matplotlib 如何更改条形图中单个条形的颜色

hwamh0ep  于 2023-10-24  发布在  其他
关注(0)|答案(3)|浏览(122)

假设,我有下面的条形图:

有什么想法如何设置不同的颜色为每个载体?作为例如,AK将是红色,GA将是绿色,等等?
我在Python中使用Pandas和matplotlib

>>> f=plt.figure()
>>> ax=f.add_subplot(1,1,1)
>>> ax.bar([1,2,3,4], [1,2,3,4])
<Container object of 4 artists>
>>> ax.get_children()
[<matplotlib.axis.XAxis object at 0x6529850>, <matplotlib.axis.YAxis object at 0x78460d0>,  <matplotlib.patches.Rectangle object at 0x733cc50>, <matplotlib.patches.Rectangle object at 0x733cdd0>, <matplotlib.patches.Rectangle object at 0x777f290>, <matplotlib.patches.Rectangle object at 0x777f710>, <matplotlib.text.Text object at 0x7836450>, <matplotlib.patches.Rectangle object at 0x7836390>, <matplotlib.spines.Spine object at 0x6529950>, <matplotlib.spines.Spine object at 0x69aef50>, <matplotlib.spines.Spine object at 0x69ae310>, <matplotlib.spines.Spine object at 0x69aea50>]
>>> ax.get_children()[2].set_color('r') #You can also try to locate the first patches.Rectangle object instead of direct calling the index.

对于上面的建议,我们到底如何才能枚举ax.get_children()并检查对象类型是否为矩形?所以如果对象是矩形,我们将分配不同的随机颜色?

irtuqstp

irtuqstp1#

简单,只需使用.set_color

>>> barlist=plt.bar([1,2,3,4], [1,2,3,4])
>>> barlist[0].set_color('r')
>>> plt.show()

对于你的新问题,也不难,只需要从你的坐标轴上找到一个横杠,一个例子:

>>> f=plt.figure()
>>> ax=f.add_subplot(1,1,1)
>>> ax.bar([1,2,3,4], [1,2,3,4])
<Container object of 4 artists>
>>> ax.get_children()
[<matplotlib.axis.XAxis object at 0x6529850>, 
 <matplotlib.axis.YAxis object at 0x78460d0>,  
 <matplotlib.patches.Rectangle object at 0x733cc50>, 
 <matplotlib.patches.Rectangle object at 0x733cdd0>, 
 <matplotlib.patches.Rectangle object at 0x777f290>, 
 <matplotlib.patches.Rectangle object at 0x777f710>, 
 <matplotlib.text.Text object at 0x7836450>, 
 <matplotlib.patches.Rectangle object at 0x7836390>, 
 <matplotlib.spines.Spine object at 0x6529950>, 
 <matplotlib.spines.Spine object at 0x69aef50>,
 <matplotlib.spines.Spine object at 0x69ae310>, 
 <matplotlib.spines.Spine object at 0x69aea50>]
>>> ax.get_children()[2].set_color('r') 
 #You can also try to locate the first patches.Rectangle object 
 #instead of direct calling the index.

如果您有一个复杂的图,并希望首先识别条形图,请添加以下内容:

>>> import matplotlib
>>> childrenLS=ax.get_children()
>>> barlist=filter(lambda x: isinstance(x, matplotlib.patches.Rectangle), childrenLS)
[<matplotlib.patches.Rectangle object at 0x3103650>, 
 <matplotlib.patches.Rectangle object at 0x3103810>, 
 <matplotlib.patches.Rectangle object at 0x3129850>, 
 <matplotlib.patches.Rectangle object at 0x3129cd0>, 
 <matplotlib.patches.Rectangle object at 0x3112ad0>]
pod7payv

pod7payv2#

我假设你正在使用Series.plot()来绘制你的数据。如果你在这里查看Series.plot()的文档:
http://pandas.pydata.org/pandas-docs/dev/generated/pandas.Series.plot.html
没有列出 color 参数,您可以在其中设置条形图的颜色。
然而,Series.plot()文档在参数列表的末尾声明了以下内容:

kwds : keywords
Options to pass to matplotlib plotting method

这意味着,当您将Series.plot()的 kind 参数指定为 bar 时,Series.plot()实际上将调用matplotlib.pyplot.bar(),并且matplotlib.pyplot.bar()将发送您在Series.plot()参数列表末尾指定的所有额外关键字参数。
如果你在这里查看matplotlib.pyplot.bar()方法的文档:
http://matplotlib.org/api/pyplot_api.html#matplotlib.pyplot.bar
..它还接受位于其参数列表末尾的关键字参数,如果仔细阅读已识别的参数名称列表,其中一个是 color,它可以是指定条形图不同颜色的序列。
综上所述,如果在Series.plot()参数列表的末尾指定 color 关键字参数,关键字参数将被转发到matplotlib.pyplot.bar()方法。下面是证明:

import pandas as pd
import matplotlib.pyplot as plt

s = pd.Series(
    [5, 4, 4, 1, 12],
    index = ["AK", "AX", "GA", "SQ", "WN"]
)

#Set descriptions:
plt.title("Total Delay Incident Caused by Carrier")
plt.ylabel('Delay Incident')
plt.xlabel('Carrier')

#Set tick colors:
ax = plt.gca()
ax.tick_params(axis='x', colors='blue')
ax.tick_params(axis='y', colors='red')

#Plot the data:
my_colors = 'rgbkymc'  #red, green, blue, black, etc.

pd.Series.plot(
    s, 
    kind='bar', 
    color=my_colors,
)

plt.show()

请注意,如果序列中的条数多于颜色数,则颜色将重复。

fnatzsnv

fnatzsnv3#

更新pandas 0.17.0

@7stud对最新Pandas版本的回答需要调用

s.plot( 
    kind='bar', 
    color=my_colors,
)

而不是

pd.Series.plot(
    s, 
    kind='bar', 
    color=my_colors,
)

绘图函数已成为Series、DataFrame对象的成员,* 实际上,使用color参数调用pd.Series.plot会产生错误 *

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