python - 具有自定义y频率python的直方图

我正在尝试绘制以下数据

+-----------+------+------+
| Duration  | Code | Seq. |
+-----------+------+------+
|    116.15 |   65 |    1 |
|    120.45 |   65 |    1 |
|    118.92 |   65 |    1 |
|      7.02 |   66 |    1 |
|     73.93 |   66 |    2 |
|    117.53 |   66 |    1 |
|       4.4 |   66 |    2 |
|    111.03 |   66 |    1 |
|      4.35 |   66 |    1 |
+-----------+------+------+

我的代码为：

x1 = df.loc[df.Code==65, 'Duration']
x2 = df.loc[df.Code==66, 'Duration']
kwargs = dict(alpha=0.5, bins=10)
plt.hist(x1, **kwargs, color='k', label='Code 65')
plt.hist(x2, **kwargs, color='g', label='Code 66')

我理想地在y轴上想要的是Seq的数目。对应于x轴上的不同Durations。但是现在，我只得到y上Durations的计数。我该如何纠正？

最佳答案

您可以使用熊猫对“ x”值进行分箱，然后使用条形图代替。

import pandas as pd
import matplotlib.pyplot as plt
df = pd.DataFrame({'Duration':[116.15, 120.45,118.92,7.02,73.93, 117.53, 4.4, 111.03, 4.35]})
df['Code'] = [65,65,65,66,66,66,66,66,66]
df['Seq.'] = [1,1,1,1,2,1,2,1,1]
df

   Duration  Code  Seq.
0    116.15    65     1
1    120.45    65     1
2    118.92    65     1
3      7.02    66     1
4     73.93    66     2
5    117.53    66     1
6      4.40    66     2
7    111.03    66     1
8      4.35    66     1

df['bin'] = pd.cut(df['Duration'],10, labels=False)
df

   Duration  Code  Seq.  bin
0    116.15    65     1    9
1    120.45    65     1    9
2    118.92    65     1    9
3      7.02    66     1    0
4     73.93    66     2    5
5    117.53    66     1    9
6      4.40    66     2    0
7    111.03    66     1    9
8      4.35    66     1    0
x1 = df.loc[df.Code==65, 'bin']
x2 = df.loc[df.Code==66, 'bin']
y1 = df.loc[df.Code==65, 'Seq.']
y2 = df.loc[df.Code==66, 'Seq.']

plt.bar(x1, y1)
plt.bar(x2, y2)
plt.show()