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我有两个熊猫数据框，我想在一个（日期）列上精确地加入/合并多个列（比如 3）和大约，即最近的邻居。我还想返回它们之间的差异（天数）。每个数据集大约有 50,000 行长。我对内部连接最感兴趣，但如果不太难掌握，“剩余”也很有趣。大多数“完全匹配”观察结果将在每个数据帧中多次存在。

我一直在尝试将difflib.get_close_matches用作字符串的所有连接（这很愚蠢，我知道！）但并不总是给出完全匹配的。我想我需要先遍历确切的匹配项，然后在该组中找到最近的匹配项，但我似乎无法做到正确......

数据框看起来像：

df1 = pd.DataFrame({'index': ['a1','a2','a3','a4'], 'col1': ['1232','432','432','123'], 'col2': ['asd','dsa12','dsa12','asd2'], 'col3': ['1','2','2','3'], 'date': ['2010-01-23','2016-05-20','2010-06-20','2008-10-21'],}).set_index('index')

df1
Out[430]: 
       col1   col2 col3        date
index                              
a1     1232    asd    1  2010-01-23
a2      432  dsa12    2  2016-05-20
a3      432  dsa12    2  2010-06-20
a4      123   asd2    3  2008-10-21

df2 = pd.DataFrame({'index': ['b1','b2','b3','b4'], 'col1': ['132','432','432','123'], 'col2': ['asd','dsa12','dsa12','sd2'], 'col3': ['1','2','2','3'], 'date': ['2010-01-23','2016-05-23','2010-06-10','2008-10-21'],}).set_index('index')

df2
Out[434]: 
      col1   col2 col3        date
index                             
b1     132    asd    1  2010-01-23
b2     432  dsa12    2  2016-05-23
b3     432  dsa12    2  2010-06-10
b4     123    sd2    3  2008-10-21

最后我想要类似的东西：

       col1   col2 col3        date diff match_index
index                              
a1     1232    asd    1  2010-01-23  nan         nan
a2      432  dsa12    2  2016-05-20   -3          b2
a3      432  dsa12    2  2010-06-20   10          b3
a4      123   asd2    3  2008-10-21  nan         nan
a5      123    sd2    3  2008-10-21  nan          b4

或者如果只使用内部连接更容易，我想：

       col1   col2 col3        date diff match_index
index                                                     
a2      432  dsa12    2  2016-05-20   -3          b2
a3      432  dsa12    2  2010-06-20   10          b3
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1 回答 1

2

我不确定这是否适合。它或多或少地实现了您想要的，但实际上并没有执行合并。它遵循与此问题相同的想法，除了不是df1仅基于一列groupby对 如果您确实想明确包含该merge命令并且对内部连接感到满意，请检查答案的最底部，它包含一个片段。

import pandas as pd
from sklearn.neighbors import NearestNeighbors

    def find_nearest(group, df2, groupname):
        try:
            match = df2.groupby(groupname).get_group(group.name)
            match['date'] = pd.to_datetime(match.date, unit = 'D')
            nbrs = NearestNeighbors(1).fit(match['date'].values[:, None])
            dist, ind = nbrs.kneighbors(group['date'].values[:, None])

            group['date1'] = group['date']
            group['date'] = match['date'].values[ind.ravel()]
            group['diff'] = (group['date1']-group['date'])
            group['match_index'] = match.index[ind.ravel()]
            return group
        except KeyError:
            return group

    #change dates from string to datetime
    df1['date'] = pd.to_datetime(df1.date, unit = 'D')
    df2['date'] = pd.to_datetime(df2.date, unit = 'D')

    #find closest dates and differences
    keys = ['col1', 'col2', 'col3']
    df1_mod = df1.groupby(keys).apply(find_nearest, df2, keys)

    #fill unmatched dates 
    df1_mod.date1.fillna(df1_mod.date, inplace=True)

    df2_mod = df2.groupby(keys).apply(find_nearest, df1, keys) 
    df2_mod.date1.fillna(df2_mod.date, inplace=True)

    #drop original column 
    df1_mod.drop('date', inplace=True, axis=1)
    df1_mod.rename(columns = {'date1':'date'}, inplace=True)

    df2_mod.drop('date', inplace=True, axis=1)
    df2_mod.rename(columns = {'date1':'date'}, inplace=True)
    df2_mod['diff'] = -df2_mod['diff']

    #drop redundant values
    df2_mod.drop(df2_mod[df2_mod.match_index.str.len()>0].index, inplace=True)

    #merge the two 
    df_final = pd.merge(df1_mod, df2_mod, how='outer')

这会产生以下结果：

In [349]: df_final
Out[349]:
   col1   col2 col3       date    diff match_index
0  1232    asd    1 2010-01-23     NaT         NaN
1   432  dsa12    2 2016-05-20 -3 days          b2
2   432  dsa12    2 2010-06-20 10 days          b3
3   123   asd2    3 2008-10-21     NaT         NaN
4   132    asd    1 2010-01-23     NaT         NaN
5   123    sd2    3 2008-10-21     NaT         NaN

使用合并命令：

In [208]: pd.merge(df1_mod, df2.drop('date', axis=1), on=['col1', 'col2', 'col3']).drop_duplicates()
Out[208]:
  col1   col2 col3       date    diff match_index
0  432  dsa12    2 2016-05-20 -3 days          b2
2  432  dsa12    2 2010-06-20 10 days          b3

评论中考虑的案例，即：

df1 = pd.DataFrame({'index': ['a1','a2','a3','a4'], 'col1': ['1232','1432','432','123'], 'col2': ['asd','dsa12','dsa12','asd2'], 'col3': ['1','2','2','3'], 'date': ['2010-01-23','2016-05-20','2010-06-20','2008-10-21'],}).set_index('index')

产生以下结果：

In [351]: df_final
Out[351]:
   col1   col2 col3       date    diff match_index
0  1232    asd    1 2010-01-23     NaT         NaN
1  1432  dsa12    2 2016-05-20     NaT         NaN
2   432  dsa12    2 2010-06-20 10 days          b3
3   123   asd2    3 2008-10-21     NaT         NaN
4   132    asd    1 2010-01-23     NaT         NaN
5   123    sd2    3 2008-10-21     NaT         NaN
于 2016-05-31T15:29:22.117 回答