Using pandas .append within for loop

Every time you call append, Pandas returns a copy of the original dataframe plus your new row. This is called quadratic copy, and it is an O(N^2) operation that will quickly become very slow (especially since you have lots of data).

In your case, I would recommend using lists, appending to them, and then calling the dataframe constructor.

a_list = []
b_list = []
for data in my_data:
    a, b = process_data(data)
    a_list.append(a)
    b_list.append(b)
df = pd.DataFrame({'A': a_list, 'B': b_list})
del a_list, b_list

Timings

%%timeit
data = pd.DataFrame([])
for i in np.arange(0, 10000):
    if i % 2 == 0:
        data = data.append(pd.DataFrame({'A': i, 'B': i + 1}, index=[0]), ignore_index=True)
else:
    data = data.append(pd.DataFrame({'A': i}, index=[0]), ignore_index=True)
1 loops, best of 3: 6.8 s per loop

%%timeit
a_list = []
b_list = []
for i in np.arange(0, 10000):
    if i % 2 == 0:
        a_list.append(i)
        b_list.append(i + 1)
    else:
        a_list.append(i)
        b_list.append(None)
data = pd.DataFrame({'A': a_list, 'B': b_list})
100 loops, best of 3: 8.54 ms per loop

You need to set the the variable data equal to the appended data frame. Unlike the append method on a python list the pandas append does not happen in place

import pandas as pd
import numpy as np

data = pd.DataFrame([])

for i in np.arange(0, 4):
    if i % 2 == 0:
        data = data.append(pd.DataFrame({'A': i, 'B': i + 1}, index=[0]), ignore_index=True)
    else:
        data = data.append(pd.DataFrame({'A': i}, index=[0]), ignore_index=True)

print(data.head())

   A    B
0  0  1.0
1  2  3.0
2  3  NaN

NOTE: This answer aims to answer the question as it was posed. It is not however the optimal strategy for combining large numbers of dataframes. For a more optimal solution have a look at Alexander's answer below


You can build your dataframe without a loop:

n = 4
data = pd.DataFrame({'A': np.arange(n)})
data['B'] = np.NaN
data.loc[data['A'] % 2 == 0, 'B'] = data['A'] + 1

For:

n = 10000

This is a bit faster:

%%timeit
data = pd.DataFrame({'A': np.arange(n)})
data['B'] = np.NaN
data.loc[data['A'] % 2 == 0, 'B'] = data['A'] + 1

100 loops, best of 3: 3.3 ms per loop

vs.

%%timeit
a_list = []
b_list = []
for i in np.arange(n):
    if i % 2 == 0:
        a_list.append(i)
        b_list.append(i + 1)
    else:
        a_list.append(i)
        b_list.append(None)
data1 = pd.DataFrame({'A': a_list, 'B': b_list})

100 loops, best of 3: 12.4 ms per loop