【Python】实现二维数据的关系

文章目录

• 1.画二维关系图
• 2.求曲线的拟合公式
• • 线性公式拟合
  • 二次多项式拟合
  • 对数函数拟合

1.画二维关系图

例如以下数据：

import matplotlib.pyplot as plt
plt.rcParams['font.sans-serif']=['SimHei']
plt.rcParams['axes.unicode_minus'] = False
plt.title("无水乙醇与正丙醇混合液的浓度与折射率")
plt.xlim(xmax=110, xmin=0)
plt.ylim(ymax=1.3850, ymin=1.3550)
plt.xlabel("正丙醇的摩尔百分数（%）")
plt.ylabel("折射率")
plt.plot([0.0, 7.70, 16.0, 24.6, 33.7, 43.2, 53.3, 63.8, 75.3, 87.3, 100.00],
         [1.3592, 1.3619, 1.3642, 1.3668, 1.3691, 1.3715, 1.3740, 1.3764, 1.3789, 1.3812, 1.3839], 'ro')
plt.show()

2.求曲线的拟合公式

例如：根据上述1的数据拟合公式

线性公式拟合

import numpy as np

Y = [0.0, 7.70, 16.0, 24.6, 33.7, 43.2, 53.3, 63.8, 75.3, 87.3, 100.00]
X = [1.3592, 1.3619, 1.3642, 1.3668, 1.3691, 1.3715, 1.3740, 1.3764, 1.3789, 1.3812, 1.3839]

z1 = np.polyfit(X, Y, 1)
p1 = np.poly1d(z1)
print(z1)
print(p1)

#[ 4069.10408341 -5535.09820153]
#4069 x - 5535

二次多项式拟合

import numpy as np
def polyfit(x, y, degree):
    results = {}
    coeffs = np.polyfit(x, y, degree)
    results['polynomial'] = coeffs.tolist()
    # r-squared
    p = np.poly1d(coeffs)
    # fit values, and mean
    yhat = p(x)  # or [p(z) for z in x]
    ybar = np.sum(y) / len(y)  # or sum(y)/len(y)
    ssreg = np.sum((yhat - ybar) ** 2)  # or sum([ (yihat - ybar)**2 for yihat in yhat])
    sstot = np.sum((y - ybar) ** 2)  # or sum([ (yi - ybar)**2 for yi in y])
    results['determination'] = ssreg / sstot  # 准确率
    return results


z1 = polyfit(X, Y, 2)
print(z1)
#结果依次为 x^2 x^1 x^0 的系数
#{'polynomial': [44925.124426109105, -119165.27286295866, 78973.45552795619], 'determination': 0.9999011699884925}

对数函数拟合

from scipy import log
from scipy.optimize import curve_fit


def func(x, a, b):
    y = a * log(x) + b
    return y


def polyfit(x, y, degree):
    results = {}
    # coeffs = numpy.polyfit(x, y, degree)
    popt, pcov = curve_fit(func, x, y)
    results['polynomial'] = popt

    # r-squared
    yhat = func(x, popt[0], popt[1])  # or [p(z) for z in x]
    ybar = np.sum(y) / len(y)  # or sum(y)/len(y)
    ssreg = np.sum((yhat - ybar) ** 2)  # or sum([ (yihat - ybar)**2 for yihat in yhat])
    sstot = np.sum((y - ybar) ** 2)  # or sum([ (yi - ybar)**2 for yi in y])
    results['determination'] = ssreg / sstot

    return results

z1 = polyfit(Y, X, 2)
print(z1)

#{'polynomial': array([1., 1.]), 'determination': inf}

于是对于以下实验数据，我们就可以带公式求出未知数啦！