如何使用 plt.loglog plot matplotlib 中的斜线

Question

Here I need to plot a frequency and a simple line with a slope of -5/3.
The problem is that the main plot is using plt.loglog() and when I want to show the line it gives me nothing or something strange.
Here are the pictures of it. I have plotted the right one and desired is the left one.

I already used np.linspace and some other things, but I was not able to solve the problem. Also, it is not clear at which points I have the first and the end of the frequency plot. That's another reason why I can not use 'np.linspace'. Can anyone help me?

Answer 1

Thanks a lot for your attention. I tried your code but I found out maybe there are better ways to do it with kind of my dataset. So I did this:

• Change the class of dataset to np.array() and had np.log() function on it:

x = ... # type(x) = list
y = ... # type(y) = list
.
.
.
x = np.log(np.array(x))
y = np.log(np.array(y))

In this case I did not have to use plt.loglog() or np.log() and np.exp() in calculations anymore.

• Locate the min, max, and mean for x and y:

ymin, ymax = ([y.min(), y.max()])
ymid = (ymin + ymax) / 2
xmin, xmax = ([x.min(), x.max()])
xmid = (xmin + xmax) / 2

• Use np.linspace() for rest:

slope = - 5 / 3
x1 = np.linspace(xmin, xmax)
y1 = slope * (x1 - xmid) + ymid
ax.plot(x1, y1, 'r')

And got the result I wanted.
the result

---

Edited: the plot in log scale.

Because of that, it is better to use plt.loglog() kind of plots in frequency spectrums, I edited these things:

• Changed back x and y to normal np.array()

 x = array(...) y = array(...)

• find the middle of x and y to have the center of the line and used simple equation of a straight line and then ploted the line using np.exp() :

 ymin, ymax = log([y.min(), y.max()]) ymid = (ymin + ymax) / 2 xmin, xmax = log([x.min(), x.max()]) xmid = (xmin + xmax) / 2 slope = - 5 / 3 y1 = slope * (xmin - xmid) + ymid y2 = slope * (xmax - xmid) + ymid ax.plot(exp([xmin, xmax]), exp([y1, y2]), 'r') plt.loglog()

the result
As you see now we have the plot in log scale.

Answer 2

Here is a possible approach. All calculations happen in logspace. The points are transformed back to linear space to be plotted. As matplotlib plots a line given two points always as straight, independent to the tranformation of the axes, only two points are needed.

Steps:

• find the lowest and highest values of the curve (y0, y1); these define the extension of the sloped line
• find a point near the center of the curve (mid_x, mid_y); this point serves as an anchor point to know where the sloped line should go
• find the x values that correspond to a sloped line through (mid_x, mid_y) and go to y0 and y1

import matplotlib.pyplot as plt
import numpy as np

fig, ax = plt.subplots(figsize=(8, 8))
# first create some toy data roughly resembling the example plot
x = np.linspace(10, 2000, 1000)
y = np.random.normal(2000 / x ** np.linspace(.7, .55, x.size), 100 / x ** .7)
ax.plot(x, y)
y0, y1 = np.log([y.min(), y.max()])
# mid_x, mid_y = np.log([x[x.size // 2],  y[y.size // 2]])
mid_x, mid_y = (np.log(x.min()) + np.log(x.max())) / 2, (y0 + y1) / 2
slope = -5 / 3
x0 = mid_x + slope * (y0 - mid_y)
x1 = mid_x + slope * (y1 - mid_y)
ax.plot(np.exp([x0, x1]), np.exp([y0, y1]), color='crimson')

plt.loglog()
plt.show()