如果我们有一些数据：

bins = pd.IntervalIndex.from_tuples([(0, 1), (2, 3), (4, 5)])
x = pd.cut([0, 0.5, 1.5, 2.5, 4.5], bins)

您可以尝试重新分配类别，例如：

In [7]: x.categories = [1,2,3]

In [8]: x   
Out[8]: 
[NaN, 1, NaN, 2, 3]
Categories (3, int64): [1 < 2 < 3]

或：

In [9]: x.categories = ["small", "medium", "big"]                         

In [10]: x                                             
Out[10]: 
[NaN, small, NaN, medium, big]
Categories (3, object): [small < medium < big]

• • 更新**：

df = pd.DataFrame([0, 0.5, 1.5, 2.5, 4.5], columns = ['col1'])
bins = pd.IntervalIndex.from_tuples([(0, 1), (2, 3), (4, 5)])
x = pd.cut(df["col1"].to_list(),bins)
x.categories = [1,2,3]
df['col1'] = x
df.col1
0    NaN
1      1
2    NaN
3      2
4      3
Name: col1, dtype: category
Categories (3, int64): [1 < 2 < 3]

• • 更新2**：

在更新版本的panda中，不使用x.categories = [1, 2, 3]重新分配类别，而是使用x.cat.rename_categories：

labels = [1, 2, 3]
x.cat.rename_categories(labels, inplace=True)

labels可以是任何类型，并且在任何情况下，将保留在创建pd.IntervalIndex时设置的原始分类顺序。

series = pd.Series([0, 0.5, 1.5, 2.5, 4.5])

bins = [(0, 1), (2, 3), (4, 5)]
index = pd.IntervalIndex.from_tuples(bins)
intervals = index.values
names = ['small', 'med', 'large']
to_name = {interval: name for interval, name in zip(intervals, names)}

named_series = pd.Series(
    pd.CategoricalIndex(pd.cut(series, bins_index)).rename_categories(to_name)
)
print(named_series)

0      NaN
1    small
2      NaN
3      med
4    large
dtype: category
Categories (3, object): ['small' < 'med' < 'large']