Python 做拟合后如何让折线变为平滑曲线，用 1d 插值拟合后曲率变大

代码如下

import matplotlib.pyplot as plt import numpy as np x = [1,5,30,200] y = [27,12,7,5] x = np.array(x) y = np.array(y) # x = np.arange(1, 17, 1) # y = np.array([4.00, 6.40, 8.00, 8.80, 9.22, 9.50, 9.70, 9.86, 10.00, 10.20, 10.32, 10.42, 10.50, 10.55, 10.58, 10.60]) z1 = np.polyfit(x, y, 3)#用 3 次多项式拟合 p1 = np.poly1d(z1) print(p1) #在屏幕上打印拟合多项式 yvals=p1(x)#也可以使用 yvals=np.polyval(z1,x) plot1=plt.plot(x, y, '*',label='original values') plot2=plt.plot(x, yvals, 'r',label='polyfit values') plt.xlabel('x axis') plt.ylabel('y axis') plt.legend(loc=1)#指定 legend 的位置,读者可以自己 help 它的用法 plt.title('polyfitting') plt.show() 

使用插值拟合后，产生一个波峰一个波谷，我只想要个平滑的曲线，不知道有啥好办法

fq1 = interp1d(x, y1, kind='quadratic') fq2 = interp1d(x, y2, kind='zero') 

下面代码是一个粘度曲线测试程序，请参考程序运行结果

 import numpy as np from scipy.interpolate import interp1d #创建待插值的数据 # x = np.linspace(0, 10*np.pi, 20) # y = np.cos(x) x = [1,5,30,200] # x = [1,100,150,200] y = [27,12,7,5] x = np.array(x) y = np.array([27,12,7,5]) # 分别用 linear 和 quadratic 插值 fl = interp1d(x, y, kind='linear') fq = interp1d(x, y, kind='quadratic') #设置 x 的最大值和最小值以防止插值数据越界 xint = np.linspace(x.max(), x.min(), 200) y1 = np.array([21.350,11,6.62,5.05]) y2 = np.array([15.630,8.31,5.18,3.53]) # x1 = np.array([50,150,200]) # xint1 = np.linspace(x1.min(), x1.max(), 110) yintl = fl(xint) yintq = fq(xint) fq1 = interp1d(x, y1, kind='quadratic') fq2 = interp1d(x, y2, kind='zero') import pylab as pl fig, ax = pl.subplots() # make ticks and tick labels # xticks = range(min(x), max(x) + 1, 3) # xticks = range(1, 200 + 50,30) xticklabels = [1,3,10,"",30,100,200] # #设置网格样式 # ax.set_xticks(xticks) ax.set_xticklabels(xticklabels, rotation=0) ax.grid(True, linestyle='-.',color="b") yfq2=fq2(xint) yfq1=fq1(xint) # pl.plot(xint,fl(xint), color="green", label = "Linear") pl.plot(xint,fq(xint), color="red", label ="Quadratic") pl.plot(xint,yfq1, "b--", label ="QuadraticL") pl.plot(xint,yfq2, color="green", label ="QuadraticH") x3= np.array([1,5,30,200]) y3 = np.array([27,12,7,5]) pl.plot(x3,y3,"b-", label ="QuadraticZ") pl.legend(loc = "best") pl.ylim(1,50,10) pl.xlim(0,200,70) pl.show()