行python 3.x的散点图

如何解决行python 3.x的散点图

我是一个本科生，为一家制造车轮的公司写论文。我必须针对零件的复杂性来分析工艺能力。 我被要求在“简单”图表中通过python可视化混乱。 我写了一个脚本，显示最大。从标称值开始的带有条的车轮的直径，请参见所附图片。

我的同事要求提供这样的图表...

我如何用橙色斑点替换连接到每个点的线（蓝色）。 我对Python没有真正的了解。...

from matplotlib import pyplot as plt
import numpy as np
import pandas as pd
from statistics import mean,median,stdev,variance
import random
import matplotlib as mpl
Start = 0
Stop = 20
limit = 60



data1=[random.randint(Start,Stop) for iter in range(limit)]
data2=[random.randint(Start,Stop) for iter in range(limit)]
data3=[random.randint(Start,Stop) for iter in range(limit)]
data4=[random.randint(Start,Stop) for iter in range(limit)]
data5=[random.randint(Start,Stop) for iter in range(limit)]

median =[median(data1),median(data2),median(data3),median(data4),median(data5)]
print(data1)


lObjectsALLcnts = data1
lObjectsALLlbls = ['median data1','median data2','median data3','median data4','median data5']

iN = len(lObjectsALLcnts)
arrCnts = np.array(lObjectsALLcnts)

theta1=np.arange(0,2*np.pi,2*np.pi/iN)
width = (2*np.pi)/iN *.9 # Breite der Balken
bottom = 10 # Durchmesser

fig = plt.figure(figsize=(20,20))
ax = fig.add_axes([0.1,0.1,0.75,0.75],polar=True)

bars = ax.bar(theta1,arrCnts,width=width,bottom=bottom)
circle = ax.scatter(5,5,c=5,s=300,cmap=colormap,norm=norm,linewidths=0)
#ax.set_theta_offset(1.57079)
plt.axis('on')

rotations = np.rad2deg(theta1)
y0,y1 = ax.get_ylim()

for x,bar,rotation,label in zip(theta1,bars,rotations,lObjectsALLlbls):
     offset = (bottom+bar.get_height())/(y1-y0)
     lab = ax.text(0,label,transform=None,ha='center',va='center',fontsize=20)
     renderer = ax.figure.canvas.get_renderer()
     bBox = lab.get_window_extent(renderer=renderer)
     invb = ax.transData.inverted().transform([[0,0],[bBox.width,0] ])
     lab.set_position((x,offset+(invb[1][0]-invb[0][0])/2.*2.7 ) )
     lab.set_transform(ax.get_xaxis_transform())
     lab.set_rotation(rotation)

draw_circle = plt.Circle((25,25),20,fill=False)
plt.gcf().gca().add_artist(draw_circle)
plt.title('Circle')
ax.set_aspect(1)
plt.show()

最诚挚的问候...