Question

我有一个问题是在同一个数据框（start_end）中将两列合并为一个，同时删除空值。我打算将“Start station”和“End station”合并到“station”中，并根据新列“station”保持“duration”。我已经尝试过pd.merge，pd.concat，pd.append，但我无法解决它。

Start_end的dataFrame：

    Duration    End station     Start station
14  1407        NaN             14th & V St NW
19  509         NaN             21st & I St NW
20  638         15th & P St NW.  NaN
27  1532        NaN              Massachusetts Ave & Dupont Circle NW
28  759         NaN              Adams Mill & Columbia Rd NW

预期产出：

    Duration    stations
14  1407        14th & V St NW
19  509         21st & I St NW
20  638         15th & P St NW
27  1532        Massachusetts Ave & Dupont Circle NW
28  759         Adams Mill & Columbia Rd NW

我到目前为止的代码：

#start_end is the dataframe, 'start station', 'end station', 'duration'
start_end = pd.concat([df_start, df_end])

这是我试图：

station = pd.merge([start_end['Start station'],start_end['End station']])

Answer 1

`fillna`

如果NaN真的为空

df.assign(**{
    'Start station': df['Start station'].fillna(df['End station'])})

    Duration      End station                         Start station
14      1407              NaN                        14th & V St NW
19       509              NaN                        21st & I St NW
20       638  15th & P St NW.                       15th & P St NW.
27      1532              NaN  Massachusetts Ave & Dupont Circle NW
28       759              NaN           Adams Mill & Columbia Rd NW

`mask`

如果NaN是字符串

df.assign(**{
    'Start station': df['Start station'].mask(
        lambda x: x == 'NaN', df['End station'])})

    Duration      End station                         Start station
14      1407              NaN                        14th & V St NW
19       509              NaN                        21st & I St NW
20       638  15th & P St NW.                       15th & P St NW.
27      1532              NaN  Massachusetts Ave & Dupont Circle NW
28       759              NaN           Adams Mill & Columbia Rd NW

Answer 2

使用ffill

df.iloc[:,2:4]=df.iloc[:,2:4].ffill(1)

Answer 3

>>> df
   Duration      End station                         Start station
0      1407              NaN                        14th & V St NW
1       509              NaN                        21st & I St NW
2       638  15th & P St NW.                                   NaN
3      1532              NaN  Massachusetts Ave & Dupont Circle NW
4       759              NaN           Adams Mill & Columbia Rd NW

为两列提供相同的名称

>>> df.columns = df.columns.str.replace('.*?station', 'station')
>>> df
   Duration          station                               station
0      1407              NaN                        14th & V St NW
1       509              NaN                        21st & I St NW
2       638  15th & P St NW.                                   NaN
3      1532              NaN  Massachusetts Ave & Dupont Circle NW
4       759              NaN           Adams Mill & Columbia Rd NW

然后堆叠然后堆叠。

>>> s = df.stack()
>>> s
0  Duration                                    1407
   station                           14th & V St NW
1  Duration                                     509
   station                           21st & I St NW
2  Duration                                     638
   station                          15th & P St NW.
3  Duration                                    1532
   station     Massachusetts Ave & Dupont Circle NW
4  Duration                                     759
   station              Adams Mill & Columbia Rd NW
dtype: object
>>> df = s.unstack()
>>> df
  Duration                               station
0     1407                        14th & V St NW
1      509                        21st & I St NW
2      638                       15th & P St NW.
3     1532  Massachusetts Ave & Dupont Circle NW
4      759           Adams Mill & Columbia Rd NW
>>>

这就是我认为的工作原理：

.stack创建一个包含MultiIndex的系列，并为您处理空值。它对齐列名称的第二级，因为列名相同，只有一个 - unstacking只生成一个列。

如果你不改变列名，这只是基于Index之间差异的猜测。

>>> # without changing column names
>>> s.index
MultiIndex(levels=[[0, 1, 2, 3, 4], ['Duration', 'End station', 'Start station']],
           labels=[[0, 0, 1, 1, 2, 2, 3, 3, 4, 4], [0, 2, 0, 2, 0, 1, 0, 2, 0, 2]])

>>> # column names the same
>>> s.index
MultiIndex(levels=[[0, 1, 2, 3, 4], ['Duration', 'station']],
           labels=[[0, 0, 1, 1, 2, 2, 3, 3, 4, 4], [0, 1, 0, 1, 0, 1, 0, 1, 0, 1]])

似乎有点棘手，也许有人会评论它。

替代方案 - 使用pd.concat和.dropna

>>> stations = pd.concat([df.iloc[:,1],df.iloc[:,2]]).dropna()
>>> stations.name = 'stations'
>>> stations
2                         15th & P St NW.
0                          14th & V St NW
1                          21st & I St NW
3    Massachusetts Ave & Dupont Circle NW
4             Adams Mill & Columbia Rd NW
Name: stations, dtype: object

>>> df2 = pd.concat([df['Duration'], stations], axis=1)
>>> df2
   Duration                              stations
0      1407                        14th & V St NW
1       509                        21st & I St NW
2       638                       15th & P St NW.
3      1532  Massachusetts Ave & Dupont Circle NW
4       759           Adams Mill & Columbia Rd NW

Answer 4

使用 combine_first。用 col2

替换 col1 中的空值

df["station"] = df["End station"].combine_first(df["Start station"])
df.drop(["End station", "Start station"], 1, inplace=True)

在pandas / python中的同一数据框中将两列合并为一列

4 个答案:

`fillna`

`mask`