根据条件合并两个 pandas 数据框

pol*_*ian 4 python join pandas

目标是df如果满足预定条件,则按行组合两行。具体来说,如果列之间的差异小于或等于 a threshold,则连接 的行df

给定两个df:df1和df2,以下代码部分实现了目标。

import pandas as pd

df1 = pd.DataFrame ( {'time': [2, 3, 4, 24, 31]} )
df2 = pd.DataFrame (  {'time': [4.1, 24.7, 31.4, 5]} )
th = 0.9
all_comb=[]
for index, row in df1.iterrows ():
    for index2, row2 in df2.iterrows ():
        diff = abs ( row ['time'] - row2 ['time'] )
        if diff <= th:
            all_comb.append({'idx_1':index,'time_1':row ['time'], 'idx_2':index2,'time_2':row2 ['time']})
df_all = pd.DataFrame(all_comb)
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输出的

       idx_1  time_1  idx_2  time_2
0      2       4      0     4.1
1      3      24      1    24.7
2      4      31      2    31.4
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然而,上述方法忽略了某些信息,即 中的 2 和 3 的值df1以及 中的 5 的值df2

预期的输出应该是这样的

idx_1  time_1  idx_2  time_2

0      2       NA    NA
1      3       NA    NA    
2       4      0     4.1
3      24      1    24.7
4      31      2    31.4
NA     NA      3     5
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感谢任何提示或任何比上述建议更紧凑和更有效的方式。

ALo*_*llz 5

您可以执行交叉合并,然后根据您的条件一次性对所有行进行子集化。然后,我们concat添加回两个 DataFrame 中不满足条件的所有行。

import pandas as pd

df1 = df1.reset_index().add_suffix('_1')
df2 = df2.reset_index().add_suffix('_2')

m = df1.merge(df2, how='cross')

# Subset to all matches: |time_diff| <= thresh
th = 0.9
m = m[(m['time_1'] - m['time_2']).abs().le(th)]

# Add back rows with no matches
res = pd.concat([df1[~df1.index_1.isin(m.index_1)],
                 m,
                 df2[~df2.index_2.isin(m.index_2)]], ignore_index=True)
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print(res)
   index_1  time_1  index_2  time_2
0      0.0     2.0      NaN     NaN
1      1.0     3.0      NaN     NaN
2      2.0     4.0      0.0     4.1
3      3.0    24.0      1.0    24.7
4      4.0    31.0      2.0    31.4
5      NaN     NaN      3.0     5.0
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