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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

Python version: 3.11.6
NumPy version: 1.26.4
Pandas version: 2.2.2
SciPy version: 1.13.0
Matplotlib version: 3.8.4
Numba version: 0.59.1
LMfit version: 1.3.1
Latexify version: Not Installed
Pastas version: 1.5.0

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(
    stresses=[prec, evap, river],
    labels=["Head\n[m]", "Prec\n[mm/d]", "Evap\n[mm/d]", "River [m3/d]"],

2. The weighting factor#

The stress is convolved 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 =, 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]")
<matplotlib.legend.Legend at 0x7f9d7c9fc7d0>

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")
ml.plots.results(figsize=(10, 6))

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

cm = ps.ChangeModel(

ml2.add_stressmodel([cm, step])
ml2.solve(report=False, tmin="2004", tmax="2017-12-31")
ml2.plots.results(figsize=(10, 6));
../_images/f5f95f27facebe284ba9eeae43e50934c3c0b8d901ee1690dc28c6720a643b4e.png ../_images/a2e8533c9f3f3f5928303e8ee114b13f7e3acfa4759e9b9ade0d1b38fa125825.png

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: 0.14
initial pmin pmax vary name dist stderr optimal
test_1_A 1.00000 1.000000e-05 100.0 True test uniform 0.014107 0.449464
test_1_a 10.00000 1.000000e-02 1000.0 True test uniform 0.137745 2.466723
test_2_A 1.00000 1.000000e-05 100.0 True test uniform 0.009413 0.851243
test_2_a 10.00000 1.000000e-02 1000.0 True test uniform 0.045022 1.124797
test_beta 0.00000 -inf inf True test uniform 6.555924 -49.877036
test_tchange 734503.00000 6.124110e+05 825914.0 False test uniform NaN 734503.000000
step_d 1.00000 NaN NaN True step uniform 0.019171 1.029063
step_tstart 734503.00000 6.124110e+05 825914.0 False step uniform NaN 734503.000000
constant_d 330.18797 NaN NaN True constant uniform 0.011379 329.301851
noise_alpha 7.00000 1.000000e-05 5000.0 True noise uniform 1.727050 28.265550

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")
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

        "Before (ChangeModel)",
        "After (ChangeModel)",
        "Before (method 2)",
        "After (method 2)",
<matplotlib.legend.Legend at 0x7f9d786c9190>


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.