Python 如何在IntervalIndex中获得一些值?
变量“lvl”是IntervalIndex类型,如下所示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',
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