Python 如何在IntervalIndex中获得一些值?

Python 如何在IntervalIndex中获得一些值?,python,pandas,Python,Pandas,变量“lvl”是IntervalIndex类型,如下所示 lvl = IntervalIndex([(1.418, 1.69], (1.69, 6.696], (6.696, 7.217], (7.217, 7.845], (7.845, 10.11] ... (17.648, 18.199], (18.199, 19.315], (19.315, 20.16], (20.16, 22.471], (22.471, 25.009]] closed='right',

变量“lvl”是IntervalIndex类型,如下所示

lvl = IntervalIndex([(1.418, 1.69], (1.69, 6.696], (6.696, 7.217], (7.217, 7.845], (7.845, 10.11] ... (17.648, 18.199], (18.199, 19.315], (19.315, 20.16], (20.16, 22.471], (22.471, 25.009]]
          closed='right',
          dtype='interval[float64]') 
现在我尝试添加一些索引值,如下所示:

for each in val_list:  
    for k in range(len(lvl)):
        lower_val = float(lvl[k][1:-1].split(',')[0])
        upper_val = float(lvl[k][1:-1].split(',')[-1])
        if each >=lower_val and each <= upper_val:
            some_var_index.append(k)
我该怎么做?
谢谢

对象不可下标意味着它的行为与普通列表不同,但您在这里做了两次:

lower_val = float(lvl[k][1:-1].split(',')[0])
upper_val = float(lvl[k][1:-1].split(',')[-1])
带有第二个方括号
[1:-1]
。相反,这两行使用了这一行:

lower_val, upper_val = lvl[k].left, lvl[k].right

你应该很好。

非常感谢你,Mkos!
lower_val, upper_val = lvl[k].left, lvl[k].right