python – 我可以对这些子图中的重叠标签做些什么?
发布时间:2020-05-23 19:43:35 所属栏目:Python 来源:互联网
导读:下面是我用matplotlib创建的图.问题非常明显 – 标签重叠,整个事情是一个难以理解的混乱. 我尝试为每个子图调用tight_layout,但这会导致我的ipython-notebook内核崩溃. 我该怎么做才能修复布局?可接受的方法包括修复每个子图的xlabel,ylabel和title,但另一种
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下面是我用matplotlib创建的图.问题非常明显 – 标签重叠,整个事情是一个难以理解的混乱. 我尝试为每个子图调用tight_layout,但这会导致我的ipython-notebook内核崩溃. 我该怎么做才能修复布局?可接受的方法包括修复每个子图的xlabel,ylabel和title,但另一种(也许更好的)方法是为整个图形提供单个xlabel,ylabel和title. 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,'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,所以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,'ro') # origin
title('Subject {0} in-Trial Gaze'.format(subnum))
xlabel('Horizontal Offsetn(degrees visual angle)')
ylabel('Vertical Offsetn(degrees visual angle)')
xlim([-.005,.005])
show()
解决方法你想要ImageGrid( tutorial).第一个例子直接从该链接解除(并轻微修改): 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,# 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.
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