Convert pandas Series to DataFrame
Question:
I have a Pandas series sf:
email
[email protected] [1.0, 0.0, 0.0]
[email protected] [2.0, 0.0, 0.0]
[email protected] [1.0, 0.0, 0.0]
[email protected] [4.0, 0.0, 0.0]
[email protected] [1.0, 0.0, 3.0]
[email protected] [1.0, 5.0, 0.0]
And I would like to transform it to the following DataFrame:
index | email | list
_____________________________________________
0 | [email protected] | [1.0, 0.0, 0.0]
1 | [email protected] | [2.0, 0.0, 0.0]
2 | [email protected] | [1.0, 0.0, 0.0]
3 | [email protected] | [4.0, 0.0, 0.0]
4 | [email protected] | [1.0, 0.0, 3.0]
5 | [email protected] | [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)
Answers:
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
to_frame():
Starting with the following Series, df:
email
[email protected] A
[email protected] B
[email protected] C
dtype: int64
I use to_frame to convert the series to DataFrame:
df = df.to_frame().reset_index()
email 0
0 [email protected] A
1 [email protected] B
2 [email protected] C
3 [email protected] 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.
One line answer would be
myseries.to_frame(name='my_column_name')
Or
myseries.reset_index(drop=True, inplace=True) # As needed
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
Series.to_frame can be used to convert a Series to DataFrame.
# The provided name (columnName) will substitute the series name
df = series.to_frame('columnName')
For example,
s = pd.Series(["a", "b", "c"], name="vals")
df = s.to_frame('newCol')
print(df)
newCol
0 a
1 b
2 c
probably graded as a non-pythonic way to do this but this’ll give the result you want in a line:
new_df = pd.DataFrame(zip(email,list))
Result:
email list
0 [email protected] [1.0, 0.0, 0.0]
1 [email protected] [2.0, 0.0, 0.0]
2 [email protected] [1.0, 0.0, 0.0]
3 [email protected] [4.0, 0.0, 3.0]
4 [email protected] [1.0, 5.0, 0.0]
Super simple way is also
df = pd.DataFrame(series)
It will return a DF of 1 column (series values) + 1 index (0….n)
I have a Pandas series sf:
email
[email protected] [1.0, 0.0, 0.0]
[email protected] [2.0, 0.0, 0.0]
[email protected] [1.0, 0.0, 0.0]
[email protected] [4.0, 0.0, 0.0]
[email protected] [1.0, 0.0, 3.0]
[email protected] [1.0, 5.0, 0.0]
And I would like to transform it to the following DataFrame:
index | email | list
_____________________________________________
0 | [email protected] | [1.0, 0.0, 0.0]
1 | [email protected] | [2.0, 0.0, 0.0]
2 | [email protected] | [1.0, 0.0, 0.0]
3 | [email protected] | [4.0, 0.0, 0.0]
4 | [email protected] | [1.0, 0.0, 3.0]
5 | [email protected] | [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)
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
to_frame():
Starting with the following Series, df:
email
[email protected] A
[email protected] B
[email protected] C
dtype: int64
I use to_frame to convert the series to DataFrame:
df = df.to_frame().reset_index()
email 0
0 [email protected] A
1 [email protected] B
2 [email protected] C
3 [email protected] 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.
One line answer would be
myseries.to_frame(name='my_column_name')
Or
myseries.reset_index(drop=True, inplace=True) # As needed
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
Series.to_frame can be used to convert a Series to DataFrame.
# The provided name (columnName) will substitute the series name
df = series.to_frame('columnName')
For example,
s = pd.Series(["a", "b", "c"], name="vals")
df = s.to_frame('newCol')
print(df)
newCol
0 a
1 b
2 c
probably graded as a non-pythonic way to do this but this’ll give the result you want in a line:
new_df = pd.DataFrame(zip(email,list))
Result:
email list
0 [email protected] [1.0, 0.0, 0.0]
1 [email protected] [2.0, 0.0, 0.0]
2 [email protected] [1.0, 0.0, 0.0]
3 [email protected] [4.0, 0.0, 3.0]
4 [email protected] [1.0, 5.0, 0.0]
Super simple way is also
df = pd.DataFrame(series)
It will return a DF of 1 column (series values) + 1 index (0….n)