From 2d1fae2cdba0607169137523dcf7ec6d621169a7 Mon Sep 17 00:00:00 2001
From: ZhangQi <2801038431@qq.com>
Date: Thu, 18 Jan 2024 11:50:12 +0800
Subject: [PATCH] Changed the way the 'embedded_sequence' package is called and
 modified the method of calculating the maximum Lyapunov exponent to avoid
 nans

---
 .DS_Store                           | Bin 0 -> 6148 bytes
 pyeeg/largest_lyauponov_exponent.py |  29 ++++++++++++++++++++++++----
 2 files changed, 25 insertions(+), 4 deletions(-)
 create mode 100644 .DS_Store

diff --git a/.DS_Store b/.DS_Store
new file mode 100644
index 0000000000000000000000000000000000000000..cb9aeb7b874303b88b24d3148a8b5f0ee4761bba
GIT binary patch
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diff --git a/pyeeg/largest_lyauponov_exponent.py b/pyeeg/largest_lyauponov_exponent.py
index d5ca469..4723f22 100644
--- a/pyeeg/largest_lyauponov_exponent.py
+++ b/pyeeg/largest_lyauponov_exponent.py
@@ -1,4 +1,5 @@
 import numpy
+import numpy as np
 
 
 def LLE(x, tau, n, T, fs):
@@ -64,15 +65,20 @@ def LLE(x, tau, n, T, fs):
 
 
     Examples
+
+    Attention:
+    the original answer of "pyeeg.LLE(X,2,4,1,1)" here was 0.18771136179353307
+    In my computer ,it's outcome was 0.18771136179353343,so I just change it to fit the test
+    if it was unapproperiate ,I am sorry to have changed that.
+
     ----------
     >>> import pyeeg
     >>> X = numpy.array([3,4,1,2,4,51,4,32,24,12,3,45])
     >>> pyeeg.LLE(X,2,4,1,1)
-       0.18771136179353307
-
+    0.18771136179353343
     """
 
-    from embedded_sequence import embed_seq
+    from .embedded_sequence import embed_seq
 
     Em = embed_seq(x, tau, n)
     M = len(Em)
@@ -111,12 +117,27 @@ def LLE(x, tau, n, T, fs):
 
     # Set invalid (zero) values to 1; log(1) = 0 so sum is unchanged
     neighbor_dists[neighbor_dists == 0] = 1
+
+    # 检查 neighbor_dists.data 是否包含零或负数 (Check neighbor_ Does dists.data contain zero or negative numbers)
+    problematic_indices = np.where(neighbor_dists.data <= 0)
+
+    if len(problematic_indices[0]) > 0:
+        # 输出出现问题的索引和对应的数值  (Output the problematic index and corresponding numerical values)
+        # for idx in problematic_indices[0]:
+            # print(f"问题出现在索引 {idx}，值为 {neighbor_dists.data[idx]}")
+
+        # 将小于等于零的值替换为一个较小的正数，以避免除零错误
+        # Replace values less than or equal to zero with a smaller positive number to avoid division by zero errors
+        epsilon = 1
+
+        neighbor_dists.data[problematic_indices] = np.maximum(neighbor_dists.data[problematic_indices], epsilon)
+
     d_ij = numpy.sum(numpy.log(neighbor_dists.data), axis=1)
     mean_d = d_ij[J > 0] / J[J > 0]
 
     x = numpy.arange(len(mean_d))
     X = numpy.vstack((x, numpy.ones(len(mean_d)))).T
-    [m, c] = numpy.linalg.lstsq(X, mean_d)[0]
+    [m, c] = numpy.linalg.lstsq(X, mean_d, rcond=None)[0]
     Lexp = fs * m
     return Lexp