选择将线绘制在图形的“边缘”上(而不是绕线),而不是在图形上绘制

时间:2020-11-08 20:54:07

标签: python matplotlib cheminformatics

所以,我有一个键的二面角图。 y轴仅为0-360,x轴为帧(请考虑时间步长)。如果值超过360,则需要将图“循环”回零,并绘制两点之间的最短距离(如果需要越过图的边缘并“循环”回而不是跨图) )。
two plots: d5 and d3

d3的图看起来还不错,但实际上需要跳过图的边缘而不是跨越它。

d5的图有一个很大的问题,对于较小的旋转,仅由于它恰好在零度以下才发生大的跳跃。

我希望这两个图都向底部绘制(朝零)并重新出现在图的顶部,从而有效地选择数据点之间的最短距离。我不希望使用涉及平移图的解决方案来消除这些伪像(它确实起作用,但我已经做到了,但是您会丢失有关角度真实值的信息)。可以绘制“低于零”(y轴从300到360 | 0到200到300)的解决方案也很好。使用其他库的解决方案非常好。如果需要,我可以提供数据集。

我想做的事示例(绿线) Example of what I'd like it to do (green line)

我试图找到类似的解决方法,但无济于事。有关周期性边界的问题使用numpy数据集掩码来隐藏某些跳跃,但它们具有连续的功能(其中“我的跳跃”)。

谢谢您的帮助,我非常感谢。

数据集(使它们比图形小一点,仅保留跳过部分):

D3:

x = [41.0, 43.0, 45.0, 47.0, 49.0, 51.0, 53.0, 55.0, 57.0, 59.0, 61.0, 63.0, 65.0, 67.0, 69.0, 71.0, 73.0, 75.0, 77.0, 79.0, 81.0, 83.0, 85.0, 87.0, 89.0, 91.0, 93.0, 95.0, 97.0, 99.0, 101.0, 103.0, 105.0, 107.0, 109.0, 111.0, 113.0, 115.0, 117.0, 119.0, 121.0, 123.0, 125.0, 127.0, 129.0, 131.0, 133.0, 135.0, 137.0, 139.0, 141.0, 143.0, 145.0, 147.0, 149.0, 151.0, 153.0, 155.0, 157.0, 159.0]

y = [45.6501, 37.4855, 40.4035, 51.4948, 55.8648, 48.9723, 60.4494, 42.7136, 20.6929, 36.7847, 44.4601, 54.04, 52.4895, 45.1991, 46.8203, 44.5827, 65.8803, 53.5398, 69.5158, 46.5372, 37.1557, 43.9031, 39.9325, 35.5248, 34.3531, 57.8377, 37.9208, 26.6508, 27.2333, 49.3798, 47.8627, 54.2795, 50.0892, 40.9849, 37.4014, 300.7947, 299.4254, 288.5113, 313.2906, 319.0095, 291.0726, 308.075, 298.451, 311.1485, 320.4832, 303.9229, 310.4584, 325.6287, 307.7328, 301.5581, 308.7813, 308.6791, 305.1343, 307.5148, 334.6374, 310.476, 315.6943, 326.0586, 298.6766, 305.6225]

最小工作示例:

import matplotlib.pyplot as plt
fig, ax = plt.subplots()
ax.plot(x, y, linewidth = 1.2, label = 'd3')
ax.set_yticks([t for t in range(0,390,30)])
ax.set_xticks([t for t in range(50,200,50)])
ax.legend(loc='lower right',prop={'size': 14})
plt.show()

2 个答案:

答案 0 :(得分:1)

使用列表中指示的基本Python,而不使用c++ number of significant digits long double 之类的高级库,您可以使用基本函数将图的两个部分分开。但是,在考虑您的特定问题时,您可能更喜欢极坐标图:

numpy

为您提供两种视图进行比较: enter image description here

答案 1 :(得分:1)

我一直在思考另一个问题-如何检测过渡。通常,信号的峰值检测很困难,但是在您的情况下,条件很简单:如果两个数据点之间的变化大于180°,则应在边界上绘制它。现在,我的建议是在循环中以相当尴尬的方式使用numpyI have been asking for suggestions of how to improve the segment extraction,可惜没有启发性的答案。至少该代码有效。

import numpy as np
from matplotlib import pyplot as plt


fig, ax = plt.subplots(figsize=(10, 5))

ymin = 0
ymax = 360
colour = "blue"

xarr, yarr = np.asarray([x, y])

#create index array for all points with jumps by more than 180 degrees
ind = list(np.where(np.abs(np.diff(yarr, append=yarr[-1])) > np.mean([ymin, ymax])))[0]

#if ind is not empty extract segment
if ind.size:
    #found breakpoints 
    for i, j in enumerate(ind):
        #first segment
        if i==0:
            #first trace
            xcurr = np.copy(xarr[:j+2])
            ycurr = np.copy(yarr[:j+2])
            xcurr[-1] = np.mean(xcurr[-2:]) 
            ycurr[-1] = (ymin, ymax)[ycurr[-1]<np.mean([ymin, ymax])]
        #all following segments  
        else:
            xcurr = np.copy(xarr[ind[i-1]:j+2])
            ycurr = np.copy(yarr[ind[i-1]:j+2])
            xcurr[0] = np.mean(xcurr[:2]) 
            ycurr[0] = (ymin, ymax)[ycurr[0]<np.mean([ymin, ymax])]
            xcurr[-1] = np.mean(xcurr[-2:]) 
            ycurr[-1] = (ymin, ymax)[ycurr[-1]<np.mean([ymin, ymax])]
        
        plt.plot(xcurr, ycurr, c=colour)
            
    #last segment
    xcurr = np.copy(xarr[j:])
    ycurr = np.copy(yarr[j:])
    xcurr[0] = np.mean(xcurr[:2]) 
    ycurr[0] = (ymin, ymax)[ycurr[0]<np.mean([ymin, ymax])]
   
else:
    #ind was empty - plot entire array
    xcurr = np.copy(xarr)
    ycurr = np.copy(yarr)

plt.plot(xcurr, ycurr, c=colour)

plt.ylim(ymin, ymax)
plt.yticks(np.linspace(ymin, ymax, 13))
plt.show()

通过以下测试数据对输出进行多次跳转

enter image description here

测试数据:

x = [41.0, 43.0, 45.0, 47.0, 49.0, 51.0, 53.0, 55.0, 57.0, 59.0, 61.0, 63.0, 65.0, 67.0, 69.0, 71.0, 73.0, 75.0, 77.0, 79.0, 81.0, 83.0, 85.0, 87.0, 89.0, 91.0, 93.0, 95.0, 97.0, 99.0, 101.0, 103.0, 105.0, 107.0, 109.0, 111.0, 113.0, 115.0, 117.0, 119.0, 121.0, 123.0, 125.0, 127.0, 129.0, 131.0, 133.0, 135.0, 137.0, 139.0, 141.0, 143.0, 145.0, 147.0, 149.0, 151.0, 153.0, 155.0, 157.0, 159.0]
y = [45.6501, 37.4855, 40.4035, 31.4948, 155.8648, 148.9723, 180.4494, 142.7136, 220.6929, 236.7847, 244.4601, 254.04, 252.4895, 245.1991, 246.8203, 244.5827, 265.8803, 253.5398, 269.5158, 246.5372, 237.1557, 243.9031, 239.9325, 235.5248, 234.3531, 257.8377, 37.9208, 26.6508, 27.2333, 349.3798, 347.8627, 354.2795, 350.0892, 340.9849, 337.4014, 300.7947, 99.4254, 88.5113, 13.2906, 19.0095, 191.0726, 208.075, 198.451, 111.1485, 120.4832, 103.9229, 110.4584, 125.6287, 107.7328, 101.5581, 108.7813, 108.6791, 105.1343, 107.5148, 134.6374, 110.476, 15.6943, 26.0586, 98.6766, 105.6225]