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)

1

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

1

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.

2

One line answer would be

myseries.to_frame(name='my_column_name')

Or

myseries.reset_index(drop=True, inplace=True) # As needed

1

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

4

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]

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