simple-interpolation 0.0.9

Brownian Bridge interpolation of timeseries, built to use with Pandas.

simple_interpolation

A Pandas implentation of the Brownian Bridge interpolation algorithm. Wiener processes are assumed to build std().

Interpolation rocks, but doing it poorly can alter the original features of your data. Brownian bridge preserves the volatibility of the original data, if done well. Mixing that with a bit theory on the stock market (Wiener processes), we built a simple interpolation library.

Read about the algorithm in the "Brownian bridge algo" section below.

Install

pip install simple_interpolation

How to use

# Example input dataframe, containing gaps
#  (i. e. X column, values 3-5)
df

	X	Y
0	0	8.089846
1	1	11.793489
2	2	9.026726
3	6	8.996177
4	7	11.221730
5	8	8.398122
6	9	8.845667
7	10	11.454700
8	11	11.431745
9	12	7.050733
10	13	10.009420
11	14	6.964674
12	15	9.541557
13	16	11.656722
14	19	11.062303
15	20	11.302763
16	21	13.042057
17	22	7.405670
18	23	8.986057
19	24	7.554964
20	25	10.467688
21	26	9.416683
22	27	10.038665
23	28	5.519665
24	45	10.184922
25	46	11.661662
26	47	9.748401
27	48	11.023116
28	49	9.298167

from simple_interpolation import core as si

# Interpolation, plot is optional (default False)
patched_df = si.interpolate_gaps( df , plot = True )
patched_df

No datetime column: assuming first column 'X' as X-axis
std() built with Wiener method
Will interpolate if X-column interval is more than 1.7675
Processed 0.00% of gaps
Ended interpolation, starting plotting the results..

Ended execution

	X	Y	interpolated
0	0.0000	8.089846	0
1	1.0000	11.793489	0
2	2.0000	9.026726	0
3	3.0000	8.291588	1
4	4.0000	8.486541	1
5	5.0000	8.736440	1
6	6.0000	8.996177	0
7	7.0000	11.221730	0
8	8.0000	8.398122	0
9	9.0000	8.845667	0
10	10.0000	11.454700	0
11	11.0000	11.431745	0
12	12.0000	7.050733	0
13	13.0000	10.009420	0
14	14.0000	6.964674	0
15	15.0000	9.541557	0
16	16.0000	11.656722	0
17	17.5000	11.359512	1
18	19.0000	11.062303	0
19	20.0000	11.302763	0
20	21.0000	13.042057	0
21	22.0000	7.405670	0
22	23.0000	8.986057	0
23	24.0000	7.554964	0
24	25.0000	10.467688	0
25	26.0000	9.416683	0
26	27.0000	10.038665	0
27	28.0000	5.519665	0
28	29.0625	6.584443	1
29	30.1250	5.504773	1
30	31.1875	5.623875	1
31	32.2500	6.275126	1
32	33.3125	6.639139	1
33	34.3750	6.394277	1
34	35.4375	6.797008	1
35	36.5000	7.885828	1
36	37.5625	8.530594	1
37	38.6250	8.921191	1
38	39.6875	8.941382	1
39	40.7500	8.900565	1
40	41.8125	9.037251	1
41	42.8750	9.360730	1
42	43.9375	9.914641	1
43	45.0000	10.184922	0
44	46.0000	11.661662	0
45	47.0000	9.748401	0
46	48.0000	11.023116	0
47	49.0000	9.298167	0

Brownian bridge algo: the theory

To render the equations on browser, install a LaTex rendering extension. Otherwise download it and open it on Jupyer.

Allows to interpolate large gaps preserving volatility of the series (as an input!). Read about it here "Brownian bridge".

Weiner method to obtain the relevant std()

In a Wiener process volatility (variance) is $$var = \Delta_t$$ so $$std = \sqrt{var} = \sqrt{\Delta_t}$$This sets how the local volatility should be analyzed.

So, if we have $std_{year}$ (or $std_{whole series}$), we can get the daily by: $$std_{year} = std_{day} \cdot \sqrt{365} \Rightarrow std_{day} = \frac{std_{year}}{\sqrt{365}}$$

So we can get the "basic building block" of the volatility by getting $std_{minute}$ in our case.

Having $std_{minute}$, we then do a "bottom-up" process building the gap:

$$ std_{gap} = std_{minute} \cdot \sqrt{number_of_mins_in_gap}$$

(Advice from Miguel, my colleague at ING)

Fixed timesteps

You can use fixed_freq argument to make the interpolated X points rounded to a certain timestep. 'fixed_freq' timesteps defaults to 'min'. Valid options from Pandas, see link: https://pandas.pydata.org/pandas-docs/stable/user_guide/timeseries.html#timeseries-offset-aliases

---

Implementation of the rounding (you probably don't need to read this)

This constraint takes us out of the brownian bridge, because for it we only interpolate the midpoints through:

\begin{cases} x_m = \frac{x_0 + x_1}{2} \ y_m = \frac{y_0 + y_1}{2} + std \end{cases}

But, if we round up to mins, this midpoint $x_m$ could be different than a minute-exact timestamp (imagine the first interpolated point on a gap of 3m: it would be 1.5m). So we round $x_m$, and search for its associated Y displacement $\Delta y$:

\begin{cases} x'm = x_m + \Delta x{toroundtomin} \ y'_m = y_m + \Delta y \end{cases}

To get the associated $\Delta y$ we must use the slope (derivative) at that straight line between points $(x_0, y_0), (x_1, y_1)$.

So:

1- Round up $x_m$ to the nearest minute (lowest, floor()-like), so we obtain: $x'm$, $\Delta x{toroundtomin}$

2- The deltas on X and Y are related by the derivative, which we are implicitly assuming linear on the brownian bridge, so it's quite straightforward to calculate $\Delta y$:

$$ \Delta y := \frac{dy}{dx} \Delta x \Rightarrow \Delta y \approx \frac{y_1 - y_0}{x_1 - x_0} \Delta x_{toroundtomin} $$

So we would have everything for the Y correction.