我做了同样的观察：所有参数的自由拟合在大多数情况下都失败了。2你可以提供一个更好的初始猜测来帮助你，没有必要修正参数。

samp = stats.lognorm(0.5,loc=0,scale=1).rvs(size=2000)

# this is where the fit gets it initial guess from
print stats.lognorm._fitstart(samp)

(1.0, 0.66628696413404565, 0.28031095750445462)

print stats.lognorm.fit(samp)
# note that the fit failed completely as the parameters did not change at all

(1.0, 0.66628696413404565, 0.28031095750445462)

# fit again with a better initial guess for loc
print stats.lognorm.fit(samp, loc=0)

(0.50146296628099118, 0.0011019321419653122, 0.99361128537912125)

您也可以创建自己的函数来计算初始猜测值，例如：

def your_func(sample):
    # do some magic here
    return guess

stats.lognorm._fitstart = your_func

我认识到了我的错误：
A）我正在绘制的样本需要来自.rvs方法。如下所示：sample_dist = sp.stats.lognorm.rvs(3, loc=0, scale=np.exp(10), size=2000)
B）拟合存在一些问题。当我们固定loc参数时，拟合成功得多。param=sp.stats.lognorm.fit(samp, floc=0)

此问题已在较新的scipy版本中得到修复。将scipy0.9升级到scipy0.14后，此问题消失。

如果你只是对绘图感兴趣，你可以使用seaborn来得到对数正态分布。

import seaborn as sns
import numpy as np
from scipy import stats

mu=0
sigma=1
n=1000

x=np.random.normal(mu,sigma,n)
sns.distplot(x, fit=stats.norm) # normal distribution

loc=0
scale=1

x=np.log(np.random.lognormal(loc,scale,n))
sns.distplot(x, fit=stats.lognorm) # log normal distribution

我在here中回答
我把代码留在这里只是为了偷懒：D

import scipy
import matplotlib.pyplot as plt
import seaborn as sns
import numpy as np

mu = 10 # Mean of sample !!! Make sure your data is positive for the lognormal example 
sigma = 1.5 # Standard deviation of sample
N = 2000 # Number of samples

norm_dist = scipy.stats.norm(loc=mu, scale=sigma) # Create Random Process
x = norm_dist.rvs(size=N) # Generate samples

# Fit normal
fitting_params = scipy.stats.norm.fit(x)
norm_dist_fitted = scipy.stats.norm(*fitting_params)
t = np.linspace(np.min(x), np.max(x), 100)

# Plot normals
f, ax = plt.subplots(1, sharex='col', figsize=(10, 5))
sns.distplot(x, ax=ax, norm_hist=True, kde=False, label='Data X~N(mu={0:.1f}, sigma={1:.1f})'.format(mu, sigma))
ax.plot(t, norm_dist_fitted.pdf(t), lw=2, color='r',
        label='Fitted Model X~N(mu={0:.1f}, sigma={1:.1f})'.format(norm_dist_fitted.mean(), norm_dist_fitted.std()))
ax.plot(t, norm_dist.pdf(t), lw=2, color='g', ls=':',
        label='Original Model X~N(mu={0:.1f}, sigma={1:.1f})'.format(norm_dist.mean(), norm_dist.std()))
ax.legend(loc='lower right')
plt.show()

# The lognormal model fits to a variable whose log is normal
# We create our variable whose log is normal 'exponenciating' the previous variable

x_exp = np.exp(x)
mu_exp = np.exp(mu)
sigma_exp = np.exp(sigma)

fitting_params_lognormal = scipy.stats.lognorm.fit(x_exp, floc=0, scale=mu_exp)
lognorm_dist_fitted = scipy.stats.lognorm(*fitting_params_lognormal)
t = np.linspace(np.min(x_exp), np.max(x_exp), 100)

# Here is the magic I was looking for a long long time
lognorm_dist = scipy.stats.lognorm(s=sigma, loc=0, scale=np.exp(mu))
# Plot lognormals
f, ax = plt.subplots(1, sharex='col', figsize=(10, 5))
sns.distplot(x_exp, ax=ax, norm_hist=True, kde=False,
             label='Data exp(X)~N(mu={0:.1f}, sigma={1:.1f})\n X~LogNorm(mu={0:.1f}, sigma={1:.1f})'.format(mu, sigma))
ax.plot(t, lognorm_dist_fitted.pdf(t), lw=2, color='r',
        label='Fitted Model X~LogNorm(mu={0:.1f}, sigma={1:.1f})'.format(lognorm_dist_fitted.mean(), lognorm_dist_fitted.std()))
ax.plot(t, lognorm_dist.pdf(t), lw=2, color='g', ls=':',
        label='Original Model X~LogNorm(mu={0:.1f}, sigma={1:.1f})'.format(lognorm_dist.mean(), lognorm_dist.std()))
ax.legend(loc='lower right')
plt.show()

诀窍在于理解这两件事：

1.如果变量的EXP是正态的，MU和STD -〉EXP（X）~ scipy.stats.lognorm（s=sigma，位置=0，比例=np.exp（mu））

• 如果变量（x）具有对数正态形式，则模型将为scipy.stats.lognorm（s=sigmaX，loc=0，scale=muX），其中：
• μ X = np平均值（np对数（x））
• σ X =非对映体标准品（非对映体对数（x））