lon]) <=0.11) & (abs(df1[df1.Lat] -df2[df2.Lat]) <=0.11))
df3 = pd.merge(df1, df2, on=[(abs(df1.Lon-df2.Lon)<=0.11), (abs(df1.Lat-df2.Lat)<=0.11)], how='inner')

Answer 1

这是可能的，但使用交叉连接，所以如果DataFrames 很大，需要大量内存：

df = pd.merge(df1.assign(A=1), df2.assign(A=1), on='A', how='outer', suffixes=('','_'))

cols = ['Lon','Lat','V1','V6']
df3 = df[(((df.Lat_ - df.Lat) <= 0.11).abs() & ((df.Lon_ - df.Lon).abs() <= 0.11))]
df3 = df3.drop_duplicates(subset=df1.columns)[cols]
print (df3)
         Lon      Lat   V1     V6
0   -94.9324  34.9099  5.0  105.9
8  -103.5240  34.4570  6.0  305.9
16  -92.5145  38.7823  4.0  206.0
25  -92.5143  37.3182  2.0  316.1
32  -92.5142  36.6965  5.0  316.1
38  -89.2187  36.4448  7.3  402.1