How to iterate over consecutive chunks of Pandas dataframe efficiently

Question

I have a large dataframe (several million rows).

I want to be able to do a groupby operation on it, but just grouping by arbitrary consecutive (preferably equal-size

Answer 1

In practice, you can't guarantee equal-sized chunks. The number of rows (N) might be prime, in which case you could only get equal-sized chunks at 1 or N. Because of this, real-world chunking typically uses a fixed size and allows for a smaller chunk at the end. I tend to pass an array to groupby. Starting from:

>>> df = pd.DataFrame(np.random.rand(15, 5), index=[0]*15)
>>> df[0] = range(15)
>>> df
    0         1         2         3         4
0   0  0.746300  0.346277  0.220362  0.172680
0   1  0.657324  0.687169  0.384196  0.214118
0   2  0.016062  0.858784  0.236364  0.963389
[...]
0  13  0.510273  0.051608  0.230402  0.756921
0  14  0.950544  0.576539  0.642602  0.907850

[15 rows x 5 columns]

where I've deliberately made the index uninformative by setting it to 0, we simply decide on our size (here 10) and integer-divide an array by it:

>>> df.groupby(np.arange(len(df))//10)

>>> for k,g in df.groupby(np.arange(len(df))//10):
...     print(k,g)
...     
0    0         1         2         3         4
0  0  0.746300  0.346277  0.220362  0.172680
0  1  0.657324  0.687169  0.384196  0.214118
0  2  0.016062  0.858784  0.236364  0.963389
[...]
0  8  0.241049  0.246149  0.241935  0.563428
0  9  0.493819  0.918858  0.193236  0.266257

[10 rows x 5 columns]
1     0         1         2         3         4
0  10  0.037693  0.370789  0.369117  0.401041
0  11  0.721843  0.862295  0.671733  0.605006
[...]
0  14  0.950544  0.576539  0.642602  0.907850

[5 rows x 5 columns]

Methods based on slicing the DataFrame can fail when the index isn't compatible with that, although you can always use .iloc[a:b] to ignore the index values and access data by position.