为什么 plt.plot 不显示图表？

Question

我正在尝试 plot 图表中以下计算的结果，x 是 N 的值，y 是计算出的误差，但是当我这样做时，我看不到图表，有人知道为什么吗？

您的支持将非常有帮助。

import numpy as np
import random 
import matplotlib.pyplot as plt 

#enter N to equal 100
N = int(input("Define a value for N: "))


Leibniz_Error = np.zeros(N)
Euler_Error = np.zeros(N)


#Leibniz
sum1 = 0
for k in range (N):       
    sum1= sum1 + 1.0/((4*k + 1)*(4*k + 3))
    Leibniz_Error = np.pi - 8*sum1
sum1 = 8*sum1
Final_Leibniz_Error = abs(np.pi - sum1)
print("Euler=", Final_Leibniz_Error)

#Euler
sum2 = 0
for k in range (1,101):
    sum2 = sum2 + 1/(k**2)
    Euler_Error = np.pi - np.sqrt (6*sum2)
sum2 = 6*sum2
sum2 = np.sqrt(sum2)
Final_Euler_Error = abs(np.pi - sum2)
print("Euler=", Final_Euler_Error)

plt.plot (N, Final_Leibniz_Error)
plt.plot (N, Final_Euler_Error, 'r-')
print (N)

这就是我在 plot 时得到的。

Answer 1

由于您只是在绘制点，因此您必须使用plt.scatter(N, Final_Leibniz_Error)代替，因为 plot 只会在点对之间绘制线

Answer 2

如果你想 plot 错误通过迭代你需要保存渐进式错误

import numpy as np
import random 
import matplotlib.pyplot as plt 

#enter N to equal 100
N = int(input("Define a value for N: "))


Leibniz_Error = np.zeros(N)
Euler_Error = np.zeros(N)


#Leibniz
sum1 = 0
leibniz_errors = []
for k in range (N):       
    sum1= sum1 + 1.0/((4*k + 1)*(4*k + 3))
    Leibniz_Error = np.pi - 8*sum1
    leibniz_errors.append(Leibniz_Error)
sum1 = 8*sum1
Final_Leibniz_Error = abs(np.pi - sum1)
print("Euler=", Final_Leibniz_Error)

#Euler
sum2 = 0
euler_errors = []
for k in range (1,101):
    sum2 = sum2 + 1/(k**2)
    Euler_Error = np.pi - np.sqrt (6*sum2)
    euler_errors.append(Euler_Error)
sum2 = 6*sum2
sum2 = np.sqrt(sum2)
Final_Euler_Error = abs(np.pi - sum2)
print("Euler=", Final_Euler_Error)

plt.plot([i for i in range(N)], leibniz_errors)
plt.plot([i for i in range(100)], euler_errors, 'r-')
plt.show()
print (N, Final_Leibniz_Error)