更新标题：Scipy.stats pdf bug？

Question

我有一个二维高斯分布的简单图

from scipy.stats import multivariate_normal
from matplotlib import pyplot as plt

means = [ 1.03872615e+00, -2.66927843e-05]
cov_matrix =  [[3.88809050e-03, 3.90737359e-06], [3.90737359e-06, 4.28819569e-09]]

# This works
a_lims = [0.7, 1.3]
b_lims = [-5, 5]

# This does not work
a_lims = [0.700006488869478, 1.2849292618191401]
b_lims =[-5.000288311285968, 5.000099437047633]

dist = multivariate_normal(mean=means, cov=cov_matrix)
a_plot, b_plot = np.mgrid[a_lims[0]:a_lims[1]:1e-2, b_lims[0]:b_lims[1]:0.1]
pos = np.empty(a_plot.shape + (2,))
pos[:, :, 0] = a_plot
pos[:, :, 1] = b_plot
z = dist.pdf(pos)

plt.figure()
plt.contourf(a_plot, b_plot, z, cmap='coolwarm',  levels=100)

如果我使用标记为“this works”的限制，我会得到以下曲线图（正确）

但是，如果我使用相同的限制，但稍作调整，它的绘图完全错误，因为本地化为不同的值（如下）

我想这是mgrid中的一个bug。有人有什么想法吗？更具体地说，为什么分布的最大值会移动

Answer 1

只关注xaxis：

In [443]: a_lims = [0.7, 1.3] 
In [444]: np.mgrid[a_lims[0]:a_lims[1]:1e-2]                                                   
Out[444]: 
array([0.7 , 0.71, 0.72, 0.73, 0.74, 0.75, 0.76, 0.77, 0.78, 0.79, 0.8 ,
       0.81, 0.82, 0.83, 0.84, 0.85, 0.86, 0.87, 0.88, 0.89, 0.9 , 0.91,
       0.92, 0.93, 0.94, 0.95, 0.96, 0.97, 0.98, 0.99, 1.  , 1.01, 1.02,
       1.03, 1.04, 1.05, 1.06, 1.07, 1.08, 1.09, 1.1 , 1.11, 1.12, 1.13,
       1.14, 1.15, 1.16, 1.17, 1.18, 1.19, 1.2 , 1.21, 1.22, 1.23, 1.24,
       1.25, 1.26, 1.27, 1.28, 1.29, 1.3 ])
In [445]: a_lims = [0.700006488869478, 1.2849292618191401]                                     
In [446]: np.mgrid[a_lims[0]:a_lims[1]:1e-2]                                                   
Out[446]: 
array([0.70000649, 0.71000649, 0.72000649, 0.73000649, 0.74000649,
       0.75000649, 0.76000649, 0.77000649, 0.78000649, 0.79000649,
       0.80000649, 0.81000649, 0.82000649, 0.83000649, 0.84000649,
       0.85000649, 0.86000649, 0.87000649, 0.88000649, 0.89000649,
       0.90000649, 0.91000649, 0.92000649, 0.93000649, 0.94000649,
       0.95000649, 0.96000649, 0.97000649, 0.98000649, 0.99000649,
       1.00000649, 1.01000649, 1.02000649, 1.03000649, 1.04000649,
       1.05000649, 1.06000649, 1.07000649, 1.08000649, 1.09000649,
       1.10000649, 1.11000649, 1.12000649, 1.13000649, 1.14000649,
       1.15000649, 1.16000649, 1.17000649, 1.18000649, 1.19000649,
       1.20000649, 1.21000649, 1.22000649, 1.23000649, 1.24000649,
       1.25000649, 1.26000649, 1.27000649, 1.28000649])
In [447]: _444.shape                                                                           
Out[447]: (61,)
In [449]: _446.shape                                                                           
Out[449]: (59,)

mgrid当a:b:c等给定范围使用np.arange(a, b, c)时arange当给定的浮点步长相对于终点不可靠时

mgrid允许您使用np.linspace，这对于浮点步骤来说是更好的选择。例如，对于第一组限制：

In [453]: a_lims = [0.7, 1.3]                                                                  
In [454]: np.mgrid[a_lims[0]:a_lims[1]:61j]                                                    
Out[454]: 
array([0.7 , 0.71, 0.72, 0.73, 0.74, 0.75, 0.76, 0.77, 0.78, 0.79, 0.8 ,
       0.81, 0.82, 0.83, 0.84, 0.85, 0.86, 0.87, 0.88, 0.89, 0.9 , 0.91,
       0.92, 0.93, 0.94, 0.95, 0.96, 0.97, 0.98, 0.99, 1.  , 1.01, 1.02,
       1.03, 1.04, 1.05, 1.06, 1.07, 1.08, 1.09, 1.1 , 1.11, 1.12, 1.13,
       1.14, 1.15, 1.16, 1.17, 1.18, 1.19, 1.2 , 1.21, 1.22, 1.23, 1.24,
       1.25, 1.26, 1.27, 1.28, 1.29, 1.3 ])

===

通过大幅缩小b_lims并生成更精细的网格，我得到了一个漂亮的倾斜椭圆

means = [ 1, 0]
a_lims = [0.7, 1.3]
b_lims = [-.0002,.0002]

dist = multivariate_normal(mean=means, cov=cov_matrix)
a_plot, b_plot = np.mgrid[ a_lims[0]:a_lims[1]:1001j, b_lims[0]:b_lims[1]:1001j]

所以我认为你的图中的差异是垂直方向上过粗的网格造成的。这可能会影响pdf生成和轮廓

原始网格点的高分辨率绘图。只有一个b级别与高概率值相交。由于椭圆是倾斜的，两个网格采样不同的部分，因此看起来不同的PDF