python - 如何找到三个参数以最佳方式将曲线拟合到python中的另一条曲线中？

如何解决python - 如何找到三个参数以最佳方式将曲线拟合到python中的另一条曲线中？

我有两个数据集 (t1,y1) 和 (t2,y2)，它们在绘制时会给出各自的曲线。

以下是我的数据：

t1 = [-2.,-1.6877551,-1.3755102,-1.06326531,-0.75102041,-0.43877551,-0.12653061,0.18571429,0.49795918,0.81020408,1.12244898,1.43469388,1.74693878,2.05918367,2.37142857,2.68367347,2.99591837,3.30816327,3.62040816,3.93265306,4.24489796,4.55714286,4.86938776,5.18163265,5.49387755,5.80612245,6.11836735,6.43061224,6.74285714,7.05510204,7.36734694,7.67959184,7.99183673,8.30408163,8.61632653,8.92857143,9.24081633,9.55306122,9.86530612,10.17755102,10.48979592,10.80204082,11.11428571,11.42653061,11.73877551,12.05102041,12.36326531,12.6755102,12.9877551,13.3       ]

y1 = [-0.21425007,-0.16045716,-0.10789988,-0.05657822,-0.00649218,0.04235823,0.08997303,0.1363522,0.18149575,0.22540368,0.26807599,0.30951268,0.34971374,0.38867919,0.42640901,0.46290321,0.49816179,0.53218474,0.56497208,0.59652379,0.62683989,0.65592036,0.6837652,0.71037443,0.73574804,0.75988602,0.78278838,0.80445512,0.82488624,0.84408174,0.86204161,0.87876586,0.8942545,0.90850751,0.92152489,0.93330666,0.94385281,0.95316333,0.96123823,0.96807751,0.97368117,0.97804921,0.98118162,0.98307842,0.98373959,0.98316514,0.98135507,0.97830937,0.97402806,0.96851112]

t2 =  [930.,930.70707071,931.41414141,932.12121212,932.82828283,933.53535354,934.24242424,934.94949495,935.65656566,936.36363636,937.07070707,937.77777778,938.48484848,939.19191919,939.8989899,940.60606061,941.31313131,942.02020202,942.72727273,943.43434343,944.14141414,944.84848485,945.55555556,946.26262626,946.96969697,947.67676768,948.38383838,949.09090909,949.7979798,950.50505051,951.21212121,951.91919192,952.62626263,953.33333333,954.04040404,954.74747475,955.45454545,956.16161616,956.86868687,957.57575758,958.28282828,958.98989899,959.6969697,960.4040404,961.11111111,961.81818182,962.52525253,963.23232323,963.93939394,964.64646465,965.35353535,966.06060606,966.76767677,967.47474747,968.18181818,968.88888889,969.5959596,970.3030303,971.01010101,971.71717172,972.42424242,973.13131313,973.83838384,974.54545455,975.25252525,975.95959596,976.66666667,977.37373737,978.08080808,978.78787879,979.49494949,980.2020202,980.90909091,981.61616162,982.32323232,983.03030303,983.73737374,984.44444444,985.15151515,985.85858586,986.56565657,987.27272727,987.97979798,988.68686869,989.39393939,990.1010101,990.80808081,991.51515152,992.22222222,992.92929293,993.63636364,994.34343434,995.05050505,995.75757576,996.46464646,997.17171717,997.87878788,998.58585859,999.29292929,1000.        ]

y2 = [-0.11235382,-0.01261221,0.08019689,0.16640183,0.24632084,0.32026221,0.3885244,0.45139624,0.50915704,0.56207678,0.61041626,0.65442723,0.69435257,0.73042644,0.76287442,0.79191369,0.81775315,0.84059362,0.86062795,0.87804121,0.89301081,0.90570669,0.91629144,0.9249205,0.93174226,0.93689826,0.94052332,0.94274571,0.94368728,0.94346364,0.94218431,0.93995286,0.93686709,0.93301915,0.92849573,0.92337819,0.91774273,0.91166053,0.90519793,0.89841654,0.89137345,0.88412133,0.87670865,0.86917975,0.86157507,0.85393127,0.84628139,0.83865501,0.83107838,0.82357461,0.81616382,0.80886326,0.8016875,0.79464857,0.78775612,0.78101757,0.77443827,0.76802165,0.76176937,0.75568149,0.74975662,0.74399206,0.73838396,0.73292749,0.72761698,0.72244608,0.71740792,0.71249524,0.70770059,0.70301643,0.69843534,0.69395011,0.68955396,0.68524067,0.68100472,0.67684144,0.67274721,0.66871956,0.66475737,0.66086099,0.65703242,0.65327544,0.64959579,0.6460013,0.64250207,0.63911062,0.63584201,0.63271404,0.62974738,0.62696574,0.62439601,0.62206842,0.62001669,0.6182782,0.61689412,0.61590959,0.61537386,0.61534045,0.61586731,0.61701695]

因此任务是通过使用依赖于在 3 个参数上，它们是：将使其与 curve1 匹配的 curve2 的高度，x 轴上 t2 中的 移位 将使 curve2 位于 curve1 的顶部，以及然后在 t2 中对曲线 2 本身进行拉伸，以使其适合曲线 1。

或者换一种方式

我正在尝试将 (t2,y2) 中的 curve2 匹配到 (t1,t2) 中的 curve1，然后当它们适当地拟合在一起时，我需要弄清楚：

1. 我需要多少时间让曲线 2“回到过去”（很清楚 t2 从 930 开始，而 t1 从 0 开始）
2. 如果曲线 2 必须在高度上拉伸（拉伸多少？）
3. t2 中的微小调整（称为拉伸），使曲线的起点/上升尽可能对齐

我做了什么：

我理解该任务与曲线拟合有些相关，但我似乎没有做对，因为我无法提出曲线拟合所需的曲线 2 的良好数学函数。 您认为我应该如何实现这一目标？

解决方法

通过观察，曲线2主要由两部分组成（如下图所示）：

对于低 t 值，函数看起来像指数函数。

对于高 t 值，该函数看起来像一个逻辑函数。

将两者线性结合，提出了一个更通用的函数：

它们是三个线性作用的参数 a,b,c 和三个非自由作用的参数 p,q,r。需要非线性回归来计算它们。

RMSE=0.017 这还不错，但没有预期的那么好。或许人们可以期待找到更方便的功能以更好地适应。

,

可能有一些库来实现建议的线性变换，可惜我不知道它们。那么，如果这只是一个例子，为什么不直接编码呢？

double maxSeconds = LocalTime.MAX.toSecondOfDay();
int seconds = (int) (maxSeconds * fraction);
LocalTime time = LocalTime.ofSecondOfDay(seconds);
LocalDateTime stamp = date.atTime(time);
System.out.println(stamp);
> 2021-02-06T20:45:14

输出：

不同的曲率，这是肯定的。如何量化这种曲率差异是另一个话题。