python – 在曲线上找到斜率变化的位置

我有时间和电压的数据点,创建如下所示的曲线.

时间数据是

array([  0.10810811,   0.75675676,   1.62162162,   2.59459459,
         3.56756757,   4.21621622,   4.97297297,   4.97297297,
         4.97297297,   4.97297297,   4.97297297,   4.97297297,
         4.97297297,   4.97297297,   5.08108108,   5.18918919,
         5.2972973 ,   5.51351351,   5.72972973,   5.94594595,
         6.27027027,   6.59459459,   7.13513514,   7.67567568,
         8.32432432,   9.18918919,  10.05405405,  10.91891892,
        11.78378378,  12.64864865,  13.51351351,  14.37837838,
        15.35135135,  16.32432432,  17.08108108,  18.16216216,
        19.02702703,  20.        ,  20.        ,  20.        ,
        20.        ,  20.        ,  20.        ,  20.        ,
        20.10810811,  20.21621622,  20.43243243,  20.64864865,
        20.97297297,  21.40540541,  22.05405405,  22.91891892,
        23.78378378,  24.86486486,  25.83783784,  26.7027027 ,
        27.56756757,  28.54054054,  29.51351351,  30.48648649,
        31.56756757,  32.64864865,  33.62162162,  34.59459459,
        35.67567568,  36.64864865,  37.62162162,  38.59459459,
        39.67567568,  40.75675676,  41.83783784,  42.81081081,
        43.89189189,  44.97297297,  46.05405405,  47.02702703,
        48.10810811,  49.18918919,  50.27027027,  51.35135135,
        52.43243243,  53.51351351,  54.48648649,  55.56756757,
        56.75675676,  57.72972973,  58.81081081,  59.89189189])

伏特数据是

array([ 4.11041056,  4.11041056,  4.11041056,  4.11041056,  4.11041056,
        4.11041056,  4.11041056,  4.10454545,  4.09794721,  4.09208211,
        4.08621701,  4.07961877,  4.07228739,  4.06568915,  4.05909091,
        4.05175953,  4.04516129,  4.03782991,  4.03123167,  4.02463343,
        4.01803519,  4.01217009,  4.00557185,  3.99970674,  3.99384164,
        3.98797654,  3.98284457,  3.97771261,  3.97331378,  3.96891496,
        3.96451613,  3.96085044,  3.95645161,  3.95205279,  3.9483871 ,
        3.94398827,  3.94032258,  3.93665689,  3.94325513,  3.94985337,
        3.95645161,  3.96378299,  3.97038123,  3.97624633,  3.98284457,
        3.98944282,  3.99604106,  4.0026393 ,  4.00923754,  4.01510264,
        4.02096774,  4.02609971,  4.02903226,  4.03196481,  4.03416422,
        4.0356305 ,  4.03709677,  4.03856305,  4.03929619,  4.04002933,
        4.04076246,  4.04222874,  4.04296188,  4.04296188,  4.04369501,
        4.04442815,  4.04516129,  4.04516129,  4.04589443,  4.04589443,
        4.04662757,  4.04662757,  4.0473607 ,  4.0473607 ,  4.04809384,
        4.04809384,  4.04809384,  4.04882698,  4.04882698,  4.04882698,
        4.04956012,  4.04956012,  4.04956012,  4.04956012,  4.05029326,
        4.05029326,  4.05029326,  4.05029326])

我想确定标记为A,B,C,D和E的点的位置.点A是斜率从零到未定义的第一个位置. B点是线不再垂直的位置. C点是曲线的最小值. D点是曲线不再垂直的地方. E点是斜率再次接近零的位置.下面的Python代码确定了A点和C点的位置.

tdiff = np.diff(time)
vdiff = np.diff(volts)

# point A
idxA = np.where(vdiff < 0)[0][0]
timeA = time[idxA]
voltA = volts[idxA]

# point C
idxc = volts.idxmin()
timeC = time[idxc]
voltC = volts[idxc]

如何确定曲线B,D和E所代表的曲线上的其他位置？

解决方法:

您正在寻找标记斜率变为或从零或无穷大的任何位置的点.我们实际上并不需要在任何地方计算斜率：yn-yn-1 == 0和yn 1-yn！= 0,反之亦然,或x相同.

我们可以采用x的差异.如果两个连续元素中的一个为零,那么diff的diff将是该点处的diff或负diff.因此,我们只想找到并标记diff(x)== diff(diff(x))和diff(x)！= 0的所有点,并根据数组之间的大小差异进行适当调整.我们也想要y所有相同的点.

在numpy术语中,这可以写成如下

def masks(vec):
    d = np.diff(vec)
    dd = np.diff(d)

    # Mask of locations where graph goes to vertical or horizontal, depending on vec
    to_mask = ((d[:-1] != 0) & (d[:-1] == -dd))
    # Mask of locations where graph comes from vertical or horizontal, depending on vec
    from_mask = ((d[1:] != 0) & (d[1:] == dd))
    return to_mask, from_mask

to_vert_mask, from_vert_mask = masks(time)
to_horiz_mask, from_horiz_mask = masks(volts)

请记住,掩码是根据二阶差异计算的,因此它们比输入短两个元素.掩码中的元素对应于输入数组中的元素,在前沿和后沿具有单元素边界(因此下面的索引为[1：-1]).您可以使用np.nonzero将蒙版转换为索引,或者您可以使用蒙版作为索引直接获取x值和y值：

def apply_mask(mask, x, y):
    return x[1:-1][mask], y[1:-1][mask]

to_vert_t, to_vert_v = apply_mask(to_vert_mask, time, volts)
from_vert_t, from_vert_v = apply_mask(from_vert_mask, time, volts)
to_horiz_t, to_horiz_v = apply_mask(to_horiz_mask, time, volts)
from_horiz_t, from_horiz_v = apply_mask(from_horiz_mask, time, volts)

plt.plot(time, volts, 'b-')
plt.plot(to_vert_t, to_vert_v, 'r^', label='Plot goes vertical')
plt.plot(from_vert_t, from_vert_v, 'kv', label='Plot stops being vertical')
plt.plot(to_horiz_t, to_horiz_v, 'r>', label='Plot goes horizontal')
plt.plot(from_horiz_t, from_horiz_v, 'k<', label='Plot stops being horizontal')
plt.legend()
plt.show()

这是结果图：

请注意,由于分类是单独进行的,因此“点A”被正确识别为垂直度开始和水平度结束的点.问题是根据这些标准,“E点”似乎不可解析.放大显示所有增殖点正确识别水平线段：

您可以通过完全丢弃from_horiz来选择“Point E”的“正确”版本,并且只选择to_horiz中的最后一个值：

to_horiz_t, to_horiz_v = apply_mask(to_horiz_mask, time, volts)
to_horiz_t, to_horiz_v = to_horiz_t[-1], to_horiz_v[-1]

plt.plot(time, volts, 'b-')
plt.plot(*apply_mask(to_vert_mask, time, volts), 'r^', label='Plot goes vertical')
plt.plot(*apply_mask(from_vert_mask, time, volts), 'kv', label='Plot stops being vertical')
plt.plot(to_horiz_t, to_horiz_v, 'r>', label='Plot goes horizontal')
plt.legend()
plt.show()

我使用它作为apply_mask结果的星形扩展的展示.结果图是：

这几乎就是你要找的情节.丢弃from_horiz也会使“点A”仅被识别为垂直下降,这很好.

由于to_horiz中的多个值显示,此方法对数据中的噪声非常敏感.您的数据非常顺利,但这种方法不太适用于未经过滤的原始测量.

python – 在曲线上找到斜率变化的位置

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