2

按c列映射后,如果A列有值,则插入A列的值;如果不是,请插入 B 列。

data1                               data2

a    b    c                      a    c    d
a1   b1   c1                     1a   c1   1d   
     b2   c2                     2a   c2   2d
a3        c3                     3a   c3   3d
                                 4a   c4   4d

我想要的结果

  result
   a    b   c     
   a1   b1  c1
   2a   b2  c2
   a3       c3

我尝试了以下方法,但我并不满意。

->>> result = data1.merge(data2, on=['c'])
Prefixes _x and _y are created. combine_first is not applied.

->>> result = data1.combine_first(data2)
It is not mapped by column c.

我怎样才能得到好的结果?我请求你的帮助。谢谢你

4

3 回答 3

1

我不是 100% 清楚你如何索引你的数据框(data1data2),但如果你在列上索引它们'c'应该可以工作。

这就是我创建数据的方式:

import pandas as pd
data1 = pd.DataFrame({'a': ['a1', None, 'a3'],
                      'b': ['b1', 'b2', None],
                      'c': ['c1', 'c2', 'c3']})

data2 = pd.DataFrame({'a': ['1a', '2a', '3a', '4a'],
                      'c': ['c1', 'c2', 'c3', 'c4'],
                      'd': ['1d', '2d', '3d', '4d']})

然后我将两者的索引设置为 column 'c'

data1 = data1.set_index('c')
data2 = data2.set_index('c')

然后我combine_first像你一样使用:

data_combined = data1.combine_first(data_2)

我明白了:

    a   b   d
c           
c1  a1  b1  1d
c2  2a  b2  2d
c3  a3  None    3d
c4  4a  NaN 4d

不知道为什么你不想要带有 index'c4'或 column的行'd',但删除它们很容易:

data_combined = data_combined.drop('d', axis=1)
data_combined = data_combined.loc[data_combined.index != 'c4']

然后我进行一些重新排序以获得您想要的结果:

data_combined = data_combined.reset_index()
data_combined = data_combined[['a', 'b', 'c']]
data_combined = data_combined.fillna('')


    a   b   c
0   a1  b1  c1
1   2a  b2  c2
2   a3      c3
于 2018-04-07T03:47:23.910 回答
1

您也可以尝试这种方式:

# set indexes
data1 = data1.set_index('c')
data2 = data2.set_index('c')

# join data on indexes
datax = data1.join(data2.drop('d', axis=1), rsuffix='_rr').reset_index()

# fill missing value in column a
datax['a'] = datax['a'].fillna(datax['a_rr'])

# drop unwanted columns
datax.drop('a_rr', axis=1, inplace=True)

# fill missing values with blank spaces
datax.fillna('', inplace=True)

# output
    a   b   c
0   a1  b1  c1
1   2a  b2  c2
2   a3      c3

# data used
data1 = pd.DataFrame({'a':['a1','','a3'],
                      'b':['b1','b2',''],
                      'c':['c1','c2','c3']})

data2 = pd.DataFrame({'a':['1a','2a','3a','4a'],
                      'c':['c1','c2','c3','c4'],
                      'd':['1d','2d','3d','4d']})
于 2018-04-07T03:55:23.360 回答
1

使用@IdoS 设置:

import pandas as pd
data1 = pd.DataFrame({'a': ['a1', None, 'a3'],
                      'b': ['b1', 'b2', None],
                      'c': ['c1', 'c2', 'c3']})

data2 = pd.DataFrame({'a': ['1a', '2a', '3a', '4a'],
                      'c': ['c1', 'c2', 'c3', 'c4'],
                      'd': ['1d', '2d', '3d', '4d']})

您可以使用set_indexcombine_first和重新索引:

df_out = data1.set_index('c').combine_first(data2.set_index('c'))\
     .reindex(data1.c)\
     .reset_index()

df_out

输出:

    c   a     b   d
0  c1  a1    b1  1d
1  c2  2a    b2  2d
2  c3  a3  None  3d
于 2018-04-07T04:19:45.187 回答