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

2020-07-03 07:59发布

问题:

Is there an easy way to plot a function which tends to infinity in the positive and negative as a single plot, without the plot joining both ends of the positive and negative?

For example, plotting y=1/x using this code gives the resulting plot:

import numpy as np
import matplotlib.pyplot as plt

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

x=np.setdiff1d(np.linspace(-10,10,100),[0]) #to remove the zero
y=f(x)
plt.plot(x, y, label=fx_name)
plt.legend(loc='upper left')
plt.show()

But I would like this output, which I achieve by plotting two separate domains:

import numpy as np
import matplotlib.pyplot as plt

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

xfn=np.setdiff1d(np.linspace(-10,0,100),[0])
xfp=np.setdiff1d(np.linspace(0,10,100),[0])
yfn=f(xfn)
yfp=f(xfp)

yf = plt.plot(xfn, yfn, label=fx_name)
plt.plot(xfp, yfp, color=yf[0].get_color())
plt.legend(loc='upper left')
plt.show()

Is there are short-cut? Many thanks.

Solution

Include zero in the domain array, and suppress the divide by zero. This forces one element of the returned co-domain array as "inf", and "inf" is not plotted.

import numpy as np
import matplotlib.pyplot as plt

def f(x):
    with np.errstate(divide='ignore', invalid='ignore'):
        return 1/x
fx_name = r'$f(x)=\frac{1}{x}$'

x=np.linspace(-10,10,101)
y=f(x)
plt.plot(x, y, label=fx_name)
plt.legend(loc='upper left')
plt.show()

I prefer this method since it avoids manual manipulation of the array, and can be easily reused for other functions which share the same domain (ex. y=1/(x+2)). Thank you all for contributions.

回答1:

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()


回答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.



回答3:

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).