这更像是一个几乎可以奏效的黑客。

#!/usr/bin/env python

from pandas import *
import matplotlib.pyplot as plt
from numpy import zeros

# Create original dataframe
df = DataFrame(np.random.rand(5,4), index=['art','mcf','mesa','perl','gcc'],
                        columns=['pol1','pol2','pol3','pol4'])
# Estimate average
average = df.mean()
average.name = 'average'

# Append dummy row with zeros and then average
row = DataFrame([dict({p:0.0 for p in df.columns}), ])

df = df.append(row)
df = df.append(average)

print df

df.plot(kind='bar')
plt.show()

并给出:
             pol1      pol2      pol3      pol4
art      0.247309  0.139797  0.673009  0.265708
mcf      0.951582  0.319486  0.447658  0.259821
mesa     0.888686  0.177007  0.845190  0.946728
perl     0.902977  0.863369  0.194451  0.698102
gcc      0.836407  0.700306  0.739659  0.265613
0        0.000000  0.000000  0.000000  0.000000
average  0.765392  0.439993  0.579993  0.487194


它给出了基准和平均值之间的视觉分隔。
有没有办法去掉X轴上的0个??
原来DataFrame不允许我这样拥有多个伪行。
我的解决办法是改变
row = pd.DataFrame([dict({p:0.0 for p in df.columns}), ])

进入之内
row = pd.Series([dict({p:0.0 for p in df.columns}), ])
row.name = ""

序列可以用空字符串命名。

最佳答案

还是挺老套的,但很管用:

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np

# Create original dataframe
df = pd.DataFrame(np.random.rand(5,4), index=['art','mcf','mesa','perl','gcc'],
                        columns=['pol1','pol2','pol3','pol4'])
# Estimate average
average = df.mean()
average.name = 'average'

# Append dummy row with zeros and then average
row = pd.DataFrame([dict({p:0.0 for p in df.columns}), ])

df = df.append(row)
df = df.reindex(np.where(df.index, df.index, ''))
df = df.append(average)
print df

df.plot(kind='bar')
plt.show()

关于python - 以可视方式分离 Pandas 中的条形图簇,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/13983498/

10-08 23:51