在matplotlib图中显示原点轴（x，y）

使用subplots不是太复杂，可能是刺。

笨，简单的方法：

%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')

我得到：

（由于x的下限为零，因此看不到垂直轴。）

选择使用简单的刺

%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))

其中system是“向外”，“轴”或“数据”和position在该坐标系中的位置。