pandas add row instead of column

Question

I'm new to pandas, but trying to simply add a row

class Security:
    def __init__(self):
        self.structure = ['timestamp', 'open', 'high', 'low', 'close', 'vol']
        self.df = pd.DataFrame(columns=self.structure)  # index =
    def whats_inside(self):
        return self.df
    """
    Some skipped code...
    """
    def add_data(self, timestamp, open, high, low, close, vol):
        data = [timestamp, open, high, low, close, vol]
        self.df = self.df.append (data)

sec = Security()
print sec.whats_inside()
sec.add_data ('2015/06/01', '1', '2', '0.5', '1', '100')
print sec.whats_inside()

but the output is:

            0 close high  low open timestamp  vol
0  2015/06/01   NaN  NaN  NaN  NaN       NaN  NaN
1           1   NaN  NaN  NaN  NaN       NaN  NaN
2           2   NaN  NaN  NaN  NaN       NaN  NaN
3         0.5   NaN  NaN  NaN  NaN       NaN  NaN
4           1   NaN  NaN  NaN  NaN       NaN  NaN
5         100   NaN  NaN  NaN  NaN       NaN  NaN

This means, I'm adding a column instead of row. Yes, I've tried to google but still didnt get the point how do make it simple pythonic way.

p.s. I know that's simple, but I'm just missing something important.

Answer 1

There are several ways to add a new row. Perhaps the easiest one is (if you want to add the row to the end) is to use loc:

df.loc[len(df)] = ['val_a', 'val_b', .... ]

loc expects an index. len(df) will return the number of rows in the dataframe so the new row will be added to the end of the dataframe.

'['val_a', 'val_b', .... ]' is a list of values of the row, in the same order of the columns, so the list's length must be equal to the number of columns, otherwise you will get a ValueError exception. An exception for this is that if you want all the columns to have the same values you are allowed to have that value as a single element in the list, for example df.loc[len(df)] = ['aa'].

NOTE: a good idea will be to always use reset_index before using this method because if you ever delete a row or work on a filtered dataframe you are not guaranteed that the rows' indexes will be in sync with the number of rows.