Python np.lognormal 为 big average 和 St Dev 提供无限结果 [英] Python np.lognormal gives infinite results for big average and St Dev

问题描述

我正在尝试为我的数据绘制对数正态分布.使用以下代码:

I am trying to draw the lognormal distribution for my data. using the following code:

mu, sigma = 136519., 50405. # mean and standard deviation
hs = np.random.lognormal(mu, sigma, 1000) #mean, s dev , Size
count, bins, ignored = plt.hist(hs, 100, normed=True)     
x = np.linspace(min(bins), max(bins), 10000)
pdf = (math.exp(-(np.log(x) - mu)**2 / (2 * sigma**2)))
#plt.axis('tight')
plt.plot(x, pdf, linewidth=2, color='r')

正如你所看到的，我的均值和西格玛是很大的值，它会产生一个问题，即 hs 趋于无穷大，从而产生错误.如果我输入 mu =3 和 sigma =1 之类的东西，它会起作用，对于大数字有什么建议吗?

As you can see, my mean and sigma are big values, it creates the problem that hs goes to infinity that gives an error. While if I put something like mu =3 and sigma =1, it works, any suggestions for big numbers?

更新 1:

我用第一个答案更正了我的代码，但现在我只得到一条直线:

I corrected my code with the first answer, but now I only get a straight line :

 mu, sigma = 136519 , 50405 # mean and standard deviation

    normal_std = np.sqrt(np.log(1 + (sigma/mu)**2))
    normal_mean = np.log(mu) - normal_std**2 / 2
    hs = np.random.lognormal(normal_mean, normal_std, 1000)
    print(hs.max())    # some finite number
    print(hs.mean())   # about 136519
    print(hs.std())    # about 50405

#    hs = np.random.lognormal(mu, sigma, 1000) #mean, s dev , Size
#    
    count, bins, ignored = plt.hist(hs, 100, normed=True) 

    x = np.linspace(min(bins), max(bins), 10000)
    pdfT = [];
    for el in range (len(x)):
        pdfTmp = (math.exp(-(np.log(x[el]) - mu)**2 / (2 * sigma**2)))
        pdfT += [pdfTmp]


    #plt.axis('tight')
    pdf = np.asarray(pdfT)
    plt.plot(x, pdf, linewidth=2, color='r')

推荐答案

np.random.lognormal 不是 对数正态分布的均值和 STD.它们是底层正态分布的均值和STD，即log(X).这意味着通过传递 136519 作为平均值，您要求 NumPy 生成大小为 exp(136519) 的数字，大约为 10**60000，远远超出双精度限制.

The parameters mu and sigma in np.random.lognormal are not the mean and STD of the lognormal distribution. They are the mean and STD of the underlying normal distribution, that is of log(X). This means that by passing 136519 for the mean you ask NumPy to generate numbers of size exp(136519) which is about 10**60000, far beyond the double precision limits.

通过一些代数，您可以从以下位置获得 np.random.lognormal 的正确参数你拥有的那些.

With a bit of algebra you can get the correct parameters for np.random.lognormal from the ones you have.

mu, sigma = 136519., 50405.
normal_std = np.sqrt(np.log(1 + (sigma/mu)**2))
normal_mean = np.log(mu) - normal_std**2 / 2
hs = np.random.lognormal(normal_mean, normal_std, 1000)
print(hs.max())    # some finite number
print(hs.mean())   # about 136519
print(hs.std())    # about 50405

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