python - 我可以对这些子图中的重叠标签做些什么？

Question

下面是我用 matplotlib 创建的图。问题很明显——标签重叠，整个事情变得难以理解。

我尝试调用tight_layout每个子图，但这会使我的 ipython-notebook 内核崩溃。

我能做些什么来修复布局？可接受的方法包括为每个子图固定 xlabel、ylabel 和标题，但另一种（也许更好）的方法是为整个图形设置一个 xlabel、ylabel 和标题。

这是我用来生成上述子图的循环：

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：

好的，这ImageGrid似乎是要走的路，但我的身材仍然看起来有点不稳定：

这是我使用的代码：

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

你想要ImageGrid（教程）。

第一个示例直接从该链接中提取（并稍作修改）：

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()