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python - 提高 for 循环性能和自定义图表

问题描述

我创建了一个代码，它返回我所追求的输出 - 2 个图形，每个图形上有多条线。但是，代码很慢而且很大（就需要多少行代码而言）。我对我可以做的任何改进很感兴趣，这将帮助我更快地获得这样的图表，并使我的代码更美观。

此外，我想在我的图表中添加更多内容（轴名称和标题是我所追求的）。通常，我会使用plt.xlabel,plt.ylabel并plt.title这样做，但是我不太明白如何在这里使用它们。这里的目的是在每个循环之后为每个图形添加一条线（我已经修改了这段代码来这样做）。

我应该注意，我需要使用 Python 来完成这项任务（所以我不能更改为其他任何东西）并且我确实需要 Sympy 库来查找在我的图表中绘制的值。

到目前为止，我的代码如下：

import matplotlib.pyplot as plt
import sympy as sym
import numpy as np

sym.init_printing()
x, y = sym.symbols('x, y') # defining our unknown probabilities
al = np.arange(20,1000,5).reshape((196,1)) # values of alpha/beta
prob_of_strA = []
prob_of_strB = []
colours=['r','g','b','k','y']
pen_values = [[0,-5,-10,-25,-50],[0,-25,-50,-125,-250]]
fig1, ax1 = plt.subplots()
fig2, ax2 = plt.subplots()
for j in range(0,len(pen_values[1])):
    for i in range(0,len(al)): # choosing the value of beta

        A = sym.Matrix([[10, 50], [int(al[i]), pen_values[0][j]]]) # defining matrix A
        B = sym.Matrix([[pen_values[1][j], 50], [int(al[i]), 10]]) # defining matrix B
        sigma_r = sym.Matrix([[x, 1-x]]) # defining the vector of probabilities 
        sigma_c = sym.Matrix([y, 1-y]) # defining the vector of probabilities 
        ts1 = A * sigma_c ; ts2 = sigma_r * B # defining our utilities 

        y_sol = sym.solvers.solve(ts1[0] - ts1[1],y,dict = True) # solving for y
        x_sol = sym.solvers.solve(ts2[0] - ts2[1],x,dict = True) # solving for x
        prob_of_strA.append(y_sol[0][y]) # adding the value of y to the vector
        prob_of_strB.append(x_sol[0][x]) # adding the value of x to the vector
    ax1.plot(al,prob_of_strA,colours[j],label = ["penalty = " + str(pen_values[0][j])]) # plotting value of y for a given penalty value
    ax2.plot(al,prob_of_strB,colours[j],label = ["penalty = " + str(pen_values[1][j])]) # plotting value of x for a given penalty value
    ax1.legend() # showing the legend
    ax2.legend() # showing the legend
    prob_of_strA = [] # emptying the vector for the next round
    prob_of_strB = [] # emptying the vector for the next round

标签： python

解决方案

您可以通过在循环内初始化空向量来节省几行代码。您不必费心在最后重新定义它们。

for j in range(0,len(pen_values[1])):
    prob_of_strA = []
    prob_of_strB = []

   for i in range(0,len(al)): # choosing the value of beta

        A = sym.Matrix([[10, 50], [int(al[i]), pen_values[0][j]]]) # defining matrix A
        B = sym.Matrix([[pen_values[1][j], 50], [int(al[i]), 10]]) # defining matrix B
        sigma_r = sym.Matrix([[x, 1-x]]) # defining the vector of probabilities 
        sigma_c = sym.Matrix([y, 1-y]) # defining the vector of probabilities 
        ts1 = A * sigma_c ; ts2 = sigma_r * B # defining our utilities 

        y_sol = sym.solvers.solve(ts1[0] - ts1[1],y,dict = True) # solving for y
        x_sol = sym.solvers.solve(ts2[0] - ts2[1],x,dict = True) # solving for x
        prob_of_strA.append(y_sol[0][y]) # adding the value of y to the vector
        prob_of_strB.append(x_sol[0][x]) # adding the value of x to the vector

    ax1.plot(al,prob_of_strA,colours[j],label = ["penalty = " + str(pen_values[0][j])]) # plotting value of y for a given penalty value
    ax2.plot(al,prob_of_strB,colours[j],label = ["penalty = " + str(pen_values[1][j])]) # plotting value of x for a given penalty value
    ax1.legend() # showing the legend
    ax2.legend() # showing the legend

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