我可以對這些子圖中的重疊標籤做些什麼？

Question

下面是我用matplotlib創建的圖。問題非常明顯 - 標籤重疊，整個事情變得難以理解。

我打過電話tight_layout每個插曲，但這個我崩潰的IPython筆記本內核。

我該如何解決佈局問題？可接受的方法包括修復每個子圖的xlabel，ylabel和標題，但另一種（也許更好）的方法是爲整個圖提供單個xlabel，ylabel和標題。

這是我用來產生上述次要情節的循環：

for i, sub in enumerate(datalist): 
    subnum = i + start_with 
    subplot(3, 4, i) 

    # format data (sub is a PANDAS dataframe) 
    xdat = sub['x'][(sub['in_trl'] == True) & (sub['x'].notnull()) & (sub['y'].notnull())] 
    ydat = sub['y'][(sub['in_trl'] == True) & (sub['x'].notnull()) & (sub['y'].notnull())] 

    # plot 
    hist2d(xdat, ydat, bins=1000) 
    plot(0, 0, 'ro') # origin 

    title('Subject {0} in-Trial Gaze'.format(subnum)) 
    xlabel('Horizontal Offset (degrees visual angle)') 
    ylabel('Vertical Offset (degrees visual angle)') 

    xlim([-.005, .005]) 
    ylim([-.005, .005]) 
    # tight_layout # crashes ipython-notebook kernel 

show() 

更新：

好了，所以ImageGrid似乎是要走的路，但我的身材還是找了一下靠不住的：

這是我使用的代碼：

fig = figure(dpi=300) 
grid = ImageGrid(fig, 111, nrows_ncols=(3, 4), axes_pad=0.1) 

for gridax, (i, sub) in zip(grid, enumerate(eyelink_data)): 
    subnum = i + start_with 

    # format data 
    xdat = sub['x'][(sub['in_trl'] == True) & (sub['x'].notnull()) & (sub['y'].notnull())] 
    ydat = sub['y'][(sub['in_trl'] == True) & (sub['x'].notnull()) & (sub['y'].notnull())] 

    # plot 
    gridax.hist2d(xdat, ydat, bins=1000) 
    plot(0, 0, 'ro') # origin 

    title('Subject {0} in-Trial Gaze'.format(subnum)) 
    xlabel('Horizontal Offset\n(degrees visual angle)') 
    ylabel('Vertical Offset\n(degrees visual angle)') 

    xlim([-.005, .005]) 
    ylim([-.005, .005]) 

show() 

Answer 1

想要ImageGrid（tutorial）。

第一示例直接從該鏈接解除（和輕度修改）：

import matplotlib.pyplot as plt 
from mpl_toolkits.axes_grid1 import ImageGrid 
import numpy as np 

im = np.arange(100) 
im.shape = 10, 10 

fig = plt.figure(1, (4., 4.)) 
grid = ImageGrid(fig, 111, # similar to subplot(111) 
       nrows_ncols = (2, 2), # creates 2x2 grid of axes 
       axes_pad=0.1, # pad between axes in inch. 
       aspect=False, # do not force aspect='equal' 
       ) 

for i in range(4): 
    grid[i].imshow(im) # The AxesGrid object work as a list of axes. 

plt.show()