一、题目描述
已知贵州茅台的2019年1月至今每天的股票收益率序列（数据见下），采用移动窗口平均法预测，计算预测误差均值。并在[2,20]范围内求使得该误差最小的移动窗口长度。
移动窗口平均法预测的计算方法说明如下：
原始序列：0.3,0.1,0.5,0.4,0.6,0.3,0.6,0.8,0.3,0.2
当窗口长度=3时，步骤如下：
- 对0.3,0.1,0.5求平均得到0.3，作为第四天的预测值,该预测值与真实值的误差为：|0.4-0.3|=0.1
- 在原始序列上移动一个位置，即求0.1，0.5,0.4的平均值得到0.33作为第五天的预测值，误差为：|0.6-0.33|=0.27
- 继续往右移动一个位置，求0.5,0.4,0.6的平均值0.5作为第六天的预测值，误差为|0.3-0.5|=0.2
- 以此类推，直到0.6,0.8,0.3，误差为|0.2-0.57|=0.37把各个步骤的误差求平均得到预测误差均值。
股票的原始为：
1. num_list = [0.02931,0.048,0.0076,0.00302, 2. 0.03759,-0.0058,-0.02738,0.00462, 3. 0.00832,0.01201,-0.0095,0.01106, 4. -0.00728,0.02506,-0.00293,0.04029, 5. 0,0.01134,0.00503,-0.00528,0.01363, 6. -0.00804,-0.02041,-0.00417,0.01918, 7. 0.02333,-0.01979,0.00668,0.03198,0.04399, 8. -0.01258,-0.01274,-0.01649,-0.03191,0.0029, 9. 0.03213,-0.02398,0.01543,0.0251,0.00361,0.03107, 10. -0.01619,-0.00253,0.00028,-0.00282,-0.00763,0.00128, 11. 0.01592,0.05239,0.02994,-0.00001,-0.0231,-0.00014, 12. 0.07143,0.00333,0,0.04983,-0.04114,0.0242,-0.02803, 13. 0.03658,0.0079,-0.00153,0.01646,-0.00886,0.03037,-0.01132, 14. -0.02321,-0.00137,0.02434,-0.0439,-0.03481,-0.03726,0.01744, 15. 0.00075,0.02649,-0.01542,0.00592,0.03231,0.00873,-0.02915, 16. -0.01667,0.01128,-0.01562,-0.01816,0.0163,-0.00126,0.01141, 17. 0.02476,-0.0104,-0.00947,0.00336,-0.01229,-0.00948,-0.02015, 18. 0.021,0.0345,0.00417,0.0011,0.12514,-0.00488,-0.03353,0.0002, 19. 0.00918,-0.01583,0.00166,0.00921,-0.00875,-0.01376,0.0079, 20. -0.00764,-0.00671,-0.00676,0.00524,0.0024,-0.00415,-0.01288, 21. 0.01691,0.00006,0.01504,-0.00154,0.00155,-0.03329,0.00106, 22. -0.01481,0.01987,0.00421,0.02622,0.03251,0.00579,0.01364, 23. -0.00196,0.02125,0.01063,-0.0566,0.01431,0.0027,0.02983, 24. 0.00724,0.00359,-0.00716,-0.00361,0.0181,0.01332, 25. 0.00001,0.0007,-0.01034,0.01373,0.00044,-0.0934, 26. -0.01329,-0.04755,0.01595,0.01016,0.01325, 27. 0.03563,0.00261,0.00521,-0.00173,0.02165, 28. -0.00429,-0.00931,-0.00086,-0.0086,0.01145, 29. -0.03361,0.02662,0.02506,-0.00506,-0.01017, 30. 0.00685,0.00255,-0.00606,0.00524,-0.00849, 31. -0.00261,-0.01,0.01102,0.01295,0.00753, 32. 0.00504,-0.01254,0.00762,0.00798,-0.00208, 33. -0.00041,0.0071,-0.0029,0.00208,0.00249, 34. 0.01657,0.00244,-0.00397,0.0025,0.00163, 35. 0.00081,-0.0065,-0.02858,-0.00115,0.00516, 36. 0.00182,-0.02466,-0.04058,0.01324,0.00618, 37. -0.01316,0.00975,0.03436,-0.01311,0.00724,-0.01027, 38. -0.00865,0.01483,-0.01144,0.02114,-0.00937, 39. -0.00071,-0.01995,0.01229,-0.00867,-0.00962, 40. 0.0159,0.01757,0.01094,-0.04649,-0.00975,-0.04131, 41. 0.0062,0.00701,0.00825,0.01371,0.00316,0.01052, 42. -0.01351,0.00889,-0.00793,0.00168,-0.02776,-0.01018, 43. 0.00561,-0.08457,0.03046,0.03448,0.00898,0.00999, 44. -0.00749,0.00094,0.02446,0.00826,-0.00688,-0.00729, 45. 0.00691,-0.0046,0.01078,0.01346,-0.00536,-0.02488, 46. -0.01484,0.01318,-0.0053,-0.01886,0.0456,0.02459, 47. 0.01005,0.02349,-0.02408,-0.01938,0.03819,-0.01236, 48. -0.04488,0.01283,-0.04437,-0.01422,-0.04424,0.01711, 49. -0.01088,0.043,0.04313,-0.01347,0.01087,-0.02281, 50. 0.02051,0.03235,-0.01164,0.03173,0.02012,-0.00856,0.01215,-0.00696]
二、题目分析
首先我们要知道什么是滑动窗口：
滑动窗口就相当于是双指针类型的题目。在left和right指针框起来的范围，就是需要进行相应判断的窗口。通过两个指针向前滑动，形成一个连续移动的矩形框窗口，对这个窗口内的元素进行相应判断。就是滑动窗口。
有了这个定义我们就很清楚啦 ，这题主要是分两步：
- 首先我们可以以窗口长度为3时举例，设定一个参数left为0，right为left+3，对股票序列进行截取，并且每截取一次left+=1，right+=1，这样就能不断截取；截取的列表求平均值，和索引为right的值取绝对值存到另一个列表fin_result中。直到right的值等于整个序列的长度，退出循环，对fin_result求和，并且除以计数器count即框选了多少次，就能求在窗口长度为3时的预测误差均值。
- 有了窗口长度为3举例，我们只要用for循环，从2到21中的每一个值作为窗格长度去求预测误差均值就好，思路都是一样的。最后用一个列表保存将值起来，并且用一个列表将窗口长度和对应的值关联起来，方便之后打印。
三、题目代码
1. #!/usr/bin/env python 2. # -*- coding: UTF-8 -*- 3. """ 4. @Project ：python 5. @File ：python_23_4.py 6. @IDE ：PyCharm 7. @Author ：咋 8. @Date ：2022/10/21 20:10 9. """ 10. #导入库 11. import matplotlib.pyplot as plt 12. import numpy as np 13. #设定画布。dpi越大图越清晰，绘图时间越久 14. num_list = [0.02931,0.048,0.0076,0.00302, 15. 0.03759,-0.0058,-0.02738,0.00462, 16. 0.00832,0.01201,-0.0095,0.01106, 17. -0.00728,0.02506,-0.00293,0.04029, 18. 0,0.01134,0.00503,-0.00528,0.01363, 19. -0.00804,-0.02041,-0.00417,0.01918, 20. 0.02333,-0.01979,0.00668,0.03198,0.04399, 21. -0.01258,-0.01274,-0.01649,-0.03191,0.0029, 22. 0.03213,-0.02398,0.01543,0.0251,0.00361,0.03107, 23. -0.01619,-0.00253,0.00028,-0.00282,-0.00763,0.00128, 24. 0.01592,0.05239,0.02994,-0.00001,-0.0231,-0.00014, 25. 0.07143,0.00333,0,0.04983,-0.04114,0.0242,-0.02803, 26. 0.03658,0.0079,-0.00153,0.01646,-0.00886,0.03037,-0.01132, 27. -0.02321,-0.00137,0.02434,-0.0439,-0.03481,-0.03726,0.01744, 28. 0.00075,0.02649,-0.01542,0.00592,0.03231,0.00873,-0.02915, 29. -0.01667,0.01128,-0.01562,-0.01816,0.0163,-0.00126,0.01141, 30. 0.02476,-0.0104,-0.00947,0.00336,-0.01229,-0.00948,-0.02015, 31. 0.021,0.0345,0.00417,0.0011,0.12514,-0.00488,-0.03353,0.0002, 32. 0.00918,-0.01583,0.00166,0.00921,-0.00875,-0.01376,0.0079, 33. -0.00764,-0.00671,-0.00676,0.00524,0.0024,-0.00415,-0.01288, 34. 0.01691,0.00006,0.01504,-0.00154,0.00155,-0.03329,0.00106, 35. -0.01481,0.01987,0.00421,0.02622,0.03251,0.00579,0.01364, 36. -0.00196,0.02125,0.01063,-0.0566,0.01431,0.0027,0.02983, 37. 0.00724,0.00359,-0.00716,-0.00361,0.0181,0.01332, 38. 0.00001,0.0007,-0.01034,0.01373,0.00044,-0.0934, 39. -0.01329,-0.04755,0.01595,0.01016,0.01325, 40. 0.03563,0.00261,0.00521,-0.00173,0.02165, 41. -0.00429,-0.00931,-0.00086,-0.0086,0.01145, 42. -0.03361,0.02662,0.02506,-0.00506,-0.01017, 43. 0.00685,0.00255,-0.00606,0.00524,-0.00849, 44. -0.00261,-0.01,0.01102,0.01295,0.00753, 45. 0.00504,-0.01254,0.00762,0.00798,-0.00208, 46. -0.00041,0.0071,-0.0029,0.00208,0.00249, 47. 0.01657,0.00244,-0.00397,0.0025,0.00163, 48. 0.00081,-0.0065,-0.02858,-0.00115,0.00516, 49. 0.00182,-0.02466,-0.04058,0.01324,0.00618, 50. -0.01316,0.00975,0.03436,-0.01311,0.00724,-0.01027, 51. -0.00865,0.01483,-0.01144,0.02114,-0.00937, 52. -0.00071,-0.01995,0.01229,-0.00867,-0.00962, 53. 0.0159,0.01757,0.01094,-0.04649,-0.00975,-0.04131, 54. 0.0062,0.00701,0.00825,0.01371,0.00316,0.01052, 55. -0.01351,0.00889,-0.00793,0.00168,-0.02776,-0.01018, 56. 0.00561,-0.08457,0.03046,0.03448,0.00898,0.00999, 57. -0.00749,0.00094,0.02446,0.00826,-0.00688,-0.00729, 58. 0.00691,-0.0046,0.01078,0.01346,-0.00536,-0.02488, 59. -0.01484,0.01318,-0.0053,-0.01886,0.0456,0.02459, 60. 0.01005,0.02349,-0.02408,-0.01938,0.03819,-0.01236, 61. -0.04488,0.01283,-0.04437,-0.01422,-0.04424,0.01711, 62. -0.01088,0.043,0.04313,-0.01347,0.01087,-0.02281, 63. 0.02051,0.03235,-0.01164,0.03173,0.02012,-0.00856,0.01215,-0.00696] 64. all_result_dic = {} 65. all_result_list = [] 66. y_list = [] 67. for i in range(2,21): 68. fin_result = [] 69. left = 0 70. count = 0 71. right = left+i 72. while True: 73. if right != len(num_list): 74. temp_1 = sum(num_list[left:right])/i 75. temp_2 = abs(num_list[right]-temp_1) 76. fin_result.append(temp_2) 77. left += 1 78. right += 1 79. count += 1 80. else: 81. break 82. all_result_dic[i] = sum(fin_result)/count 83. all_result_list.append(sum(fin_result)/count) 84. k2 = [k for k, v in all_result_dic.items() if v ==min(all_result_list)] 85. for i in range(2,21): 86. y_list.append(all_result_dic[i]) 87. print("当移动窗口长度为:",i,"对应的误差为：",all_result_dic[i]) 88. print("在[2:20]范围内使得该误差最小的移动窗格长度为：",k2[0],"对应的误差为：",all_result_dic[k2[0]]) 89. fig=plt.figure(figsize=(4, 4), dpi=300) 90. #导入数据 91. x=list(np.arange(2, 21)) 92. y= y_list 93. #绘图命令 94. plt.plot(x, y, lw=4, ls='-', c='b', alpha=0.1) 95. plt.plot() 96. #show出图形 97. plt.show() 98. #保存图片 99. fig.savefig("画布")
在最后，我用matplotlib.pyplot做了简单的可视化处理，使结果更加直观，运行结果为：
1. 当移动窗口长度为: 2 对应的误差为： 0.020459651567944236 2. 当移动窗口长度为: 3 对应的误差为： 0.01908844988344989 3. 当移动窗口长度为: 4 对应的误差为： 0.018718245614035088 4. 当移动窗口长度为: 5 对应的误差为： 0.018167260563380288 5. 当移动窗口长度为: 6 对应的误差为： 0.01809260895170789 6. 当移动窗口长度为: 7 对应的误差为： 0.01771991894630194 7. 当移动窗口长度为: 8 对应的误差为： 0.017835907473309605 8. 当移动窗口长度为: 9 对应的误差为： 0.017389440476190475 9. 当移动窗口长度为: 10 对应的误差为： 0.017071688172043017 10. 当移动窗口长度为: 11 对应的误差为： 0.0169103008502289 11. 当移动窗口长度为: 12 对应的误差为： 0.01677201564380265 12. 当移动窗口长度为: 13 对应的误差为： 0.016765702341137122 13. 当移动窗口长度为: 14 对应的误差为： 0.016638602597402602 14. 当移动窗口长度为: 15 对应的误差为： 0.016635518248175184 15. 当移动窗口长度为: 16 对应的误差为： 0.016579548992673998 16. 当移动窗口长度为: 17 对应的误差为： 0.01662127162629757 17. 当移动窗口长度为: 18 对应的误差为： 0.01668552685526856 18. 当移动窗口长度为: 19 对应的误差为： 0.0166987680311891 19. 当移动窗口长度为: 20 对应的误差为： 0.016531403345724907 20. 在[2:20]范围内使得该误差最小的移动窗格长度为： 20 对应的误差为： 0.016531403345724907
四、总结
这道题目的难点在于left和right双变量的运用以及临界点的判断，需要熟悉掌握字典与列表的各项操作，如列表的添加，查找，字典根据值查找键。在最后，用数据可视化作图，使结果更加直观！
