Rolling mean of a sorted subarray of a Dataframe

user_15481403 0
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Given a large mutli-column Pandas dataframe, I would like to compute the rolling "k-mean" over a window of N elements as quickly as possible.
Here "k-mean" is defined as the mean of the N-2k elements of N which exclude the k largest and k smallest elements.

Example

Given the dataframe:

df = pandas.DataFrame(
{'A': [34, 78, -2, -96, 58, -34, 44, -50, 42],
 'B': [-82, 28, 96, 46, 36, -34, -20, 10, -40]})

    A   B
0  34 -82
1  78  28
2  -2  96
3 -96  46
4  58  36
5 -34 -34
6  44 -20
7 -50  10
8  42 -40

With N=6 and k=1 the expected output is:

      A     B
0   NaN   NaN
1   NaN   NaN
2   NaN   NaN
3   NaN   NaN
4   NaN   NaN
5  14.0  19.0
6  16.5  22.5
7 -10.5  18.0
8   0.5  -2.0

Attempt

My code that seems to do the required is:

def k_mean(s: pandas.Series, trim: int) -> float:
    assert trim >= 0, f"Trim must not be negative, {trim} provided."
    if trim == 0:
        return s.mean()
    return s.sort_values()[trim:-trim].mean()

df.rolling(window=6, axis=0).apply(k_mean, kwargs={'trim': 1})

My question: Is my code correct and if so, is there a faster way to achieve the same result, especially considering large multicolumn dataframes?
Perhaps there is a neat mathematical trick to help?

I'm not too concered about the handling of the starting period if it helps speed up performance, either is can be NaN until N or can grow to N once 2k+1 elements are in the window.

pandas python performance
1 Answers

You can use Numba JIT to speed up the code significantly. The main idea is to convert each column to a Numpy array and then iterate over them using a sliding window.

import pandas
import numpy
import numba

# Note:
# You can declare the Numba function parameters types to reduce compilation time:
# @numba.njit('float64[::1](int64[::1], int64, int64)')

@numba.njit
def col_k_mean(arr: numpy.array, window: int, trim: int):
    out = numpy.full(len(arr), numpy.nan)
    if trim == 0:
        localSum = arr[0:window].sum()
        windowInv = 1.0 / window
        for i in range(window-1, len(arr)-1):
            out[i] = localSum * windowInv
            localSum += arr[i+1] - arr[i-window+1]
        if window-1 <= len(arr)-1:
            out[len(arr)-1] = localSum * windowInv
    else:
        for i in range(window-1, len(arr)):
            out[i] = numpy.sort(arr[i-window+1:i+1])[trim:-trim].mean()
    return out

def apply_k_mean(df: pandas.DataFrame, window: int, trim: int) -> pandas.DataFrame:
    assert trim >= 0, f"Trim must not be negative, {trim} provided."
    return pandas.DataFrame({col: col_k_mean(df[col].to_numpy(), window, trim) for col in df})

apply_k_mean(df, window=6, trim=1)

Note that this method is efficient only when the windows is not huge. With huge windows, it is better to use a more advanced sorting strategy, like the one based on priority queues (using heaps) or more generally incremental sorts. Alternatively, if trim is very small and window is huge, then 2 partitions can be used rather than a full sort.

On my machine, with a random dataframe of size (2, 10000) and with window=10 as well as trim=2, the above code is 300 times faster than the reference implementation (excluding the JIT compilation time)! With trim=0, it is 5800 times faster!

The computation can be even faster on huge dataframes using parallelism (supported in Numba using both parallel=True and prange).

user_129395570
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