Convert pandas Series to DataFrame

I have a Pandas series sf:

email [1.0, 0.0, 0.0] [2.0, 0.0, 0.0] [1.0, 0.0, 0.0] [4.0, 0.0, 0.0] [1.0, 0.0, 3.0] [1.0, 5.0, 0.0] 

And I would like to transform it to the following DataFrame:

index | email | list _____________________________________________ 0 | | [1.0, 0.0, 0.0] 1 | | [2.0, 0.0, 0.0] 2 | | [1.0, 0.0, 0.0] 3 | | [4.0, 0.0, 0.0] 4 | | [1.0, 0.0, 3.0] 5 | | [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) 

7 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 A B C dtype: int64 

I use to_frame to convert the series to DataFrame:

df = df.to_frame().reset_index() email 0 0 A 1 B 2 C 3 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 

Super simple way is also

df = pd.DataFrame(series) 

It will return a DF of 1 column (series values) + 1 index (0....n)

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 [1.0, 0.0, 0.0] 1 [2.0, 0.0, 0.0] 2 [1.0, 0.0, 0.0] 3 [4.0, 0.0, 3.0] 4 [1.0, 5.0, 0.0]