我有两组数据，它们有一点重叠（见下图）。我需要找到这两组数据之间的一个点，这个点可以让我们猜测一个未知数据点应该属于哪个特定类别。

如果我有一个新的数据点（假设是5020），并且必须在它属于A组还是B组上押注$$$，我该如何计算一个点来使我的押注最有把握？

请看下面的样本数据集和相应的图表，其中包含了这两组之间的近似点（通过目测计算得出）。

GROUP A[385,515,975,1136,2394,2436,4051,4399,4484,4768,4768,4849,4856,4954,5020,5020,5020,5020,5020,5020,5020,5020,5020,5052,5163,5200,5271,5421,5421,5442,5746,5765,5903,5992,5992,6046,6122,6205,6208,6239,6310,6360,6416,6512,6536,6543,6581,6609,6696,6699,6752,6796,6806,6855,6859,6886,6906,6911,6923,6953,7016,7072,7086,7089,7110,7232,7278,7293,7304,7309,7348,7367,7378,7380,7419,7453,7454,7492,7506,7549,7563,7721,7723,7731,7745,7750,7751,7783,7791,7813,7813,7814,7818,7833,7863,7875,7886,7887,7902,7907,7935,7942,7942,7948,7973,7995,8002,8013,8013,8015,8024,8025,8030,8038,8041,8050,8056,8060,8064,8071,8081,8082,8085,8093,8124,8139,8142,8167,8179,8204,8214,8223,8225,8247,8248,8253,8258,8264,8265,8265,8269,8277,8278,8289,8300,8312,8314,8323,8328,8334,8363,8369,8390,8397,8399,8399,8401,8436,8442,8456,8457,8471,8474,8483,8503,8511,8516,8533,8560,8571,8575,8583,8592,8593,8626,8635,8635,8644,8659,8685,8695,8695,8702,8714,8715,8717,8729,8732,8740,8743,8750,8756,8772,8772,8778,8797,8828,8840,8840,8843,8856,8865,8874,8876,8878,8885,8887,8893,8896,8905,8910,8955,8970,8971,8991,8995,9014,9016,9042,9043,9063,9069,9104,9106,9107,9116,9131,9157,9227,9359,9471]GROUP B[12,16,29,32,33,35,39,42,44,44,44,45,45,45,45,45,45,45,45,45,47,51,51,51,57,57,60,61,61,62,71,75,75,75,75,75,75,76,76,76,76,76,76,79,84,84,85,89,93,93,95,96,97,98,100,100,100,100,100,102,102,103,105,108,109,109,109,109,109,109,109,109,109,109,109,109,110,110,112,113,114,114,116,116,118,119,120,121,122,124,125,128,129,130,131,132,133,133,137,138,144,144,146,146,146,148,149,149,150,150,150,151,153,155,157,159,164,164,164,167,169,170,171,171,171,171,173,174,175,176,176,177,178,179,180,181,181,183,184,185,187,191,193,199,203,203,205,205,206,212,213,214,214,219,224,224,224,225,225,226,227,227,228,231,234,234,235,237,240,244,245,245,246,246,246,248,249,250,250,251,255,255,257,264,264,267,270,271,271,281,282,286,286,291,291,292,292,294,295,299,301,302,304,304,304,304,304,306,308,314,318,329,340,344,345,356,359,363,368,368,371,375,379,386,389,390,392,394,408,418,438,440,456,456,458,460,461,467,491,503,505,508,524,557,558,568,591,609,622,656,665,668,687,705,728,817,839,965,1013,1093,1126,1512,1935,2159,2384,2424,2426,2484,2738,2746,2751,3006,3184,3184,3184,3184,3184,4023,5842,5842,6502,7443,7781,8132,8237,8501]

数组统计数据：

                      A组       B组总数                     231        286平均值                 7534.71     575.56标准差                 1595.04    1316.03

回答：

我想指出另一种使用密度估计的方法。

根据你的数据，很容易使用核密度估计来拟合一个平滑的概率密度函数。下面的Python代码展示了如何使用scipy中的kde模块。

from scipy.stats.kde import gaussian_kdefrom numpy import linspaceimport matplotlib.pyplot as pltdata1 = [385,515,975,1136,2394,2436,4051,4399,4484,4768,4768,4849,4856,4954,5020,5020,5020,5020,5020,5020,5020,5020,5020,5052,5163,5200,5271,5421,5421,5442,5746,5765,5903,5992,5992,6046,6122,6205,6208,6239,6310,6360,6416,6512,6536,6543,6581,6609,6696,6699,6752,6796,6806,6855,6859,6886,6906,6911,6923,6953,7016,7072,7086,7089,7110,7232,7278,7293,7304,7309,7348,7367,7378,7380,7419,7453,7454,7492,7506,7549,7563,7721,7723,7731,7745,7750,7751,7783,7791,7813,7813,7814,7818,7833,7863,7875,7886,7887,7902,7907,7935,7942,7942,7948,7973,7995,8002,8013,8013,8015,8024,8025,8030,8038,8041,8050,8056,8060,8064,8071,8081,8082,8085,8093,8124,8139,8142,8167,8179,8204,8214,8223,8225,8247,8248,8253,8258,8264,8265,8265,8269,8277,8278,8289,8300,8312,8314,8323,8328,8334,8363,8369,8390,8397,8399,8399,8401,8436,8442,8456,8457,8471,8474,8483,8503,8511,8516,8533,8560,8571,8575,8583,8592,8593,8626,8635,8635,8644,8659,8685,8695,8695,8702,8714,8715,8717,8729,8732,8740,8743,8750,8756,8772,8772,8778,8797,8828,8840,8840,8843,8856,8865,8874,8876,8878,8885,8887,8893,8896,8905,8910,8955,8970,8971,8991,8995,9014,9016,9042,9043,9063,9069,9104,9106,9107,9116,9131,9157,9227,9359,9471]data2 = [12,16,29,32,33,35,39,42,44,44,44,45,45,45,45,45,45,45,45,45,47,51,51,51,57,57,60,61,61,62,71,75,75,75,75,75,75,76,76,76,76,76,76,79,84,84,85,89,93,93,95,96,97,98,100,100,100,100,100,102,102,103,105,108,109,109,109,109,109,109,109,109,109,109,109,109,110,110,112,113,114,114,116,116,118,119,120,121,122,124,125,128,129,130,131,132,133,133,137,138,144,144,146,146,146,148,149,149,150,150,150,151,153,155,157,159,164,164,164,167,169,170,171,171,171,171,173,174,175,176,176,177,178,179,180,181,181,183,184,185,187,191,193,199,203,203,205,205,206,212,213,214,214,219,224,224,224,225,225,226,227,227,228,231,234,234,235,237,240,244,245,245,246,246,246,248,249,250,250,251,255,255,257,264,264,267,270,271,271,281,282,286,286,291,291,292,292,294,295,299,301,302,304,304,304,304,304,306,308,314,318,329,340,344,345,356,359,363,368,368,371,375,379,386,389,390,392,394,408,418,438,440,456,456,458,460,461,467,491,503,505,508,524,557,558,568,591,609,622,656,665,668,687,705,728,817,839,965,1013,1093,1126,1512,1935,2159,2384,2424,2426,2484,2738,2746,2751,3006,3184,3184,3184,3184,3184,4023,5842,5842,6502,7443,7781,8132,8237,8501]pdf1 = gaussian_kde(data1)pdf2 = gaussian_kde(data2)x = linspace(0, 9500, 1000)plt.plot(x, pdf1(x),'r')plt.plot(x, pdf2(x),'g')plt.legend(['data1 pdf', 'data2 pdf'])plt.show()

在图中，绿色是第二个数据集的概率密度函数；红色是第一个数据集的概率密度函数。显然，决策边界是通过绿色与红色相交的点所经过的垂直线。

要数值上找到边界，我们可以执行类似下面的操作（假设只有一个交点，否则没有意义）：

min_diff = 10000min_diff_x = -1for x in linspace(3600, 4000, 400):    diff = abs(pdf1(x) - pdf2(x))    if diff < min_diff:        min_diff = diff        min_diff_x = xprint min_diff, min_diff_x

我们发现边界大约位于3762处。

如果两个概率密度函数有多个交点，要预测数据点x属于哪个类别，我们计算pdf1(x)和pdf2(x)，最大值对应的类别是贝叶斯风险最小化的类别。有关贝叶斯风险和预测错误概率评估的更多详细信息，请参见这里。

下图展示了一个包含三个概率密度函数的示例，在任何查询点x处，我们应该分别询问这三个概率密度函数，并选择pdf(x)值最大的那个作为预测类别。