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

[qile11]

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()

[qile1]

没人知道？写个函数计算是不是可以实现

[daya0576]

```python
import matplotlib.pyplot as plt
import numpy as np

x = np.array([1,5,30,200])
y = np.array([27,12,7,5])
plot1=plt.plot(x, y, 'x',label='original values')

# calculate polynomial
z = np.polyfit(x, y, 2)
f = np.poly1d(z)
x_new = np.linspace(x[0], x[-1], 200)
y_new = f(x_new)
plot2=plt.plot(x_new, y_new, 'r',label='polyfit values')

plt.show()
```
都拿到 3 元函数了, 那就是画图的问题, 多取几个点就好了.
---
但是拟合的曲线会有 overfitting 的问题, 不知道怎么解决.

[Ballacuda]

https://stackoverflow.com/questions/20618804/how-to-smooth-a-curve-in-the-right-way
http://scipy.github.io/old-wiki/pages/Cookbook/SavitzkyGolay

[qile11]

@daya0576
@Ballacuda
感谢回复，可能我方向不对，我画的是黏度曲线，不可能出现负数，上面画出来的都有负数，
[]( https://imgchr.com/i/PE2kbF)
[img]https://s1.ax1x.com/2018/07/03/PE2kbF.png[/img]