问题
I have a Pandas series sf:
email
email1@email.com [1.0, 0.0, 0.0]
email2@email.com [2.0, 0.0, 0.0]
email3@email.com [1.0, 0.0, 0.0]
email4@email.com [4.0, 0.0, 0.0]
email5@email.com [1.0, 0.0, 3.0]
email6@email.com [1.0, 5.0, 0.0]
And I would like to transform it to the following DataFrame:
index | email | list
_____________________________________________
0 | email1@email.com | [1.0, 0.0, 0.0]
1 | email2@email.com | [2.0, 0.0, 0.0]
2 | email3@email.com | [1.0, 0.0, 0.0]
3 | email4@email.com | [4.0, 0.0, 0.0]
4 | email5@email.com | [1.0, 0.0, 3.0]
5 | email6@email.com | [1.0, 5.0, 0.0]
I found a way to do it, but I doubt it\'s the more efficient one:
df1 = pd.DataFrame(data=sf.index, columns=[\'email\'])
df2 = pd.DataFrame(data=sf.values, columns=[\'list\'])
df = pd.merge(df1, df2, left_index=True, right_index=True)
回答1:
Rather than create 2 temporary dfs you can just pass these as params within a dict using the DataFrame constructor:
pd.DataFrame({'email':sf.index, 'list':sf.values})
There are lots of ways to construct a df, see the docs
回答2:
to_frame():
Starting with the following Series, df:
email
email1@email.com A
email2@email.com B
email3@email.com C
dtype: int64
I use to_frame to convert the series to DataFrame:
df = df.to_frame().reset_index()
email 0
0 email1@email.com A
1 email2@email.com B
2 email3@email.com C
3 email4@email.com D
Now all you need is to rename the column name and name the index column:
df = df.rename(columns= {0: 'list'})
df.index.name = 'index'
Your DataFrame is ready for further analysis.
Update: I just came across this link where the answers are surprisingly similar to mine here.
回答3:
One line answer would be
myseries.to_frame(name='my_column_name')
myseries.reset_index(drop=True, inplace=True) # As needed
回答4:
Series.reset_index with name
argument
Often the use case comes up where a Series needs to be promoted to a DataFrame. But if the Series has no name, then reset_index
will result in something like,
s = pd.Series([1, 2, 3], index=['a', 'b', 'c']).rename_axis('A')
s
A
a 1
b 2
c 3
dtype: int64
s.reset_index()
A 0
0 a 1
1 b 2
2 c 3
Where you see the column name is "0". We can fix this be specifying a name
parameter.
s.reset_index(name='B')
A B
0 a 1
1 b 2
2 c 3
s.reset_index(name='list')
A list
0 a 1
1 b 2
2 c 3
Series.to_frame
If you want to create a DataFrame without promoting the index to a column, use Series.to_frame
, as suggested in this answer. This also supports a name parameter.
s.to_frame(name='B')
B
A
a 1
b 2
c 3
pd.DataFrame Constructor
You can also do the same thing as Series.to_frame
by specifying a columns
param:
pd.DataFrame(s, columns=['B'])
B
A
a 1
b 2
c 3
回答5:
Why not series_obj.to_frame()?
It gets my job done.
来源:https://stackoverflow.com/questions/26097916/convert-pandas-series-to-dataframe