我的代码是从1D中的高斯绘制的:
import numpy as np
from scipy.stats import norm
from scipy.optimize import curve_fit
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
import gauss
# Beginning in one dimension:
mean = 0; Var = 1; N = 1000
scatter = np.random.normal(mean,np.sqrt(Var),N)
scatter = np.sort(scatter)
mu,sigma = norm.fit(scatter)
我使用norm.fit()
获得mu和sigma现在我想使用
获取参数xdata = np.linspace(-5,5,N)
pop, pcov = curve_fit(gauss.gauss_1d,xdata,scatter)
问题是我不知道如何将我的散点(从1D高斯绘制)映射到x线以便使用curve_fit。
另外,假设我只是使用和mu和sigma一样早。
我使用绘图:
n, bins, patches = plt.hist(scatter,50,facecolor='green')
y = 2*max(n)*mlab.normpdf(bins,mu,sigma)
l = plt.plot(bins,y,'r--')
plt.xlabel('x-coord')
plt.ylabel('Occurrences')
plt.grid(True)
plt.show()
但我必须猜测幅度为2 * max(n)。它有效,但它不健壮。如何在不猜测的情况下找到振幅?
答案 0 :(得分:1)
为避免猜测幅度,请使用hist()
拨打normed=True
,然后幅度对应normpdf()
。
为了进行曲线拟合,我建议不使用密度而是使用累积分布:每个样本的高度为1/N
,连续总计为1.这样做的好处是您不需要将样品分组。
import numpy as np
from scipy.stats import norm
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
# Beginning in one dimension:
mean = 0; Var = 1; N = 100
scatter = np.random.normal(mean,np.sqrt(Var),N)
scatter = np.sort(scatter)
mu1,sigma1 = norm.fit(scatter) # classical fit
scat_sum = np.cumsum(np.ones(scatter.shape))/N # cumulative samples
[mu2,sigma2],Cx = curve_fit(norm.cdf, scatter, scat_sum, p0=[0,1]) # curve fit
print(u"norm.fit(): µ1= {:+.4f}, σ1={:.4f}".format(mu1, sigma1))
print(u"curve_fit(): µ2= {:+.4f}, σ2={:.4f}".format(mu2, sigma2))
fg = plt.figure(1); fg.clf()
ax = fg.add_subplot(1, 1, 1)
t = np.linspace(-4,4, 1000)
ax.plot(t, norm.cdf(t, mu1, sigma1), alpha=.5, label="norm.fit()")
ax.plot(t, norm.cdf(t, mu2, sigma2), alpha=.5, label="curve_fit()")
ax.step(scatter, scat_sum, 'x-', where='post', alpha=.5, label="Samples")
ax.legend(loc="best")
ax.grid(True)
ax.set_xlabel("$x$")
ax.set_ylabel("Cumulative Probability Density")
ax.set_title("Fit to Normal Distribution")
fg.canvas.draw()
plt.show()
打印
norm.fit(): µ1= +0.1534, σ1=1.0203
curve_fit(): µ2= +0.1135, σ2=1.0444
和情节