我想构建一个图形,在其底部x轴上具有一组刻度线,在其顶部x轴上具有与底部刻度线对齐的另一组刻度线。特别是在我的情况下,这些是批次和时期。对于底部的每个n
批处理点(不一定是刻度),我希望在顶部有一个纪元刻度。考虑以下示例:
import numpy as np
import matplotlib.pyplot as plt
batches = np.arange(1,101)
epoch_ends = batches[[(i*10)-1 for i in range(1,11)]]
accuracy = np.apply_along_axis(arr=batches, axis=0, func1d=lambda x : x/len(batches))
loss = np.apply_along_axis(arr=batches, axis=0, func1d=lambda x : 1 - (x/len(batches)))
fig, ax1 = plt.subplots( nrows=1, ncols=1 )
ax2 = ax1.twinx()
ax3 = ax1.twiny()
ax1.set_xlabel('batches')
ax1.set_xticks(np.arange(1, len(batches)+1, 9))
ax1.set_ylabel('accuracy')
ax1.grid()
ax2.set_ylabel('loss')
ax2.set_yticklabels(np.linspace(3, 10, 9))
ax3.set_xlabel('epochs')
ax3.set_xticks(epoch_ends)
ax3.set_xticklabels(range(1, len(epoch_ends)+1))
acc_plt = ax1.plot(batches, accuracy, color='red', label='acc')
loss_plt = ax2.plot(batches, loss, color='blue', label='loss')
lns = acc_plt+loss_plt
labs = [l.get_label() for l in lns]
ax1.legend(lns, labs, loc=2)
plt.show()
batches
和epoch_ends
分别看起来像这样
[ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72
73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
91 92 93 94 95 96 97 98 99 100]
[ 10 20 30 40 50 60 70 80 90 100]
因此,我希望纪元刻度1与批处理x-coorrdiante 10、2与20等对齐。
但是,正如您在图片中看到的那样,它们没有对齐。
要使此功能生效,我需要更改什么代码?
答案 0 :(得分:2)
这是对齐它们的一种方法。想法如下:
ax1
)在较低的x轴上绘制数据ax3.set_xlim(ax1.get_xlim())
将x轴的上限设置为与x轴的下限相同。ax3.set_xticklabels()
重命名刻度标签。代码是这里:我用注释#
替换了代码中已有的部分。
# imports and define data and compute accuracy and loss here
# Initiate figure and axis objects here
ax1.set_xlabel('batches')
ax1.set_xticks(np.arange(1, len(batches)+1, 9))
ax1.set_ylabel('accuracy')
ax1.grid()
acc_plt = ax1.plot(batches, accuracy, color='red', label='acc')
loss_plt = ax2.plot(batches, loss, color='blue', label='loss')
ax2.set_ylabel('loss')
ax2.set_yticklabels(np.linspace(3, 10, 9))
new_tick_locations = np.arange(1, 11)*10
new_tick_labels = range(1, 11)
ax3.set_xlabel('epochs')
ax3.set_xlim(ax1.get_xlim())
ax3.set_xticks(new_tick_locations)
ax3.set_xticklabels(new_tick_labels)
# Set legends and show the plot
答案 1 :(得分:1)
这使您无法控制刻度线(您没有说是否需要),但是会像您想要的那样将两个x轴对齐: (注释新行)
import numpy as np
import matplotlib.pyplot as plt
batches = np.arange(1,101)
epoch_ends = batches[[(i*10)-1 for i in range(1,11)]]
accuracy = np.apply_along_axis(arr=batches, axis=0, func1d=lambda x : x/len(batches))
loss = np.apply_along_axis(arr=batches, axis=0, func1d=lambda x : 1 - (x/len(batches)))
fig, ax1 = plt.subplots( nrows=1, ncols=1 )
ax2 = ax1.twinx()
ax3 = ax1.twiny()
ax1.set_xlabel('batches')
#ax1.set_xticks(np.arange(1, len(batches)+1, 9))
ax1.set_xlim(0, 100) # Set xlim
ax1.set_ylabel('accuracy')
ax1.grid()
ax2.set_ylabel('loss')
ax2.set_yticklabels(np.linspace(3, 10, 9))
ax3.set_xlabel('epochs')
#ax3.set_xticks(epoch_ends)
#ax3.set_xticklabels(range(1, len(epoch_ends)+1))
ax3.set_xlim(0, 10) # Set xlim
acc_plt = ax1.plot(batches, accuracy, color='red', label='acc')
loss_plt = ax2.plot(batches, loss, color='blue', label='loss')
lns = acc_plt+loss_plt
labs = [l.get_label() for l in lns]
ax1.legend(lns, labs, loc=2)
plt.show()