我使用下面的代码绘制一条有两个斜率的线,如图所示。斜率应该在一定限度[limit = 5]后下降。我正在使用矢量化方法来设置斜率值。是否有其他方法来设置斜率值。有人可以帮助我吗?
import matplotlib.pyplot as plt
import numpy as np
#Setting the condition
L=5 #Limit
m=1 #Slope
c=0 #Intercept
x=np.linspace(0,10,1000)
#Calculate the y value
y=m*x+c
#plot the line
plt.plot(x,y)
#Set the slope values using vectorisation
m[(x<L)] = 1.0
m[(x>L)] = 0.75
# plot the line again
plt.plot(x,y)
#Display with grids
plt.grid()
plt.show()
答案 0 :(得分:1)
按照您的代码,您应该像这样修改主要部分:
x=np.linspace(0,10,1000)
m = np.empty(x.shape)
c = np.empty(x.shape)
m[(x<L)] = 1.0
c[x<L] = 0
m[(x>L)] = 0.75
c[x>L] = L*(1.0 - 0.75)
y=m*x+c
plt.plot(x,y)
答案 1 :(得分:1)
你可能正在过度思考这个问题。图中有两个线段:
您知道A = 5
,m = 1
,所以A' = 5
。你也知道B = 10
。鉴于(B' - A') / (B - A) = 0.75
,我们有B' = 8.75
。因此,您可以按如下方式制作图:
from matplotlib import pyplot as plt
m0 = 1
m1 = 0.75
x0 = 0 # Intercept
x1 = 5 # A
x2 = 10 # B
y0 = 0 # Intercept
y1 = y0 + m0 * (x1 - x0) # A'
y2 = y1 + m1 * (x2 - x1) # B'
plt.plot([x0, x1, x2], [y0, y1, y2])
希望您能看到针对给定限制集计算y值的模式。结果如下:
现在让我们说你确实想要使用矢量化这个不明原因。您可能希望预先计算所有y值并绘制一次,否则您将得到奇怪的结果。以下是对原始代码的一些修改:
from matplotlib import pyplot as plt
import numpy as np
#Setting the condition
L = 5 #Limit
x = np.linspace(0, 10, 1000)
lMask = (x<=L) # Avoid recomputing this mask
# Compute a vector of slope values for each x
m = np.zeros_like(x)
m[lMask] = 1.0
m[~lMask] = 0.75
# Compute the y-intercept for each segment
b = np.zeros_like(x)
#b[lMask] = 0.0 # Already set to zero, so skip this step
b[~lMask] = L * (m[0] - 0.75)
# Compute the y-vector
y = m * x + b
# plot the line again
plt.plot(x, y)
#Display with grids
plt.grid()
plt.show()