如何在python中隐藏刻度标签但保持刻度到位？
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How to hide ticks label in python but keep the ticks in place?
提问by Yotam
I want to hide my ticks label on a plot I created, but keep this tick itself (the little marks on the axis). When I try to use what I've found here, for example, the entire tick is removed, and not just the labels. How can I remove only the labels then?
我想在我创建的图上隐藏我的刻度标签，但保留这个刻度本身（轴上的小标记）。例如，当我尝试使用我在此处找到的内容时，会删除整个刻度线，而不仅仅是标签。我怎样才能只删除标签呢？
采纳答案by Hooked
Set the tick labels not to be an empty array, but to be a list of empty strings. In the example below, I've replaced the xtick labels with empty strings, leaving the y axis untouched. From here you can selectively keep certain labels if you wished.
将刻度标签设置为不是空数组，而是空字符串列表。在下面的例子中，我用空字符串替换了 xtick 标签，y 轴保持不变。如果您愿意，您可以从这里有选择地保留某些标签。
import pylab as plt
fig, ax = plt.subplots()
ax.plot([1,2,3],[4,5,6])
labels = [item.get_text() for item in ax.get_xticklabels()]
empty_string_labels = ['']*len(labels)
ax.set_xticklabels(empty_string_labels)
plt.show()
This code is adapted from a previous answer.
此代码改编自以前的答案。
回答by divenex
Here is a slightly simpler answer, using ax.tick_params
这是一个稍微简单的答案，使用 ax.tick_params
import matplotlib.pylab as plt
fig, ax = plt.subplots()
plt.plot([1,2,3],[4,5,6])
ax.tick_params(labelbottom=False)
plt.show()
回答by jbplasma
I didn't find divenex's answer to work for me. This answer creates ticks without triggering any automatic labeling, hence no requirement to use "labelbottom= False":
我没有找到适合我的 Divenex 答案。此答案会在不触发任何自动标记的情况下创建刻度，因此不需要使用“labelbottom= False”：
if the bottom and left ticks already exist:
如果底部和左侧刻度已经存在：
import matplotlib.pyplot as plt
fig, ax = plt.subplots()
plt.plot([1,2,3],[4,5,6])
ax.tick_params(right= True,top= True)
if no ticks already exist:
如果没有蜱已经存在：
import matplotlib.pyplot as plt
fig, ax = plt.subplots()
plt.plot([1,2,3],[4,5,6])
ax.tick_params(right= True,top= True,left= True, bottom= True)
