在序列数组中找到最佳匹配数字序列

Question

假设我有这两个数组：

float arr[] = {40.4357,40.6135,40.2477,40.2864,39.3449,39.8901,40.103,39.9959,39.7863,39.9102,39.2652,39.2688,39.5147,38.2246,38.5376,38.4512,38.9951,39.0999,39.3057,38.53,38.2761,38.1722,37.8816,37.6521,37.8306,38.0853,37.9644,38.0626,38.0567,38.3518,38.4044,38.3553,38.4978,38.3768,38.2058,38.3175,38.3123,38.262,38.0093,38.3685,38.0111,38.4539,38.8122,39.1413,38.9409,39.2043,39.3538,39.4123,39.3628,39.2825,39.1898,39.0431,39.0634,38.5993,38.252,37.3793,36.6334,36.4009,35.2822,34.4262,34.2119,34.1552,34.3325,33.9626,33.2661,32.3819,35.1959,36.7602,37.9039,37.8103,37.5832,37.9718,38.3111,38.9323,38.6763,39.1163,38.8469,39.805,40.2627,40.3689,40.4064,40.0558,40.815,41.0234,41.0128,41.0296,41.0927,40.7046,40.6775,40.2711,40.1283,39.7518,40.0145,40.0394,39.8461,39.6317,39.5548,39.1996,38.9861,38.8507,38.8603,38.483,38.4711,38.4214,38.4286,38.5766,38.7532,38.7905,38.6029,38.4635,38.1403,36.6844,36.616,36.4053,34.7934,34.0226,33.0505,33.4978,34.6106,35.284,35.7535,35.3541,35.5481,35.4086,35.7096,36.0526,36.1222,35.9408,36.1007,36.7952,36.99,37.1024,37.0993,37.3144,36.6951,37.1213,38.0026,38.1266,39.2538,38.8963,39.0158,38.6235,38.7908,38.6041,38.4489,38.3207,37.7398,38.5304,38.925,38.7249,38.9221,39.1704,39.5113,40.0613,39.3602,39.8689,39.973,40.0524,40.0025,40.7584,40.9714,40.9106,40.9685,40.6554,39.7314,39.0044,38.7183,38.5163,38.6101,38.2004,38.7606,38.7532,37.8903,37.8403,38.5368,39.0462,38.8279,39.0748,39.2907,38.5447,38.423,38.5624,38.476,38.5784,39.0905,39.379,39.4739,39.5774,40.7036,40.3044,39.6162,39.9967,40.0562,39.3426,38.666,38.7561,39.2823,38.8548,37.6214,37.8188,38.1086,38.3619,38.5472,38.1357,38.1422,37.95,37.1837,37.4636,36.8852,37.1617,37.5051,37.7724,38.0879,37.7197,38.0422,37.8551,38.5688,38.8388};
float pattern[] = {38.6434,38.1409,37.3391,37.5457,37.7487,37.7499,37.6121,37.4789,37.5821,37.6541,38.0365,37.7907,37.9932,37.9945,37.7032,37.3556,37.6359,37.5412,37.5296,37.8829,38.3797,38.4452,39.0929,39.1233,39.3014,39.0317,38.903,38.8221,39.045,38.6944,39.0699,39.0978,38.9877,38.8123,38.7491,38.5888,38.7875,38.2086,37.7484,37.3961,36.8663,36.2607,35.8838,35.3297,35.5574,35.7239};

Ives上传了这个示例图：

正如您在图表模式中看到的那样，几乎适合索引17处的数组

找到这个指数的最好和最快的方法是什么？有没有办法对匹配保险丝有信心，价值不相同，你可以看到？

Answer 1

如果起始索引是您唯一的自由度，您可以尝试每个索引并计算每个数据点的平方误差总和。在Python中，这可能是这样的：

data = [40.4357,40.6135,40.2477,...]
pattern = [38.6434,38.1409,37.3391,37.5457,37.7487,...]

best_ind, best_err = 0, 1e9999
for i in range(len(data) - len(pattern)):
    subdata = data[i : i + len(pattern)]
    err = sum((d-p)**2 for (d, p) in zip(subdata, pattern))
    if err < best_err:
        best_ind, best_err = i, err

结果：

>>> print best_ind, best_err
17 21.27929269

Answer 2

简单的算法是选择收敛度量（如何描述相似度，这可能是误差的平均值，或者它们的平方值或适合您目的的任何其他函数）并应用步骤

1. 让 i = 0 为整数索引， M 为 size = length（data） - length（pattern）+ 1 存储测量值
2. 如果 i＆lt;尺寸然后按 i 移动您的模式，否则转到第5步
3. 计算相似度并存储到 M
4. i = i + 1 ，转到2并重复
5. 选择 M
中的最小值索引

Answer 3

它是Python中的一个单行程序，使用元组按字典顺序排序的事实：

In [1]:

import numpy as np
arr = np.array( [ 40.4357,40.6135,40.2477,40.2864,39.3449,39.8901,40.103,39.9959,39.7863,39.9102,39.2652,39.2688,39.5147,38.2246,38.5376,38.4512,38.9951,39.0999,39.3057,38.53,38.2761,38.1722,37.8816,37.6521,37.8306,38.0853,37.9644,38.0626,38.0567,38.3518,38.4044,38.3553,38.4978,38.3768,38.2058,38.3175,38.3123,38.262,38.0093,38.3685,38.0111,38.4539,38.8122,39.1413,38.9409,39.2043,39.3538,39.4123,39.3628,39.2825,39.1898,39.0431,39.0634,38.5993,38.252,37.3793,36.6334,36.4009,35.2822,34.4262,34.2119,34.1552,34.3325,33.9626,33.2661,32.3819,35.1959,36.7602,37.9039,37.8103,37.5832,37.9718,38.3111,38.9323,38.6763,39.1163,38.8469,39.805,40.2627,40.3689,40.4064,40.0558,40.815,41.0234,41.0128,41.0296,41.0927,40.7046,40.6775,40.2711,40.1283,39.7518,40.0145,40.0394,39.8461,39.6317,39.5548,39.1996,38.9861,38.8507,38.8603,38.483,38.4711,38.4214,38.4286,38.5766,38.7532,38.7905,38.6029,38.4635,38.1403,36.6844,36.616,36.4053,34.7934,34.0226,33.0505,33.4978,34.6106,35.284,35.7535,35.3541,35.5481,35.4086,35.7096,36.0526,36.1222,35.9408,36.1007,36.7952,36.99,37.1024,37.0993,37.3144,36.6951,37.1213,38.0026,38.1266,39.2538,38.8963,39.0158,38.6235,38.7908,38.6041,38.4489,38.3207,37.7398,38.5304,38.925,38.7249,38.9221,39.1704,39.5113,40.0613,39.3602,39.8689,39.973,40.0524,40.0025,40.7584,40.9714,40.9106,40.9685,40.6554,39.7314,39.0044,38.7183,38.5163,38.6101,38.2004,38.7606,38.7532,37.8903,37.8403,38.5368,39.0462,38.8279,39.0748,39.2907,38.5447,38.423,38.5624,38.476,38.5784,39.0905,39.379,39.4739,39.5774,40.7036,40.3044,39.6162,39.9967,40.0562,39.3426,38.666,38.7561,39.2823,38.8548,37.6214,37.8188,38.1086,38.3619,38.5472,38.1357,38.1422,37.95,37.1837,37.4636,36.8852,37.1617,37.5051,37.7724,38.0879,37.7197,38.0422,37.8551,38.5688,38.8388] )
pattern = np.array( [ 38.6434,38.1409,37.3391,37.5457,37.7487,37.7499,37.6121,37.4789,37.5821,37.6541,38.0365,37.7907,37.9932,37.9945,37.7032,37.3556,37.6359,37.5412,37.5296,37.8829,38.3797,38.4452,39.0929,39.1233,39.3014,39.0317,38.903,38.8221,39.045,38.6944,39.0699,39.0978,38.9877,38.8123,38.7491,38.5888,38.7875,38.2086,37.7484,37.3961,36.8663,36.2607,35.8838,35.3297,35.5574,35.7239 ] )

min( ( ( ( arr[i:i+len(pattern)] - pattern ) ** 2 ).mean(), i ) for i in xrange(len(arr)-len(pattern)) )

Out[5]:
(0.46259331934782588, 17) 

其中0.46是最小均方误差，17是arr中最小值的位置。