Pandas 中的区间数据类型 - 查找中点、左侧、中心等
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Interval datatype in Pandas - find midpoint, left, center etc
提问by penguin
In pandas 20.1, with the interval type, is it possible to find the midpoint, left or center values in a series.
在 pandas 20.1 中，使用区间类型，是否可以找到一系列中的中点、左值或中心值。
For example:
例如：
Create an interval datatype column, and perform some aggregation calculations over these intervals:
df_Stats = df.groupby(['month',pd.cut(df['Distances'], np.arange(0, 135,1))]).agg(aggregations)
创建一个区间数据类型列，并对这些区间执行一些聚合计算：
df_Stats = df.groupby(['month',pd.cut(df['Distances'], np.arange(0, 135,1))]).agg(aggregations)
This returns df_Stats with an interval column datatype : df['Distances']
这将返回具有间隔列数据类型的 df_Stats ： df['Distances']
Now I want to associate the left end of the interval to the result of these aggregations using a series function:
df['LeftEnd'] = df['Distances'].left
现在我想使用系列函数将区间的左端与这些聚合的结果相关联：
df['LeftEnd'] = df['Distances'].left
However, I can run this element wise:
但是，我可以明智地运行这个元素：
df.loc[0]['LeftEnd'] = df.loc[0]['Distances'].left
This works. Thoughts?
这有效。想法？
采纳答案by Jeff
So pd.cut()actually creates a CategoricalIndex, with an IntervalIndexas the categories.
所以pd.cut()实际上创建了一个CategoricalIndex,IntervalIndex作为类别。
In [13]: df = pd.DataFrame({'month': [1, 1, 2, 2], 'distances': range(4), 'value': range(4)})
In [14]: df
Out[14]:
distances month value
0 0 1 0
1 1 1 1
2 2 2 2
3 3 2 3
In [15]: result = df.groupby(['month', pd.cut(df.distances, 2)]).value.mean()
In [16]: result
Out[16]:
month distances
1 (-0.003, 1.5] 0.5
2 (1.5, 3.0] 2.5
Name: value, dtype: float64
You can simply coerce them to an IntervalIndex(this also works if they are a column), then access.
您可以简单地将它们强制转换为IntervalIndex（如果它们是列，这也适用），然后访问。
In [17]: pd.IntervalIndex(result.index.get_level_values('distances')).left
Out[17]: Float64Index([-0.003, 1.5], dtype='float64')
In [18]: pd.IntervalIndex(result.index.get_level_values('distances')).right
Out[18]: Float64Index([1.5, 3.0], dtype='float64')
In [19]: pd.IntervalIndex(result.index.get_level_values('distances')).mid
Out[19]: Float64Index([0.7485, 2.25], dtype='float64')
回答by Mahesh Babu J
Say 'cut' is the column nameafter performing pd.cut.
说'cut' 是执行 pd.cut 后的列名。
instead of ->
而不是 ->
df['LeftEnd'] = df['Distances'].left
perform one of the following -->
执行以下操作之一 -->
df['LeftEnd'] = df['cut'].apply(lambda x: x.left)
df['LeftEnd'] = df['cut'].apply(lambda x: x.left).astype(str)
