1 
from patsy import *
from pandas import *
dta =  DataFrame([["lo", 1],["hi", 2.4],["lo", 1.2],["lo", 1.4],["very_high",1.8]], columns=["carbs", "score"])
dmatrix("carbs + score", dta)
DesignMatrix with shape (5, 4)
Intercept  carbs[T.lo]  carbs[T.very_high]  score
        1            1                   0    1.0
        1            0                   0    2.4
        1            1                   0    1.2
        1            1                   0    1.4
        1            0                   1    1.8
Terms:
'Intercept' (column 0), 'carbs' (columns 1:3), 'score' (column 3)

问题：不是使用 Designinfo 指定列的“名称”（这基本上使我的代码的可重用性降低），我可以不读取此 DesignMatrix 给出的名称，以便稍后将其输入 DataFrame，而无需知道预先“参考水平/对照组”水平是什么？

IE。当我做 dmatrix("C(carbs, Treatment(reference='lo')) + score", dta)

"""
# How can I get something like this with dmatrix's output without hardcoding ?
names = obtained from dmatrix's output above 
This should give names = ['Intercept' ,'carbs[T.lo]', 'carbs[T.very_high]', 'score']
"""
g=DataFrame(dmatrix("carbs + score", dta),columns=names)

    Intercept  carbs[T.lo]  carbs[T.very_high]  score
   0  1  2    3
0  1  1  0  1.0
1  1  0  0  2.4
2  1  1  0  1.2
3  1  1  0  1.4
4  1  0  1  1.8

type(g)=<class 'pandas.core.frame.DataFrame'>

所以 g 将是转换后的数据框，我可以在不需要记录（或硬编码）列名及其参考级别的情况下对其进行逻辑建模。

4

1 回答 1

4

我认为您要查找的信息位于design_info.column_names：

>>> dm = dmatrix("carbs + score", dta)
>>> dm.design_info
DesignInfo(['Intercept', 'carbs[T.lo]', 'carbs[T.very_high]', 'score'],
           term_slices=OrderedDict([(Term([]), slice(0, 1, None)), (Term([EvalFactor('carbs')]), slice(1, 3, None)), (Term([EvalFactor('score')]), slice(3, 4, None))]),
           builder=<patsy.build.DesignMatrixBuilder at 0xb03f8cc>)
>>> dm.design_info.column_names
['Intercept', 'carbs[T.lo]', 'carbs[T.very_high]', 'score']

所以

>>> DataFrame(dm, columns=dm.design_info.column_names)
   Intercept  carbs[T.lo]  carbs[T.very_high]  score
0          1            1                   0    1.0
1          1            0                   0    2.4
2          1            1                   0    1.2
3          1            1                   0    1.4
4          1            0                   1    1.8

[5 rows x 4 columns]
于 2014-05-09T11:44:02.570 回答