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

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.

标签：python,pandas,numpy,scipy