show origin axis (x,y) in matplotlib plot

Question

I have following simple plot, and I would like to display the origin axis (x, y). I already have grid, but I need the x, y axis to be emphasized.

this is my code:

x = linspace(0.2,10,100) plot(x, 1/x) plot(x, log(x)) axis('equal') grid() 

I have seen this question. The accepted answer suggests to use "Axis spine" and just links to some example. The example is however too complicated, using subplots. I am unable to figure out, how to use "Axis spine" in my simple example.

Answer 1

Using subplots is not too complicated, the spines might be.

Dumb, simple way:

%matplotlib inline import numpy as np import matplotlib.pyplot as plt  x = np.linspace(0.2,10,100) fig, ax = plt.subplots() ax.plot(x, 1/x) ax.plot(x, np.log(x)) ax.set_aspect('equal') ax.grid(True, which='both')  ax.axhline(y=0, color='k') ax.axvline(x=0, color='k') 

And I get:

(you can't see the vertical axis since the lower x-limit is zero.)

Alternative using simple spines

%matplotlib inline import numpy as np import matplotlib.pyplot as plt  x = np.linspace(0.2,10,100) fig, ax = plt.subplots() ax.plot(x, 1/x) ax.plot(x, np.log(x)) ax.set_aspect('equal') ax.grid(True, which='both')  # set the x-spine (see below for more info on `set_position`) ax.spines['left'].set_position('zero')  # turn off the right spine/ticks ax.spines['right'].set_color('none') ax.yaxis.tick_left()  # set the y-spine ax.spines['bottom'].set_position('zero')  # turn off the top spine/ticks ax.spines['top'].set_color('none') ax.xaxis.tick_bottom() 

Alternative using seaborn (my favorite)

import numpy as np import matplotlib.pyplot as plt import seaborn seaborn.set(style='ticks')  x = np.linspace(0.2,10,100) fig, ax = plt.subplots() ax.plot(x, 1/x) ax.plot(x, np.log(x)) ax.set_aspect('equal') ax.grid(True, which='both') seaborn.despine(ax=ax, offset=0) # the important part here 

Using the set_position method of a spine

Here are the docs for a the set_position method of spines:

Spine position is specified by a 2 tuple of (position type, amount). The position types are:

• 'outward' : place the spine out from the data area by the specified number of points. (Negative values specify placing the
  spine inward.)

• 'axes' : place the spine at the specified Axes coordinate (from 0.0-1.0).

• 'data' : place the spine at the specified data coordinate.

Additionally, shorthand notations define a special positions:

• 'center' -> ('axes',0.5)
• 'zero' -> ('data', 0.0)

So you can place, say the left spine anywhere with:

ax.spines['left'].set_position((system, poisition))

where system is 'outward', 'axes', or 'data' and position in the place in that coordinate system.

Answer 2

Let me answer to this (rather old) question for those who will search for it as I just did. Although it suggested working solutions, I consider the (only) provided answer as way too complex, when it comes to such a simple situation like that described in the question (note: this method requires you to specify all axes endpoints).

I found a simple working solution in one of the first tutorials on matplotlib's pyplot. It is sufficient to add the following line after the creation of the plot

plt.axis([xmin, xmax, ymin, ymax])

as in the following example:

from matplotlib import pyplot as plt  xs = [1,2,3,4,5] ys = [3,5,1,2,4]  plt.scatter(xs, ys) plt.axis([0,6,0,6])  #this line does the job plt.show() 

which produces the following result:

matplotlib plot with axes endpoints specified