Changing response functions

R.A. Collenteur, University of Graz, 2021

In this notebook the new ChangeModel is tested, based on the work by Obergfjell et al. (2019). The main idea is to apply different response functions for two different periods. As an example we look at the the groundwater levels measured near the river the Mur in Austria, where a dam was recently built.

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import pastas as ps

ps.set_log_level("ERROR")
ps.show_versions()

Python version: 3.10.12
NumPy version: 1.23.5
Pandas version: 2.0.3
SciPy version: 1.11.1
Matplotlib version: 3.7.2
Numba version: 0.57.1
LMfit version: 1.2.2
Latexify version: Not Installed
Pastas version: 1.1.0

1. Load the data

prec = pd.read_csv("data_step/prec.csv", index_col=0, parse_dates=True).squeeze()
evap = pd.read_csv("data_step/evap.csv", index_col=0, parse_dates=True).squeeze()
head = pd.read_csv("data_step/head.csv", index_col=0, parse_dates=True).squeeze()
river = pd.read_csv("data_step/river.csv", index_col=0, parse_dates=True).squeeze()
river -= river.min()

axes = ps.plots.series(
    head=head,
    stresses=[prec, evap, river],
    tmin="2000",
    labels=["Head\n[m]", "Prec\n[mm/d]", "Evap\n[mm/d]", "River [m3/d]"],
)

2. The weighting factor

The stress is convoluted two times with different response functions. Then, a weighting function is used to add the two contributions together and compute the final contribution.

npoints = 100

tchange = 50 / npoints
t = np.linspace(0, 1, npoints)
color = plt.cm.viridis(np.linspace(0, 1, 10))

for beta, c in zip(np.linspace(-1, 1, 10), color):
    beta1 = beta * npoints
    omega = 1 / (np.exp(beta1 * (t - tchange)) + 1)
    plt.plot(omega, color=c, label="$beta$={:.2f}".format(beta))

plt.ylabel("$\omega$ [-]")
plt.xlabel("Time [t]")
plt.legend()

<matplotlib.legend.Legend at 0x7efdd11da110>

3. Make a model

We now make two models:

• one model where we assume the response of the heads to the river level remains the same

• and one model where the response to the river levels changes.

# Normal Model
ml = ps.Model(head, name="linear")

sm = ps.StressModel(river, ps.Exponential(), name="test")
step = ps.StepModel("2012-01-01", rfunc=ps.One(), name="step")

ml.add_stressmodel([sm, step])
ml.solve(report=False, tmin="2004", tmax="2017-12-31", noise=True)
ml.plots.results(figsize=(10, 6))

# ChangeModel
ml2 = ps.Model(head, name="linear")

cm = ps.ChangeModel(
    river,
    ps.Exponential(),
    ps.Exponential(),
    name="test",
    tchange="2012-01-01",
)

ml2.add_stressmodel([cm, step])
ml2.solve(report=False, tmin="2004", tmax="2017-12-31", noise=True)
ml2.plots.results(figsize=(10, 6));

/home/docs/checkouts/readthedocs.org/user_builds/pastas/envs/latest/lib/python3.10/site-packages/pastas/stressmodels.py:1687: RuntimeWarning: overflow encountered in exp
  omega = 1 / (np.exp(beta * (t - sigma)) + 1)
/home/docs/checkouts/readthedocs.org/user_builds/pastas/envs/latest/lib/python3.10/site-packages/pastas/stressmodels.py:1687: RuntimeWarning: overflow encountered in exp
  omega = 1 / (np.exp(beta * (t - sigma)) + 1)
/home/docs/checkouts/readthedocs.org/user_builds/pastas/envs/latest/lib/python3.10/site-packages/pastas/stressmodels.py:1687: RuntimeWarning: overflow encountered in exp
  omega = 1 / (np.exp(beta * (t - sigma)) + 1)

The second model shows a better fit, but also the step trend changed.

print("RMSE for the first model:", ml.stats.rmse().round(2))
print("RMSE for the second model:", ml2.stats.rmse().round(2))

RMSE for the first model: 0.17
RMSE for the second model: 0.13

/home/docs/checkouts/readthedocs.org/user_builds/pastas/envs/latest/lib/python3.10/site-packages/pastas/stressmodels.py:1687: RuntimeWarning: overflow encountered in exp
  omega = 1 / (np.exp(beta * (t - sigma)) + 1)

ml2.parameters

	initial	name	optimal	pmin	pmax	vary	stderr
test_1_A	1.00000	test	0.881948	1.000000e-05	100.0	True	9.400205e-03
test_1_a	10.00000	test	1.272077	1.000000e-02	1000.0	True	4.395089e-02
test_2_A	1.00000	test	0.428230	1.000000e-05	100.0	True	1.161800e-02
test_2_a	10.00000	test	1.930610	1.000000e-02	1000.0	True	1.026581e-01
test_beta	0.00000	test	781566.435302	-inf	inf	True	1.394655e-16
test_tchange	734503.00000	test	734503.000000	6.124110e+05	825914.0	False	NaN
step_d	1.00000	step	1.130024	0.000000e+00	NaN	True	1.750598e-02
step_tstart	734503.00000	step	734503.000000	6.124110e+05	825914.0	False	NaN
constant_d	330.18797	constant	329.256815	NaN	NaN	True	1.046840e-02
noise_alpha	7.00000	noise	24.476228	1.000000e-05	5000.0	True	1.370048e+00

4. Compare the response functions

We can also look at the response to the river before and after,

cm_rf1 = cm.rfunc1.step(
    p=ml2.parameters.loc[["test_1_A", "test_1_a"], "optimal"].values
)
cm_rf2 = cm.rfunc2.step(
    p=ml2.parameters.loc[["test_2_A", "test_2_a"], "optimal"].values
)

plt.plot(np.arange(1, len(cm_rf1) + 1), cm_rf1)
plt.plot(np.arange(1, len(cm_rf2) + 1), cm_rf2)

plt.legend(["Before", "After"]);

5. Another way

We can also add the stress twice, saving one parameter that needs to be estimated.

ml3 = ps.Model(head, name="linear")

river1 = river.copy()
river1.loc["2012":] = 0

river2 = river.copy()
river2.loc[:"2011"] = 0

r1 = ps.StressModel(river1, rfunc=ps.Exponential(), name="river")
r2 = ps.StressModel(river2, rfunc=ps.Exponential(), name="river2")
step = ps.StepModel("2012-01-01", rfunc=ps.One(), name="step")

ml3.add_stressmodel([r1, r2, step])
ml3.solve(report=False, tmin="2004", tmax="2017-12-31", noise=True)
ml3.plots.results(figsize=(10, 6));

How do the results compare?

# change model
cm_rf1 = cm.rfunc1.step(
    p=ml2.parameters.loc[["test_1_A", "test_1_a"], "optimal"].values
)
cm_rf2 = cm.rfunc2.step(
    p=ml2.parameters.loc[["test_2_A", "test_2_a"], "optimal"].values
)
plt.plot(np.arange(1, len(cm_rf1) + 1), cm_rf1)
plt.plot(np.arange(1, len(cm_rf2) + 1), cm_rf2)

# 2 stressmodels
ml3.get_step_response("river").plot()
ml3.get_step_response("river2").plot()

plt.legend(
    [
        "Before (ChangeModel)",
        "After (ChangeModel)",
        "Before (method 2)",
        "After (method 2)",
    ]
)

<matplotlib.legend.Legend at 0x7efdc8d5fbe0>

References

Obergfell, C., Bakker, M. and Maas, K. (2019), Identification and Explanation of a Change in the Groundwater Regime using Time Series Analysis. Groundwater, 57: 886-894. https://doi.org/10.1111/gwat.12891