如何在 Python 的直方图中显示百分比标签

Question

我设法创建了一个图表，显示我的 Pandas 数据框中每个年龄的每个类别标签的记录数。但我也希望看到每个年龄组中“非功能性”类别的百分比标签。

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()

结果如下图

Answer 1

规范：布尔值，可选 如果True，返回元组的第一个元素将 将计数归一化以形成概率密度，即 n/(len(x)`dbin)，即直方图积分求和 为 1. 如果 stacked 也是 True，则直方图的总和为 归一化为 1。 默认为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'])