如何将y轴缩放成y**constan

Question

我想做一个空白的Stüve diagram。
我的代码是：

R=287.04  #Jkg^-1K^-1
cp=1005   #Jkg^-1K^-1
p0=1000 #hPa
L=2.5*10**6 #J kg^-1
temp_celsius = np.array(range(-80,41))
temp_kelvin=temp_celsius+273.15
pressure_hPa=np.array(range(1050,90,-10))

#make a grid of the data
tempdata,pressdata=np.meshgrid(temp_kelvin,pressure_hPa*100.)

#Initialise the arrays
pot_temp_kelvin=np.zeros(tempdata.shape)
es=pot_temp_kelvin
ms=es
pseudo_pot_temp_kelvin=es

#Get the potential temperature                         
pot_temp_kelvin=tempdata*(p0*100/pressdata)**(R/cp)

#Get the saturation mix ratio
#first the saturation vapor pressure after Magnus
#Definition of constants for the Magnus-formula
c1=17.62
c2=243.12

es=6.112*np.exp((17.62*(tempdata-273.15))/(243.12+tempdata-273.15)) #hPa
#Now I'm calculation the saturation mixing ratio
ms=622*(es/(pressdata/100-es)) #g/kg

#At least I need the pseudo-adiabatic potential temperature
pseudo_pot_temp_kelvin=pot_temp_kelvin*np.exp(L*ms/1000./cp/tempdata)

#define the levels which should be plotted in the figure
levels_theta=np.array(range(200,405,5))
levels_ms=np.array([0.1,0.2,0.5, 1.0, 1.5, 2.0, 3.0, 4.0, 6.0, 8.0, 10.0, 12.0, 15.0, 20.0, 25.0, 30.0])
levels_theta_e=np.array(range(220,410,10))

#The plot
fig = plt.figure(figsize=(15,15))
theta=plt.contour(temp_celsius,pressure_hPa,pot_temp_kelvin,levels_theta,colors='blue')
plt.clabel(theta,levels_theta[0::2], fontsize=10, fmt='%1.0f')

sat_mix_ratio=plt.contour(temp_celsius,pressure_hPa,ms,levels_ms,colors='green')
plt.clabel(sat_mix_ratio,fontsize=10,fmt='%1.1f')
    theta_e=plt.contour(temp_celsius,pressure_hPa,pseudo_pot_temp_kelvin,levels_theta_e,colors='red')
plt.clabel(theta_e,levels_theta_e[1::2],fontsize=10,fmt='%1.0f')

plt.xlabel('Temperature [$^\circ$C]')
plt.ylabel('Pressure [hPa]')
plt.xticks(range(-80,45,5))
plt.xlim((-80,50))
plt.yticks(range(1050,50,-50))
plt.gca().invert_yaxis()
plt.grid(color='black',linestyle='-')
plt.show()

一切正常，但真正的Stüve图应该看起来像Stüve diagram with sounding

如您所见，y-轴有一个特定的标度：p**（R/cp）。。。。（=p**（287.04）/1005）

为了使我的y轴看起来像示例中的轴，我必须对我的程序做些什么？你知道吗

Answer 1

您必须为此定义自己的轴比例。我的答案是基于this answer和custom scale example。你知道吗

这是自定义缩放类：

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.ticker import FormatStrFormatter
from matplotlib import scale as mscale
from matplotlib import transforms as mtransforms

class CustomScale(mscale.ScaleBase):
    name = 'custom'

    def __init__(self, axis, **kwargs):
        mscale.ScaleBase.__init__(self)
        self.thresh = None #thresh

    def get_transform(self):
        return self.CustomTransform(self.thresh)

    def set_default_locators_and_formatters(self, axis):
        pass

    class CustomTransform(mtransforms.Transform):
        input_dims = 1
        output_dims = 1
        is_separable = True

        def __init__(self, thresh):
            mtransforms.Transform.__init__(self)
            self.thresh = thresh

        def transform_non_affine(self, a):
            return a**(R/cp)

        def inverted(self):
            return CustomScale.InvertedCustomTransform(self.thresh)

    class InvertedCustomTransform(mtransforms.Transform):
        input_dims = 1
        output_dims = 1
        is_separable = True

        def __init__(self, thresh):
            mtransforms.Transform.__init__(self)
            self.thresh = thresh

        def transform_non_affine(self, a):
            return a**(cp/R)

        def inverted(self):
            return CustomScale.CustomTransform(self.thresh)

# Now that the Scale class has been defined, it must be registered so
# that ``matplotlib`` can find it.
mscale.register_scale(CustomScale)

现在，您可以使用以下选项为y=轴提供自定义比例：

plt.gca().set_yscale('custom')

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下图将自定义比例与对数比例（plt.gca().set_yscale('log')）和默认无比例进行比较：