Question

我正在尝试空白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()

一切都运作良好，但真正的Stüve图应该看起来像Stüve diagram with sounding

如您所见，y - 轴具有特定比例：p **（R / cp）....（= p **（287.04）/ 1005）

我该怎么处理我的程序，以便我的y - 轴看起来像示例中的轴？

Answer 1

您必须为此定义自己的轴刻度。我的回答基于this answer和custom scale example。

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您的y =轴提供自定义比例：

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

以下图表将自定义比例与对数比例（plt.gca().set_yscale('log')）和默认值无比例进行比较：

example

如何将y轴缩放为y **常数

1 个答案: