How to plot y=1/x as a single graph [duplicate]

Answer 1

based on Rutger Kassies ides:

n_points = 100
x=np.setdiff1d(np.linspace(-10,10,n_points),[0]) #to remove the zero

y=f(x)
y[n_points//2-1:n_points//2+1] = np.nan

use your original plot an set the points around 0 to np.nan. that way too many points get set to None but it's symmetric.

you could also setup your linspace to includ 0 such that f(x) = np.nan: n_points = 101. (this answer and 2 comments stated that right before i did... please credit there).

Answer 2

Not necessary easier as your workaround, but you could insert a 'nan' element at the index where the sign flips, for example:

idx = np.argmax(np.diff(np.sign(y)))+1

x = np.insert(x, idx, np.nan)
y = np.insert(y, idx, np.nan)

The 'nan' causes Matplotlib to interrupt the line.

Answer 3

Actually you want to include x = 0 because this results in y = nan, forming a gap in the plot.

import numpy as np
import matplotlib.pyplot as plt

def f(x):
    return 1/x
fx_name = r'$f(x)=\frac{1}{x}$'

# using 101 steps results in in array including the value 0
x=np.linspace(-10,10,101)
# f(0) = nan -> a nan value creates a gap
y=f(x)
plt.plot(x, y, label=fx_name)
plt.legend(loc='upper left')
plt.show()