如何在Python中的直方图中显示百分比标签

Question

我设法创建了一个图表，显示了我的Pandas数据框中每个年龄段的每个类标签的记录数。但我也希望看到每个年龄组中“非功能”课程的百分比标签。

图表的Python代码是

train['age_wpt'] = train.date_recorded.str.split('-').str.get(0).apply(int) - train.construction_year

figure = plt.figure(figsize=(15,8))
plt.hist([
        train[(train.status_group=='functional') & (train.age_wpt < 60.0) & (train.age_wpt >= 0.0)]['age_wpt'],
        train[(train.status_group=='non functional') & (train.age_wpt < 60.0) & (train.age_wpt >= 0.0)]['age_wpt'],
        train[(train.status_group=='functional needs repair') & (train.age_wpt < 60.0) & (train.age_wpt >= 0.0)]['age_wpt']
        ], 
         stacked=True, color = ['b','r','y'],
         bins = 30,label = ['functional','non functional', 'functional needs repair'])
plt.xlabel('Age')
plt.ylabel('Number of records')
plt.legend()

这导致以下图表

Answer 1

  

normed：布尔值，可选       如果True，返回元组的第一个元素将       是规范化以形成概率密度的计数，即       n/(len(x)`dbin)，即直方图的积分将求和       如果堆叠也是 True ，则直方图的总和为       归一化为1。       默认值为False

plt.hist([
        train[(train.status_group=='functional') & (train.age_wpt < 60.0) & (train.age_wpt >= 0.0)]['age_wpt'],
        train[(train.status_group=='non functional') & (train.age_wpt < 60.0) & (train.age_wpt >= 0.0)]['age_wpt'],
        train[(train.status_group=='functional needs repair') & (train.age_wpt < 60.0) & (train.age_wpt >= 0.0)]['age_wpt']
        ], 
         stacked=False, color = ['b','r','y'], normed=True
         bins = 30,label = ['functional','non functional', 'functional needs repair'])