Interval datatype in Pandas - find midpoint, left, center etc

Question

In pandas 20.1, with the interval type, is it possible to find the midpoint, left or center values in a series.

For example:

1. 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)
   

This returns df_Stats with an interval column datatype : df['Distances']

2. 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
   

However, I can run this element wise:

    df.loc[0]['LeftEnd'] = df.loc[0]['Distances'].left

This works. Thoughts?

Quinlan

Answer 1

So pd.cut() actually creates a CategoricalIndex, with an IntervalIndex as the categories.

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.

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')

Chaves

Answer 2

Say 'cut' is the column name after performing pd.cut.

instead of ->

 df['LeftEnd'] = df['Distances'].left

perform one of the following -->

 df['LeftEnd'] = df['cut'].apply(lambda x: x.left)

 df['LeftEnd'] = df['cut'].apply(lambda x: x.left).astype(str)

Blindworm