Append an empty row in dataframe using pandas

You can add it by appending a Series to the dataframe as follows. I am assuming by blank you mean you want to add a row containing only “Nan”.
You can first create a Series object with Nan. Make sure you specify the columns while defining ‘Series’ object in the -Index parameter.
The you can append it to the DF. Hope it helps!

from numpy import nan as Nan
import pandas as pd

>>> df1 = pd.DataFrame({'A': ['A0', 'A1', 'A2', 'A3'],
...                     'B': ['B0', 'B1', 'B2', 'B3'],
...                     'C': ['C0', 'C1', 'C2', 'C3'],
...                     'D': ['D0', 'D1', 'D2', 'D3']},
...                     index=[0, 1, 2, 3])

>>> s2 = pd.Series([Nan,Nan,Nan,Nan], index=['A', 'B', 'C', 'D'])
>>> result = df1.append(s2)
>>> result
     A    B    C    D
0   A0   B0   C0   D0
1   A1   B1   C1   D1
2   A2   B2   C2   D2
3   A3   B3   C3   D3
4  NaN  NaN  NaN  NaN

The code below worked for me.

df.append(pd.Series([np.nan]), ignore_index = True)

Assuming df is your dataframe,

df_prime = pd.concat([df, pd.DataFrame([[np.nan] * df.shape[1]], columns=df.columns)], ignore_index=True)

where df_prime equals df with an additional last row of NaN’s.

Note that pd.concat is slow so if you need this functionality in a loop, it’s best to avoid using it.
In that case, assuming your index is incremental, you can use

df.loc[df.iloc[-1].name + 1,:] = np.nan

You can add a new series, and name it at the same time. The name will be the index of the new row, and all the values will automatically be NaN.

df.append(pd.Series(name='Afterthought'))

Assuming your df.index is sorted you can use:

df.loc[df.index.max() + 1] = None

It handles well different indexes and column types.

[EDIT] it works with pd.DatetimeIndex if there is a constant frequency, otherwise we must specify the new index exactly e.g:

df.loc[df.index.max() + pd.Timedelta(milliseconds=1)] = None

long example:

df = pd.DataFrame([[pd.Timestamp(12432423), 23, 'text_field']], 
                    columns=["timestamp", "speed", "text"],
                    index=pd.DatetimeIndex(start='2111-11-11',freq='ms', periods=1))
df.info()

<class 'pandas.core.frame.DataFrame'>
DatetimeIndex: 1 entries, 2111-11-11 to 2111-11-11
Freq: L
Data columns (total 3 columns):
timestamp 1 non-null datetime64[ns]
speed 1 non-null int64
text 1 non-null object
dtypes: datetime64[ns](1), int64(1), object(1)
memory usage: 32.0+ bytes

df.loc[df.index.max() + 1] = None
df.info()

<class 'pandas.core.frame.DataFrame'>
DatetimeIndex: 2 entries, 2111-11-11 00:00:00 to 2111-11-11 00:00:00.001000
Data columns (total 3 columns):
timestamp 1 non-null datetime64[ns]
speed 1 non-null float64
text 1 non-null object
dtypes: datetime64[ns](1), float64(1), object(1)
memory usage: 64.0+ bytes

df.head()

                            timestamp                   speed      text
2111-11-11 00:00:00.000 1970-01-01 00:00:00.012432423   23.0    text_field
2111-11-11 00:00:00.001 NaT NaN NaN

You can also use:

your_dataframe.insert(loc=0, value=np.nan, column="")

where loc is your empty row index.

Append “empty” row to data frame and fill selected cells:

Generate empty data frame (no rows just columns a and b):

import pandas as pd    
col_names =  ["a","b"]
df  = pd.DataFrame(columns = col_names)

Append empty row at the end of the data frame:

df = df.append(pd.Series(), ignore_index = True)

Now fill the empty cell at the end (len(df)-1) of the data frame in column a:

df.loc[[len(df)-1],'a'] = 123

Result:

     a    b
0  123  NaN

And of course one can iterate over the rows and fill cells:

col_names =  ["a","b"]
df  = pd.DataFrame(columns = col_names)
for x in range(0,5):
    df = df.append(pd.Series(), ignore_index = True)
    df.loc[[len(df)-1],'a'] = 123

Result:

     a    b
0  123  NaN
1  123  NaN
2  123  NaN
3  123  NaN
4  123  NaN

@Dave Reikher‘s answer is the best solution.

df.loc[df.iloc[-1].name + 1,:] = np.nan

Here’s a similar answer without the NumPy library

df.loc[len(df.index)] = ['' for x in df.columns.values.tolist()]

• len(df.index) = number of rows. Always 1 more than index count.
• By using df.loc[len(df.index)] you are selecting the next index number (row) available.
• df.iloc[-1].name + 1 equals df.loc[len(df.index)]
• Instead of using NumPy, you can also use a python comprehension
• Create a list from the column names: df.columns.values.tolist()
• Create a new list of empty strings ” based on the number of columns.
• ['' for x in df.columns.values.tolist()]