BCKN
BCKN

Reputation: 331

Merge two columns into one within the same dataframe

I have a question to merge two columns into one in the same dataframe(start_end), also remove null value. I intend to merge 'Start station' and 'End station' into 'station', and keep 'duration' according to the new column 'station'. I have tried pd.merge, pd.concat, pd.append, but I cannot work it out.

dataFrame of Start_end:

    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

Expected output:

    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

Code i have so far:

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

This is what I attempted to:

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

Upvotes: 15

Views: 25758

Answers (4)

sjd
sjd

Reputation: 1401

Using combine_first. replaces null values in col1 with col2

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

Upvotes: 8

wwii
wwii

Reputation: 23743

>>> 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

Give the two columns the same name

>>> 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

Stack then unstack.

>>> 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
>>> 

This is how I think this works:

.stack creates a series with a MultiIndex and takes care of the null values for you. It aligns the second level on the column names and because the column names are the same there is only one - unstacking just produces a single column.

That's really just a guess based on the differences between Index's if you don't change the column names.

>>> # 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]])

Seems a bit tricky, maybe someone will comment on it.


Alternative - Using pd.concat and .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

Upvotes: 5

BENY
BENY

Reputation: 323226

Using ffill

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

Upvotes: 1

piRSquared
piRSquared

Reputation: 294218

fillna

If NaN are truly nulls

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

If NaN are strings

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

Upvotes: 16

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