因此,我正在尝试创建一个包含 9 个图形的 3x3 框,虽然我已经通过手动为每个单独的框编写代码来做到这一点,但我想学习如何使用循环来完成它。我似乎无法弄清楚。目前,我正在使用以下内容:
from matplotlib import gridspec
f, ((ax1, ax2, ax3), (ax4, ax5, ax6), (ax7, ax8, ax9)) = plt.subplots(3, 3, sharex='col', sharey='row')
gs = gridspec.GridSpec(3, 3)
fig = plt.figure(figsize=(20,20))
fig.text(0.5, .95, 'Constant Slope for [O/Fe]/[Fe/H] for Various R and Z', ha='center', va='center', size = 50)
fig.text(0.5, 0.08, '[Fe/H]', ha='center', va='center', size = 60)
fig.text(0.09, 0.5, '[O/Fe]', ha='center', va='center', rotation='vertical', size = 60)
ax1 = plt.subplot(gs[0])
histogram1 = ax1.hist2d(fehsc, ofesc, bins=nbins, range=[[-1,.5],[0.2,0.4]])
counts = histogram1[0]
xpos = histogram1[1]
ypos = histogram1[2]
image = histogram1[3]
newcounts = counts #we're going to iterate over this
for i in range (nbins):
xin = xpos[i]
yin = ypos
yline = m*xin + b
reset = np.where(yin < yline) #anything less than yline we want to be 0
#index = index[0:len(index)-1]
countout = counts[i]
countout[reset] = 0
newcounts[i] = countout
ax1.plot(xarr2, yarr2, color='w', linewidth='5', alpha = 0.3)
ax1.plot(xarr, yarr, color='r')
ax1.set_title('R in [5,7] kpc | Z in [1,2] kpc', size = 20)
ax2 = plt.subplot(gs[1])
histogram2 = ax2.hist2d(fehsc2, ofesc2, bins=nbins, range=[[-1,.5],[0.2,0.4]])
counts = histogram2[0]
xpos = histogram2[1]
ypos = histogram2[2]
image = histogram2[3]
newcounts = counts #we're going to iterate over this
for i in range (nbins):
xin = xpos[i]
yin = ypos
yline = m*xin + b
reset = np.where(yin < yline) #anything less than yline we want to be 0
#index = index[0:len(index)-1]
countout = counts[i]
countout[reset] = 0
newcounts[i] = countout
ax2.plot(xarr2, yarr2, color='w', linewidth='5', alpha = 0.3)
ax2.plot(xarr, yarr, color='r')
ax2.set_title('R in [7,9] kpc | Z in [1,2] kpc', size = 20)
以此类推,直到 ax9。
我尝试做的是以下内容:
for k in range(1,10):
ax[k] = plt.subplot(gs[0])
histogram1 = ax[k].hist2d(fehsc, ofesc, bins=nbins, range=[[-1,.5],[0.2,0.4]])
counts = histogram1[0]
xpos = histogram1[1]
ypos = histogram1[2]
image = histogram1[3]
newcounts = counts #we're going to iterate over this
for i in range (nbins):
xin = xpos[i]
yin = ypos
yline = m*xin + b
reset = np.where(yin < yline) #anything less than yline we want to be 0
countout = counts[i]
countout[reset] = 0
newcounts[i] = countout
ax[k].plot(xarr2, yarr2, color='w', linewidth='5', alpha = 0.3)
ax[k].plot(xarr, yarr, color='r')
ax[k].set_title('R in [5,7] kpc | Z in [1,2] kpc', size = 20)
因为我想在循环中使用 ax(k) 并一次运行所有九次迭代。但显然这不是这样做的方法,或者只是行不通。调用 ax_ 时是否可以在循环中从 1 到 9 迭代它?
最佳答案
迭代多个子图的一种简单方法是通过
创建它们fig, axes = plt.subplots(nrows=n, ncols=m)
axes
是 n
行和 m
列的数组。然后可以独立地遍历行和列,或者遍历轴数组的扁平版本。
后者显示在这个例子中:
import matplotlib.pyplot as plt
import numpy as np
x = np.arange(11)
y = np.random.rand(len(x), 9)*10
fig, axes = plt.subplots(3,3, sharex=True, sharey=True)
for i, ax in enumerate(axes.flatten()):
ax.bar(x, y[:,i], color=plt.cm.Paired(i/10.))
plt.show()
关于python - 如何在 python 中为子图运行智能循环,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42845887/