Python 连接两个数据帧，并使用一些重叠的日期索引，结果数据帧取；“左”；默认情况下，除了；“左”；是楠吗

给出下面的DF1和DF2，如何得到DF结果

DF1:                        DF2:                        DFresult:       
Date | Value                Date | Value                Date | Value
------------                ------------                ------------
1-01-2019 | 1                                           1-01-2019 | 1       (no overlap, take the one that exists)
1-02-2019 | 1                                           1-02-2019 | 1       (no overlap, take the one that exists)
1-03-2019 | np.NaN          1-03-2019 | 2               1-03-2019 | 2       (left is NaN, take right)
1-04-2019 | 1               1-04-2019 | np.NaN          1-04-2019 | 1       (left is not NaN, take left)
1-05-2019 | np.NaN          1-05-2019 | np.NaN          1-05-2019 | np.NaN  (both NaN, keep it)
1-06-2019 | 1               1-06-2019 | 2               1-06-2019 | 1       (left is not NaN, take left)
                            1-07-2019 | 2               1-07-2019 | 2       (no overlap, take the one that exists)
                            1-08-2019 | 2               1-08-2019 | 2       (no overlap, take the one that exists)
                            1-09-2019 | 2               1-09-2019 | 2       (no overlap, take the one that exists)
                            1-10-2019 | 2               1-10-2019 | 2       (no overlap, take the one that exists)
                            1-11-2019 | 2               1-11-2019 | 2       (no overlap, take the one that exists)

如果我想使用一个函数来确定重叠决策呢？例如，如果left高于right，或者left为NaN，则以left为例：

DF1:                        DF2:                        DFresult:       
Date | Value                Date | Value                Date | Value
------------                ------------                ------------
1-01-2019 | 1                                           1-01-2019 | 1       (no overlap, take the one that exists)
1-02-2019 | 1                                           1-02-2019 | 1       (no overlap, take the one that exists)
1-03-2019 | np.NaN          1-03-2019 | 2               1-03-2019 | 2       (left is NaN, take right)
1-04-2019 | 1               1-04-2019 | np.NaN          1-04-2019 | 1       (right is NaN, take left)
1-05-2019 | np.NaN          1-05-2019 | np.NaN          1-05-2019 | np.NaN  (both NaN, keep it)
1-06-2019 | 1               1-06-2019 | 2               1-06-2019 | 2       (left is not higher, take right)
1-06-2019 | 3               1-07-2019 | 2               1-07-2019 | 3       (left is higher, take left)
1-06-2019 | 1               1-08-2019 | 2               1-08-2019 | 2       (left is not higher, take right)
                            1-09-2019 | 2               1-09-2019 | 2       (no overlap, take the one that exists)
                            1-10-2019 | 2               1-10-2019 | 2       (no overlap, take the one that exists)
                            1-11-2019 | 2               1-11-2019 | 2       (no overlap, take the one that exists)

试一试

out = pd.concat([DF1,DF2]).groupby('Date',as_index=False).max()
# for your original one 

#out = pd.concat([DF1,DF2]).groupby('Date',as_index=False).first()

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很好，谢谢！如果条件不是“max（）”或“first”，而是类似于如果DF1-DF2>4，该怎么办？