Know the period of a stationary series

Question

I am studying time series, I am using Python, and I need to know the period of the stationary part (The decomposition is the following, I am interested in the seasonal part).

What I have done is take a number (any) and count the number of steps (until I find it again, So I find the period). This is very outdated (in my short perspective).

Do you know any function that calculates the period of a series? Or maybe .. do you know any set of instructions in Pandas, to avoid the use of loops and conditionals? How can I perform this task?

PS: The data I get is similar to this: If counting is done, the data is repeated every twelve steps.

import pandas as pd
import matplotlib.pyplot as plt

seasonal = [-0.0012477419628991032, -0.0042654910887713745, -0.006234490214646844, 0.0007106773261453963, 1.1533604530851796e-08, -0.004141904258934777, 0.0006148978972421542, 0.0017480068715999646, 0.0011169491792932956, 0.002641724820318341, 0.005415250461344693, 0.003642109435703726, -0.0012477419628991032, -0.0042654910887713745, -0.006234490214646844, 0.0007106773261453963, 1.1533604530851796e-08, -0.004141904258934777, 0.0006148978972421542, 0.0017480068715999646, 0.0011169491792932956, 0.002641724820318341, 0.005415250461344693, 0.003642109435703726, -0.0012477419628991032, -0.0042654910887713745, -0.006234490214646844, 0.0007106773261453963, 1.1533604530851796e-08, -0.004141904258934777, 0.0006148978972421542, 0.0017480068715999646, 0.0011169491792932956, 0.002641724820318341, 0.005415250461344693, 0.003642109435703726, -0.0012477419628991032, -0.0042654910887713745, -0.006234490214646844, 0.0007106773261453963, 1.1533604530851796e-08, -0.004141904258934777, 0.0006148978972421542, 0.0017480068715999646, 0.0011169491792932956, 0.002641724820318341, 0.005415250461344693, 0.003642109435703726, -0.0012477419628991032, -0.0042654910887713745, -0.006234490214646844, 0.0007106773261453963, 1.1533604530851796e-08, -0.004141904258934777, 0.0006148978972421542, 0.0017480068715999646, 0.0011169491792932956, 0.002641724820318341, 0.005415250461344693, 0.003642109435703726, -0.0012477419628991032, -0.0042654910887713745, -0.006234490214646844, 0.0007106773261453963, 1.1533604530851796e-08, -0.004141904258934777, 0.0006148978972421542, 0.0017480068715999646, 0.0011169491792932956, 0.002641724820318341, 0.005415250461344693, 0.003642109435703726, -0.0012477419628991032, -0.0042654910887713745, -0.006234490214646844, 0.0007106773261453963, 1.1533604530851796e-08, -0.004141904258934777, 0.0006148978972421542, 0.0017480068715999646, 0.0011169491792932956, 0.002641724820318341, 0.005415250461344693, 0.003642109435703726, -0.0012477419628991032, -0.0042654910887713745, -0.006234490214646844, 0.0007106773261453963, 1.1533604530851796e-08, -0.004141904258934777, 0.0006148978972421542, 0.0017480068715999646, 0.0011169491792932956, 0.002641724820318341, 0.005415250461344693, 0.003642109435703726, -0.0012477419628991032, -0.0042654910887713745, -0.006234490214646844, 0.0007106773261453963, 1.1533604530851796e-08, -0.004141904258934777, 0.0006148978972421542, 0.0017480068715999646, 0.0011169491792932956, 0.002641724820318341, 0.005415250461344693, 0.003642109435703726, -0.0012477419628991032, -0.0042654910887713745, -0.006234490214646844, 0.0007106773261453963, 1.1533604530851796e-08, -0.004141904258934777, 0.0006148978972421542, 0.0017480068715999646, 0.0011169491792932956, 0.002641724820318341, 0.005415250461344693, 0.003642109435703726]
indice = pd.date_range("2019-07-31 23:55:00", periods=len(seasonal), freq="T")
seasonal = pd.Series(data=seasonal, index=indice)

periodo = 0                                 ### 
valor = seasonal.iloc[0]                      #    All this part ...  
                                              # can it be changed
for item in seasonal:                         # for a better structured function,
  if periodo != 0 and item == valor:          # which looks for the period
    break                                     # of a group of data?
                                              # 
  periodo += 1                              ###    Thanks

print("Periodo: {}".format(periodo))
seasonal.plot()
plt.show()

Corralien · Accepted Answer · 2021-07-07T22:42:31.850

The provided answer comes essentially from here. Use auto-correlation to solve your problem.

def find_period(signal):
    acf = np.correlate(signal, signal, 'full')[-len(signal):]
    inflection = np.diff(np.sign(np.diff(acf)))
    peaks = (inflection < 0).nonzero()[0] + 1
    return peaks[acf[peaks].argmax()]

>>> find_period(seasonal)
12

Keep in mind that this is easy because your signal is duplicated ten times. If you have noise in your signal, you have to preprocess your data.

Know the period of a stationary series

1 Answers1