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Question

我有一组对数正态的分布式样本。我可以使用具有线性或对数x轴的自然图来可视化样本。我可以对直方图进行拟合以获得PDF，然后将其缩放到具有线性x轴的图中的histrogram，另请参阅this previously posted question。

但是，我无法使用对数x轴将PDF正确绘制到绘图中。

不幸的是，将PDF区域缩放到直方图不仅存在问题，而且PDF也会向左移动，如下图所示。

我现在的问题是，我在这里做错了什么？使用CDF绘制预期的直方图as suggested in this answer，有效。我想知道我在这段代码中做错了什么，因为我的理解它也应该有用。

这是python代码（我很抱歉它很长但我想发布一个＆＃34;完整的独立版本＆＃34;）：

import numpy as np
import matplotlib.pyplot as plt
import scipy.stats

# generate log-normal distributed set of samples
np.random.seed(42)
samples   = np.random.lognormal( mean=1, sigma=.4, size=10000 )

# make a fit to the samples
shape, loc, scale = scipy.stats.lognorm.fit( samples, floc=0 )
x_fit       = np.linspace( samples.min(), samples.max(), 100 )
samples_fit = scipy.stats.lognorm.pdf( x_fit, shape, loc=loc, scale=scale )

# plot a histrogram with linear x-axis
plt.subplot( 1, 2, 1 )
N_bins = 50
counts, bin_edges, ignored = plt.hist( samples, N_bins, histtype='stepfilled', label='histogram' )
# calculate area of histogram (area under PDF should be 1)
area_hist = .0
for ii in range( counts.size):
    area_hist += (bin_edges[ii+1]-bin_edges[ii]) * counts[ii]
# oplot fit into histogram
plt.plot( x_fit, samples_fit*area_hist, label='fitted and area-scaled PDF', linewidth=2)
plt.legend()

# make a histrogram with a log10 x-axis
plt.subplot( 1, 2, 2 )
# equally sized bins (in log10-scale)
bins_log10 = np.logspace( np.log10( samples.min()  ), np.log10( samples.max() ), N_bins )
counts, bin_edges, ignored = plt.hist( samples, bins_log10, histtype='stepfilled', label='histogram' )
# calculate area of histogram
area_hist_log = .0
for ii in range( counts.size):
    area_hist_log += (bin_edges[ii+1]-bin_edges[ii]) * counts[ii]
# get pdf-values for log10 - spaced intervals
x_fit_log       = np.logspace( np.log10( samples.min()), np.log10( samples.max()), 100 )
samples_fit_log = scipy.stats.lognorm.pdf( x_fit_log, shape, loc=loc, scale=scale )
# oplot fit into histogram
plt.plot( x_fit_log, samples_fit_log*area_hist_log, label='fitted and area-scaled PDF', linewidth=2 )

plt.xscale( 'log' )
plt.xlim( bin_edges.min(), bin_edges.max() )
plt.legend()
plt.show()

更新1 ：

我忘了提到我正在使用的版本：

python      2.7.6
numpy       1.8.2
matplotlib  1.3.1
scipy       0.13.3

更新2 ：

正如@Christoph和@zaxliu所指出的那样（感谢两者），问题在于缩放PDF。当我使用与直方图相同的箱子时，它可以工作，就像@ zaxliu的解决方案一样，但是当我为PDF使用更高的分辨率时仍然存在一些问题（如上例所示）。如下图所示：

右边图中的代码是（我省略了导入和数据样本生成的东西，你可以在上面的例子中找到）：

# equally sized bins in log10-scale
bins_log10 = np.logspace( np.log10( samples.min()  ), np.log10( samples.max() ), N_bins )
counts, bin_edges, ignored = plt.hist( samples, bins_log10, histtype='stepfilled', label='histogram' )

# calculate length of each bin (required for scaling PDF to histogram)
bins_log_len = np.zeros( bins_log10.size )
for ii in range( counts.size):
    bins_log_len[ii] = bin_edges[ii+1]-bin_edges[ii]

# get pdf-values for same intervals as histogram
samples_fit_log = scipy.stats.lognorm.pdf( bins_log10, shape, loc=loc, scale=scale )

# oplot fitted and scaled PDF into histogram
plt.plot( bins_log10, np.multiply(samples_fit_log,bins_log_len)*sum(counts), label='PDF using histogram bins', linewidth=2 )

# make another pdf with a finer resolution
x_fit_log       = np.logspace( np.log10( samples.min()), np.log10( samples.max()), 100 )
samples_fit_log = scipy.stats.lognorm.pdf( x_fit_log, shape, loc=loc, scale=scale )
# calculate length of each bin (required for scaling PDF to histogram)
# in addition, estimate middle point for more accuracy (should in principle also be done for the other PDF)
bins_log_len       = np.diff( x_fit_log )
samples_log_center = np.zeros( x_fit_log.size-1 )
for ii in range( x_fit_log.size-1 ):
    samples_log_center[ii] = .5*(samples_fit_log[ii] + samples_fit_log[ii+1] )

# scale PDF to histogram
# NOTE: THIS IS NOT WORKING PROPERLY (SEE FIGURE)
pdf_scaled2hist = np.multiply(samples_log_center,bins_log_len)*sum(counts)

# oplot fit into histogram
plt.plot( .5*(x_fit_log[:-1]+x_fit_log[1:]), pdf_scaled2hist, label='PDF using own bins', linewidth=2 )

plt.xscale( 'log' )
plt.xlim( bin_edges.min(), bin_edges.max() )
plt.legend(loc=3)

Answer 1

根据我在@Warren Weckesser的原始答案中的理解，您reffered to“您需要做的一切”是：

写出cdf(b) - cdf(a)的近似值 cdf(b) - cdf(a) = pdf(m)*(b - a) 其中m是，例如，区间[a，b]
的中点

我们可以尝试遵循他的建议，并根据垃圾箱的中心点绘制两种获取pdf值的方法：

PDF功能
具有CDF功能：

import numpy as np
import matplotlib.pyplot as plt
from scipy import stats 

# generate log-normal distributed set of samples
np.random.seed(42)
samples = np.random.lognormal(mean=1, sigma=.4, size=10000)
N_bins = 50

# make a fit to the samples
shape, loc, scale = stats.lognorm.fit(samples, floc=0)
x_fit       = np.linspace(samples.min(), samples.max(), 100)
samples_fit = stats.lognorm.pdf(x_fit, shape, loc=loc, scale=scale)

# plot a histrogram with linear x-axis
fig, (ax1, ax2) = plt.subplots(1,2, figsize=(10,5), gridspec_kw={'wspace':0.2})
counts, bin_edges, ignored = ax1.hist(samples, N_bins, histtype='stepfilled', alpha=0.4,
                                      label='histogram')

# calculate area of histogram (area under PDF should be 1)
area_hist = ((bin_edges[1:] - bin_edges[:-1]) * counts).sum()

# plot fit into histogram
ax1.plot(x_fit, samples_fit*area_hist, label='fitted and area-scaled PDF', linewidth=2)
ax1.legend()

# equally sized bins in log10-scale and centers
bins_log10 = np.logspace(np.log10(samples.min()), np.log10(samples.max()), N_bins)
bins_log10_cntr = (bins_log10[1:] + bins_log10[:-1]) / 2

# histogram plot
counts, bin_edges, ignored = ax2.hist(samples, bins_log10, histtype='stepfilled', alpha=0.4,
                                      label='histogram')

# calculate length of each bin and its centers(required for scaling PDF to histogram)
bins_log_len = np.r_[bin_edges[1:] - bin_edges[: -1], 0]
bins_log_cntr = bin_edges[1:] - bin_edges[:-1]

# get pdf-values for same intervals as histogram
samples_fit_log = stats.lognorm.pdf(bins_log10, shape, loc=loc, scale=scale)

# pdf-values for centered scale
samples_fit_log_cntr = stats.lognorm.pdf(bins_log10_cntr, shape, loc=loc, scale=scale)

# pdf-values using cdf 
samples_fit_log_cntr2_ = stats.lognorm.cdf(bins_log10, shape, loc=loc, scale=scale)
samples_fit_log_cntr2 = np.diff(samples_fit_log_cntr2_)

# plot fitted and scaled PDFs into histogram
ax2.plot(bins_log10, 
         samples_fit_log * bins_log_len * counts.sum(), '-', 
         label='PDF with edges',  linewidth=2)

ax2.plot(bins_log10_cntr, 
         samples_fit_log_cntr * bins_log_cntr * counts.sum(), '-', 
         label='PDF with centers', linewidth=2)

ax2.plot(bins_log10_cntr, 
         samples_fit_log_cntr2 * counts.sum(), 'b-.', 
         label='CDF with centers', linewidth=2)


ax2.set_xscale('log')
ax2.set_xlim(bin_edges.min(), bin_edges.max())
ax2.legend(loc=3)
plt.show()

你可以看到第一个（使用pdf）和第二个（使用cdf）方法得到几乎相同的结果，并且两者都不完全匹配用bin的边缘计算的pdf。

如果放大，您会清楚地看到差异：

现在可以问的问题是：使用哪一个？我想答案将取决于但如果我们看一下累积概率：

print 'Cumulative probabilities:'
print 'Using edges:         {:>10.5f}'.format((samples_fit_log * bins_log_len).sum())
print 'Using PDF of centers:{:>10.5f}'.format((samples_fit_log_cntr * bins_log_cntr).sum())
print 'Using CDF of centers:{:>10.5f}'.format(samples_fit_log_cntr2.sum())

您可以从输出中看到哪个方法更接近1.0：

Cumulative probabilities:
Using edges:            1.03263
Using PDF of centers:   0.99957
Using CDF of centers:   0.99991

CDF似乎给出了最接近的近似值。

这很长，但我希望这是有道理的。

<强>更新

我已调整代码以说明如何平滑PDF行。注意s变量，它定义了线的平滑程度。我在变量中添加了_s后缀，以指示调整需要发生的位置。

# generate log-normal distributed set of samples
np.random.seed(42)
samples = np.random.lognormal(mean=1, sigma=.4, size=10000)
N_bins = 50

# make a fit to the samples
shape, loc, scale = stats.lognorm.fit(samples, floc=0)

# plot a histrogram with linear x-axis
fig, ax2 = plt.subplots()#1,2, figsize=(10,5), gridspec_kw={'wspace':0.2})

# equally sized bins in log10-scale and centers
bins_log10 = np.logspace(np.log10(samples.min()), np.log10(samples.max()), N_bins)
bins_log10_cntr = (bins_log10[1:] + bins_log10[:-1]) / 2

# smoother PDF line
s = 10 # mulpiplier to N_bins - the bigger s is the smoother the line
bins_log10_s = np.logspace(np.log10(samples.min()), np.log10(samples.max()), N_bins * s)
bins_log10_cntr_s = (bins_log10_s[1:] + bins_log10_s[:-1]) / 2

# histogram plot
counts, bin_edges, ignored = ax2.hist(samples, bins_log10, histtype='stepfilled', alpha=0.4,
                                      label='histogram')

# calculate length of each bin and its centers(required for scaling PDF to histogram)
bins_log_len = np.r_[bins_log10_s[1:] - bins_log10_s[: -1], 0]
bins_log_cntr = bins_log10_s[1:] - bins_log10_s[:-1]

# smooth pdf-values for same intervals as histogram
samples_fit_log_s = stats.lognorm.pdf(bins_log10_s, shape, loc=loc, scale=scale)

# pdf-values for centered scale
samples_fit_log_cntr = stats.lognorm.pdf(bins_log10_cntr_s, shape, loc=loc, scale=scale)

# smooth pdf-values using cdf 
samples_fit_log_cntr2_s_ = stats.lognorm.cdf(bins_log10_s, shape, loc=loc, scale=scale)
samples_fit_log_cntr2_s = np.diff(samples_fit_log_cntr2_s_)

# plot fitted and scaled PDFs into histogram
ax2.plot(bins_log10_cntr_s, 
         samples_fit_log_cntr * bins_log_cntr * counts.sum() * s, '-', 
         label='Smooth PDF with centers', linewidth=2)

ax2.plot(bins_log10_cntr_s, 
         samples_fit_log_cntr2_s * counts.sum() * s, 'k-.', 
         label='Smooth CDF with centers', linewidth=2)

ax2.set_xscale('log')
ax2.set_xlim(bin_edges.min(), bin_edges.max())
ax2.legend(loc=3)
plt.show)

这会产生这个情节：

如果您放大平滑版本与非平滑版本，您会看到：

希望这有帮助。

Answer 2

由于我遇到了同样的问题并且弄明白了，我想解释一下是什么，并为原始问题提供不同的解决方案。

当您使用对数区间进行直方图时，这相当于更改变量，其中x是原始样本（或用于绘制它们的网格），t是一个新变量，这些箱是线性间隔的。因此，实际对应于直方图的PDF是

我们仍在使用x变量作为PDF的输入，因此这变为

您需要将PDF乘以x！

它修复了PDF的形状，但我们仍然需要缩放PDF，以使曲线下面积等于直方图。事实上，PDF下的区域不等于1，因为我们正在整合x，并且

因为我们正在处理对数正态变量。由于根据scipy documentation，分布参数对应shape = sigma和scale = exp(mu)，因此我们可以轻松地将代码中的右侧计算为scale * np.exp(shape**2/2.)。

实际上，单行代码修复了原始脚本，将计算出的PDF值乘以x并除以上面计算的面积：

samples_fit_log *= x_fit_log / (scale * np.exp(shape**2/2.))

导致以下情节：

或者，您可以通过在日志空间中集成直方图来更改直方图“区域”的定义。请记住，在日志空间（t变量）中，PDF具有区域1.因此，您可以跳过缩放因子，并将上面的行替换为：

area_hist_log = np.dot(np.diff(np.log(bin_edges)), counts)
samples_fit_log *= x_fit_log

后一种解决方案可能更为可取，因为它不依赖于手头分布的任何信息。它适用于任何分发，而不仅仅是log-normal。

作为参考，这是添加了我的行的原始脚本：

import numpy as np
import matplotlib.pyplot as plt
import scipy.stats

# generate log-normal distributed set of samples
np.random.seed(42)
samples   = np.random.lognormal( mean=1, sigma=.4, size=10000 )

# make a fit to the samples
shape, loc, scale = scipy.stats.lognorm.fit( samples, floc=0 )
x_fit       = np.linspace( samples.min(), samples.max(), 100 )
samples_fit = scipy.stats.lognorm.pdf( x_fit, shape, loc=loc, scale=scale )

# plot a histrogram with linear x-axis
plt.subplot( 1, 2, 1 )
N_bins = 50
counts, bin_edges, ignored = plt.hist( samples, N_bins, histtype='stepfilled', label='histogram' )
# calculate area of histogram (area under PDF should be 1)
area_hist = .0
for ii in range( counts.size):
    area_hist += (bin_edges[ii+1]-bin_edges[ii]) * counts[ii]
# oplot fit into histogram
plt.plot( x_fit, samples_fit*area_hist, label='fitted and area-scaled PDF', linewidth=2)
plt.legend()

# make a histrogram with a log10 x-axis
plt.subplot( 1, 2, 2 )
# equally sized bins (in log10-scale)
bins_log10 = np.logspace( np.log10( samples.min()  ), np.log10( samples.max() ), N_bins )
counts, bin_edges, ignored = plt.hist( samples, bins_log10, histtype='stepfilled', label='histogram' )
# calculate area of histogram
area_hist_log = .0
for ii in range( counts.size):
    area_hist_log += (bin_edges[ii+1]-bin_edges[ii]) * counts[ii]
# get pdf-values for log10 - spaced intervals
x_fit_log       = np.logspace( np.log10( samples.min()), np.log10( samples.max()), 100 )
samples_fit_log = scipy.stats.lognorm.pdf( x_fit_log, shape, loc=loc, scale=scale )
# scale pdf output:
samples_fit_log *= x_fit_log / (scale * np.exp(shape**2/2.))
# alternatively you could do:
#area_hist_log = np.dot(np.diff(np.log(bin_edges)), counts)
#samples_fit_log *= x_fit_log

# oplot fit into histogram
plt.plot( x_fit_log, samples_fit_log*area_hist_log, label='fitted and area-scaled PDF', linewidth=2 )

plt.xscale( 'log' )
plt.xlim( bin_edges.min(), bin_edges.max() )
plt.legend()
plt.show()

Answer 3

正如@Christoph指出的那样，问题在于您缩放采样pdf的方式。

因为pdf是概率密度的密度，如果你想要一个bin中的预期频率，你应该首先将密度乘以bin长度来得到样本落入这个bin的概率，然后你可以乘以这个概率乘以样本总数来估计落入这个箱子的样本数量。

换句话说，每个bin应该以对数标度不均匀地缩放，而你使用“hist下的区域”统一缩放它们。作为修复，您可以执行以下操作：

# make a histrogram with a log10 x-axis
plt.subplot( 1, 2, 2 )
# equally sized bins (in log10-scale)
bins_log10 = np.logspace( np.log10( samples.min()  ), np.log10( samples.max() ), N_bins )
counts, bin_edges, ignored = plt.hist( samples, bins_log10, histtype='stepfilled', label='histogram' )
# calculate length of each bin
len_bin_log = np.zeros([bins_log10.size,])
for ii in range( counts.size):
    len_bin_log[ii] = (bin_edges[ii+1]-bin_edges[ii])

# get pdf-values for log10 - spaced intervals
# x_fit_log       = np.logspace( np.log10( samples.min()), np.log10( samples.max()), N_bins )
samples_fit_log = scipy.stats.lognorm.pdf( bins_log10, shape, loc=loc, scale=scale )

# oplot fit into histogram
plt.plot(bins_log10 , np.multiply(samples_fit_log,len_bin_log)*sum(counts), label='fitted and area-scaled PDF', linewidth=2 )
plt.xscale( 'log' )
plt.xlim( bin_edges.min(), bin_edges.max() )
# plt.legend()
plt.show()

此外，您可能还需要考虑以类似的方式修改线性比例的缩放方法。实际上，您不需要累积面积，只需要按箱尺寸，然后按样本总数计算多个密度。

更新

我认为目前用于估算箱内概率的方法可能不是最准确的方法。由于pdf曲线是凹的，因此使用中点上的样本进行估计可能更准确。

使用python

3 个答案:

更新