的功能變化的變量對同一軸上

Question

目前我正在試圖繪製它描述了宇宙學線性擾動增長不同的世界模型的函數創建多個地塊。我希望能夠將所有曲線放在同一組軸上，但是我很難設置它。

我的目標是繪製該功能d，相對於Z，但有不同的密度參數（$ \ $歐米茄）多條曲線。

我已成功兩種解決方案但兩者都沒有完美的工作，首先是效率非常低（添加新的功能，每個組參數）：

z = np.arange(0.0,10,0.1) 

#density parameters 
MOm = 1.0 
MOv = 0.0 

COm = 0.3 
COv = 0.7 

H0 = 70 

def Mf(z): 
    A = (5/2)*MOm*(H0**2) 
    H = H0 * np.sqrt(MOm*((1+z)**3) + MOv) 
    return A * ((1+z)/(H**3)) 

def MF(z): 
    res = np.zeros_like(z) 
    for i,val in enumerate(z): 
     y,err = integrate.quad(Mf,val,np.inf) 
     res[i] = y 
    return res 

def MD(z): 
    return (H0 * np.sqrt(MOm*((1+z)**3) + MOv)) * MF(z) 

def Cf(z): 
    A = (5/2)*COm*(H0**2) 
    H = H0 * np.sqrt(COm*((1+z)**3) + COv) 
    return A * ((1+z)/(H**3)) 

def CF(z): 
    res = np.zeros_like(z) 
    for i,val in enumerate(z): 
     y,err = integrate.quad(Cf,val,np.inf) 
     res[i] = y 
    return res 

def CD(z): 
    return (H0 * np.sqrt(COm*((1+z)**3) + COv)) * CF(z) 


plt.plot(z,MD(z),label="Matter Dominated") 
plt.plot(z,CD(z),label="Current Epoch") 

所以我試圖使它具有更簡單的循環，但一直未能解決如何標籤添加到每個小區內環路：

Om = (1.0,0.3) 
Ov = (0.0,0.7) 

for param1,param2 in zip(Om,Ov): 
    def f(z): 
     A = (5/2)*param1*(H0**2) 
     H = H0 * np.sqrt(param1*((1+z)**3) + param2) 
     return A * ((1+z)/(H**3)) 

    def F(z): 
     res = np.zeros_like(z) 
     for i,val in enumerate(z): 
      y,err = integrate.quad(f,val,np.inf) 
      res[i] = y 
     return res 

    def D(z): 
     return (H0 * np.sqrt(param1*((1+z)**3) + param2)) * F(z) 

    plt.plot(z,D(z)) 

可能有人請幫忙解釋一下這樣做的一個有效的方法？或者如何使用for循環在標籤上添加標籤。任何幫助將不勝感激。

Answer 1

所以我試圖使其與一個for循環簡單，但一直未能解決如何標籤添加到每個小區內環路

from scipy import integrate 
from matplotlib import pyplot as plt 

MOm = 1.0 
MOv = 0.0 

COm = 0.3 
COv = 0.7 

z = np.arange(0.0,10,0.1) 
H0 = 70 
Om = (1.0,0.3) 
Ov = (0.0,0.7) 

fig = plt.figure(1) 
for param1,param2 in zip(Om,Ov): 
    def f(z): 
     A = (5/2)*param1*(H0**2) 
     H = H0 * np.sqrt(param1*((1+z)**3) + param2) 
     return A * ((1+z)/(H**3)) 

    def F(z): 
     res = np.zeros_like(z) 
     for i,val in enumerate(z): 
      y,err = integrate.quad(f,val,np.inf) 
      res[i] = y 
     return res 

    def D(z): 
     return (H0 * np.sqrt(param1*((1+z)**3) + param2)) * F(z) 

## Now define labels as you need and labels as follows: 

    plt.plot(z,D(z),label = 'Om: {}, Ov: {}'.format(param1,param2)) 
    plt.legend() 

Answer 2

你可以使用預定義的字符串基於這兩個參數在循環中創建一個標籤，該字符串使用相應的值進行格式化。

label="Om {}, Ov {}".format(param1, param2) 

總體而言，這將給予：

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

z = np.arange(0.0,10,0.1) 
MOm = 1.0 
MOv = 0.0 
COm = 0.3 
COv = 0.7 
H0 = 70 

Om = (1.0,0.3) 
Ov = (0.0,0.7) 
plt.figure(figsize=(3.8,2.4)) 
for param1,param2 in zip(Om,Ov): 
    def f(z): 
     A = (5/2)*param1*(H0**2) 
     H = H0 * np.sqrt(param1*((1+z)**3) + param2) 
     return A * ((1+z)/(H**3)) 

    def F(z): 
     res = np.zeros_like(z) 
     for i,val in enumerate(z): 
      y,err = integrate.quad(f,val,np.inf) 
      res[i] = y 
     return res 

    def D(z): 
     return (H0 * np.sqrt(param1*((1+z)**3) + param2)) * F(z) 

    plt.plot(z,D(z), label="Om {}, Ov {}".format(param1, param2)) 

plt.legend() 
plt.show()