Question

我使用pandas.cut和由IntervalIndex.from_tuples创建的bin离散了数据框中的一列。

剪切按预期工作，但是类别显示为我在IntervalIndex中指定的元组。有什么办法可以将类别重命名为其他标签，例如（小，中，大）？

示例：

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

结果类别为：

[NaN, (0, 1], NaN, (2, 3], (4, 5]]
Categories (3, interval[int64]): [(0, 1] < (2, 3] < (4, 5]]

我正在尝试将[(0, 1] < (2, 3] < (4, 5]]更改为1, 2 ,3或small, medium ,large之类的内容。

遗憾的是，使用IntervalIndex时会忽略pd.cut的label参数参数。

谢谢！

更新：

由于@SergeyBushmanov，我注意到只有在尝试更改数据框内的类别标签时才存在此问题（这是我要尝试的操作）。更新的示例：

In [1]: df = pd.DataFrame([0, 0.5, 1.5, 2.5, 4.5], columns = ['col1'])
In [2]: bins = pd.IntervalIndex.from_tuples([(0, 1), (2, 3), (4, 5)])
In [3]: df['col1'] = pd.cut(df['col1'], bins)
In [4]: df['col1'].categories = ['small','med','large']

In [5]: df['col1']

Out [5]:
0       NaN
1    (0, 1]
2       NaN
3    (2, 3]
4    (4, 5]
Name: col1, dtype: category
Categories (3, interval[int64]): [(0, 1] < (2, 3] < (4, 5]]

Answer 1

假设我们碰巧有一些数据：

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]

Answer 2

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']

在将Pandas.cut与IntervalIndex结合使用后，如何重命名类别？

2 个答案: