Approximate entropy

Algorithm

A comprehensive step-by-step tutorial with an explanation of the theoretical foundations of Approximate Entropy is available. The algorithm is:

; Step 1: Assume a time series of data u(1), u(2),\ldots, u(N). These are N raw data values from measurements equally spaced in time. ; Step 2: Let m \in \mathbb{Z}^+ be a positive integer, with m \leq N, which represents the length of a run of data (essentially a window). Let r \in \mathbb{R}^+ be a positive real number, which specifies a filtering level. Let n=N-m+1. ; Step 3: Define \mathbf{x}(i) = \big[u(i),u(i+1),\ldots,u(i+m-1)\big] for each i where 1 \leq i \leq n. In other words, \mathbf{x}(i) is an m-dimensional vector that contains the run of data starting with u(i). Define the distance between two vectors \mathbf{x}(i) and \mathbf{x}(j) as the maximum of the distances between their respective components, given by :: \begin{align} d[\mathbf{x}(i),\mathbf{x}(j) ] & = \max_k \big(|\mathbf{x}(i)_k - \mathbf{x}(j)k| \big) \ & = \max_k \big(|u(i+k-1) - u(j+k-1)| \big) \ \end{align} : for 1 \leq k \leq m. ; Step 4: Define a count C^m_i as :: C_i^m (r)={(\text{number of } j \text { such that } d[\mathbf{x}(i),\mathbf{x}(j)] \leq r) \over n} : for each i where 1 \leq i,j \leq n. Note that since j takes on all values between 1 and n, the match will be counted when j=i (i.e. when the test subsequence, \mathbf{x}(j), is matched against itself, \mathbf{x}(i)). ; Step 5: Define :: \phi ^m (r) = {1 \over n} \sum{i=1}^{n}\log (C_i^m (r)) : where \log is the natural logarithm, and for a fixed m , r , and n as set in Step 2. ; Step 6: Define approximate entropy (\mathrm{ApEn}) as :: \mathrm{ApEn}(m,r,N)(u) = \phi ^m (r) - \phi^{m+1} (r) ;Parameter selection: Typically, choose m=2 or m=3 , whereas r depends greatly on the application.

An implementation on Physionet, which is based on Pincus, use d[\mathbf{x}(i), \mathbf{x}(j)] instead of d[\mathbf{x}(i), \mathbf{x}(j)] \le r in Step 4. While a concern for artificially constructed examples, it is usually not a concern in practice.

Example

Consider a sequence of N=51 samples of heart rate equally spaced in time:

: \ S_N = {85, 80, 89, 85, 80, 89, \ldots}

Note the sequence is periodic with a period of 3. Let's choose m=2 and r=3 (the values of m and r can be varied without affecting the result).

Form a sequence of vectors: :\begin{align} \mathbf{ x}(1) & = [u(1) \ u(2)]=[85 \ 80]\ \mathbf{ x}(2) & = [u(2) \ u(3)]=[80 \ 89]\ \mathbf{ x}(3) & = [u(3) \ u(4)]=[89 \ 85]\ \mathbf{ x}(4) & = [u(4) \ u(5)]=[85 \ 80]\ & \ \ \vdots \end{align}

Distance is calculated repeatedly as follows. In the first calculation,

:\ d[\mathbf{x}(1), \mathbf{x}(1)]=\max_k |\mathbf{x}(1)_k - \mathbf{x}(1)_k|=0 which is less than r .

In the second calculation, note that |u(2)-u(3)| |u(1)-u(2)|, so

:\ d[\mathbf{x}(1), \mathbf{x}(2)]=\max_k |\mathbf{x}(1)_k-\mathbf{x}(2)_k|=|u(2)-u(3)|=9 which is greater than r . Similarly, :\begin{align} d[\mathbf{x}(1) &, \mathbf{x}(3)] = |u(2)-u(4)| = 5r\ d[\mathbf{x}(1) &, \mathbf{x}(4)] = |u(1)-u(4)| = |u(2)-u(5)| = 0 & \vdots \ d[\mathbf{x}(1) &, \mathbf{x}(j)] = \cdots \ & \vdots \ \end{align} The result is a total of 17 terms \mathbf{ x}(j) such that d[\mathbf{x}(1), \mathbf{x}(j)]\le r . These include \mathbf{x}(1), \mathbf{x}(4), \mathbf{x}(7),\ldots,\mathbf{x}(49). In these cases, C^m_i(r) is

:\ C_1^2 (3)=\frac{17}{50} :\ C_2^2 (3)=\frac{17}{50} :\ C_3^2 (3)=\frac{16}{50} :\ C_4^2 (3)=\frac{17}{50}\ \cdots

Note in Step 4, 1 \leq i \leq n for \mathbf{x}(i) . So the terms \mathbf{x}(j) such that d[\mathbf{x}(3), \mathbf{x}(j)] \leq r include \mathbf{x}(3), \mathbf{x}(6), \mathbf{x}(9),\ldots,\mathbf{x}(48), and the total number is 16.

At the end of these calculations, we have

:\phi^2 (3) = {1 \over 50} \sum_{i=1}^{50}\log(C_i^2(3))\approx-1.0982

Then we repeat the above steps for m=3 . First form a sequence of vectors: :\begin{align} \mathbf{x}(1) & = [u(1) \ u(2) \ u(3)]=[85 \ 80 \ 89]\ \mathbf{x}(2) & = [u(2) \ u(3) \ u(4)]=[80 \ 89 \ 85]\ \mathbf{x}(3) & = [u(3) \ u(4) \ u(5)]=[89 \ 85 \ 80]\ \mathbf{x}(4) & = [u(4) \ u(5) \ u(6)]=[85 \ 80 \ 89]\ &\ \ \vdots \end{align}

By calculating distances between vector \mathbf{x}(i), \mathbf{x}(j), 1 \le i \le 49 , we find the vectors satisfying the filtering level have the following characteristic: :d[\mathbf{x}(i), \mathbf{x}(i+3)]=0 Therefore, :\ C_1^3 (3)=\frac{17}{49} :\ C_2^3 (3)=\frac{16}{49} :\ C_3^3 (3)=\frac{16}{49} :\ C_4^3 (3)=\frac{17}{49}\ \cdots At the end of these calculations, we have :\phi^3 (3)={1 \over 49} \sum_{i=1}^{49}\log(C_i^3(3))\approx-1.0982

Finally, : \mathrm{ ApEn}=\phi^2 (3)-\phi^3 (3)\approx0.000010997

The value is very small, so it implies the sequence is regular and predictable, which is consistent with the observation.

Python implementation

import math


def approx_entropy(time_series, run_length, filter_level) -> float:
    """
    Approximate entropy

    >>> import random
    >>> regularly = [85, 80, 89] * 17
    >>> print(f"{approx_entropy(regularly, 2, 3):e}")
    1.099654e-05
    >>> randomly = [random.choice([85, 80, 89]) for _ in range(17*3)]
    >>> 0.8 < approx_entropy(randomly, 2, 3) < 1
    True
    """

    def _maxdist(x_i, x_j):
        return max(abs(ua - va) for ua, va in zip(x_i, x_j))

    def _phi(m):
        n = len(time_series) - m + 1
        x = [
            [time_series[j] for j in range(i, i + m)]
            for i in range(n)
        ]
        counts = [
            sum(1 for x_j in x if _maxdist(x_i, x_j) <= filter_level) / n for x_i in x
        ]
        return sum(math.log(c) for c in counts) / n

    return abs(_phi(run_length + 1) - _phi(run_length))


if __name__ == "__main__":
    import doctest

    doctest.testmod()

MATLAB implementation

• Fast Approximate Entropy from MatLab Central
• approximateEntropy

Interpretation

The presence of repetitive patterns of fluctuation in a time series renders it more predictable than a time series in which such patterns are absent. ApEn reflects the likelihood that similar patterns of observations will not be followed by additional similar observations. A time series containing many repetitive patterns has a relatively small ApEn; a less predictable process has a higher ApEn.

Advantages

The advantages of ApEn include:

• Lower computational demand. ApEn can be designed to work for small data samples ( N points) and can be applied in real time.
• Less effect from noise. If data is noisy, the ApEn measure can be compared to the noise level in the data to determine what quality of true information may be present in the data.

Limitations

The ApEn algorithm counts each sequence as matching itself to avoid the occurrence of \log(0) in the calculations. This step might introduce bias in ApEn, which causes ApEn to have two poor properties in practice:

1. ApEn is heavily dependent on the record length and is uniformly lower than expected for short records.
2. It lacks relative consistency. That is, if ApEn of one data set is higher than that of another, it should, but does not, remain higher for all conditions tested.

Applications

ApEn has been applied to classify electroencephalography (EEG) in psychiatric diseases, such as schizophrenia,{{cite journal epilepsy,{{cite journal and addiction.{{cite journal

References

References

1. (2020-01-22). "Analyzing changes in the complexity of climate in the last four decades using MERRA-2 radiation data". Scientific Reports.
2. (June 2019). "Approximate Entropy and Sample Entropy: A Comprehensive Tutorial". Entropy.
3. "PhysioNet".