matplotlib 在子区网格中重新定位子区

6kkfgxo0  于 2023-03-19  发布在  其他
关注(0)|答案(2)|浏览(143)

我正在尝试绘制一个包含7个子图的图。目前我正在绘制两列,一列包含4个图,另一列包含3个图,即:

我以如下方式构建这一情节:

#! /usr/bin/env python
    import numpy as plotting
    import matplotlib
    from pylab import *
    x = np.random.rand(20)
    y = np.random.rand(20)
    fig = figure(figsize=(6.5,12))
    subplots_adjust(wspace=0.2,hspace=0.2)
    iplot = 420
    for i in range(7):
       iplot += 1
       ax = fig.add_subplot(iplot)
       ax.plot(x,y,'ko')
       ax.set_xlabel("x")
       ax.set_ylabel("y")
    savefig("subplots_example.png",bbox_inches='tight')

然而,对于出版物来说,我认为这看起来有点难看--我想做的是将最后一个子情节移到两列之间的中心。那么,调整最后一个子情节的位置使其居中的最佳方法是什么呢?也就是说,将前6个子情节放在3X 2的网格中,将下面的最后一个子情节放在两列之间居中。如果可能的话,我希望能够保留for循环,这样我就可以简单地用途:

if i == 6:
       # do something to reposition/centre this plot
mwg9r5ms

mwg9r5ms1#

使用网格规格(doc)和4x4网格,并使每个图跨越2列,如下所示:

import matplotlib.gridspec as gridspec
gs = gridspec.GridSpec(4, 4)
ax1 = plt.subplot(gs[0, 0:2])
ax2 = plt.subplot(gs[0,2:])
ax3 = plt.subplot(gs[1,0:2])
ax4 = plt.subplot(gs[1,2:])
ax5 = plt.subplot(gs[2,0:2])
ax6 = plt.subplot(gs[2,2:])
ax7 = plt.subplot(gs[3,1:3])
fig = gcf()
gs.tight_layout(fig)
ax_lst = [ax1,ax2,ax3,ax4,ax5,ax6,ax7]
oknrviil

oknrviil2#

如果你想保留for循环,你可以用subplot2grid来安排你的绘图,它允许一个colspan参数:

import numpy as np
import matplotlib.pyplot as plt

x = np.random.rand(20)
y = np.random.rand(20)
fig = plt.figure(figsize=(6.5,12))
plt.subplots_adjust(wspace=0.2,hspace=0.2)
iplot = 420
for i in range(7):
    iplot += 1
    if i == 6:
        ax = plt.subplot2grid((4,8), (i//2, 2), colspan=4)
    else:
        # You can be fancy and use subplot2grid for each plot, which doesn't
        # require keeping the iplot variable:
        # ax = plt.subplot2grid((4,2), (i//2,i%2))

        # Or you can keep using add_subplot, which may be simpler:
        ax = fig.add_subplot(iplot)
    ax.plot(x,y,'ko')
    ax.set_xlabel("x")
    ax.set_ylabel("y")
plt.savefig("subplots_example.png",bbox_inches='tight')

相关问题