计算 matplotlib 中曲线 python 的每个点的切线

Question

I made one curve with a series of point. I want to calculate gradient of the jieba curve.

plt.loglog(jieba_ranks, jieba_counts, linestyle='-', label='jieba')
plt.loglog([1,jieba_counts[0]],[jieba_counts[0],1],color='r', linestyle='--', label='zipf\'s law a =1')
plt.legend(loc="best")

plt.title("Zipf plot for jieba")
plt.xlabel("Frequency rank of token")
plt.ylabel("Absolute frequency of token")
plt.grid(True,axis="y",ls="-", which="both")

Edit: I used np.gradient to produce slope of jieba curve and plot it with jieba_ranks

slope_Y = np.gradient(np.log(jieba_counts), np.log(jieba_ranks))
fig1, ax1 = plt.subplots()
ax1.plot(np.log(jieba_ranks), slope_Y)

But, the gradient curve created didn't describe the relationship between the zipf and the jieba. Maybe, I need calculate the distance of each point on zipf and jieba.

Answer 1

numpy makes gradient available, this function would probably be useful for solving your problem

if you add data/code to the question I can try and suggest something more sensible!

Answer 2

Next time when you ask a question, it would be helpful to include a little bit more information, maybe even show us your plot. Because the following example will give bad results if your sampling is not dense enough. (if you have bad sampling you might want to fit a spline to your points and get the tangent of that curve, etc..)

But lets go with the easiest case, when your sampling is dense enough, and your function is not noisy. You can calculate the (forward) derivatives very easily the following way:

import numpy as np
import matplotlib.pyplot as plt

X = np.arange(0,np.pi*2,np.pi*2/100)
Y = np.sin(X)

slope_Y = np.diff(Y)/np.diff(X)

plt.plot(X,Y)
plt.plot(X[:-1],slope_Y)

The original curve is a sin(x), and the derivative of that curve is a cos(x), what you can easily see if you run this code.

If this does not solve your problem, please include more information.

As Sam Mason suggested above you can simply use the gradient function of numpy as well.