Question

在这里的熊猫笔记本上

http://nbviewer.ipython.org/urls/raw.github.com/carljv/Will_it_Python/master/ARM/ch5/arsenic_wells_switching.ipynb

我看到qcut的结果被指定为DataFrame的新列。 Dataframe有两列，但以某种方式将qcut输出分配给一个新列神奇地找到了“var”变量所在的正确级别 - 未检查另一个变量。这里的熊猫语义是什么？样本输出如下：

In [2]:
from pandas import *
from statsmodels.formula.api import logit
from statsmodels.nonparametric import KDE
from patsy import dmatrix, dmatrices

In [3]:
df = read_csv('wells.dat', sep = ' ', header = 0, index_col = 0)
print df.head()
   switch  arsenic       dist  assoc  educ
1       1     2.36  16.826000      0     0
2       1     0.71  47.321999      0     0
3       0     2.07  20.966999      0    10
4       1     1.15  21.486000      0    12
5       1     1.10  40.874001      1    14


In [4]:
model_form = ('switch ~ center(I(dist / 100.)) + center(arsenic) + ' +
              'center(I(educ / 4.)) + ' +
              'center(I(dist / 100.)) : center(arsenic) + ' + 
              'center(I(dist / 100.)) : center(I(educ / 4.)) + ' + 
              'center(arsenic) : center(I(educ / 4.))'
             )
model4 = logit(model_form, df = df).fit()   

In [20]:
resid_df = DataFrame({'var': df['arsenic'], 'resid': model4.resid})
resid_df[:10]
Out [20]:
       resid   var
1   0.842596  2.36
2   1.281417  0.71
3  -1.613751  2.07
4   0.996195  1.15
5   1.005102  1.10
6   0.592056  3.90
7   0.941372  2.97
8   0.640139  3.24
9   0.886626  3.28
10  1.130149  2.52

In [15]:
qcut(df['arsenic'], 40)
Out [15]:
Categorical: arsenic
array([(2.327, 2.47], (0.68, 0.71], (1.953, 2.07], ..., [0.51, 0.53],
       (0.62, 0.64], (0.64, 0.68]], dtype=object)
Levels (40): Index([[0.51, 0.53], (0.53, 0.56], (0.56, 0.59],
                    (0.59, 0.62], (0.62, 0.64], (0.64, 0.68],
                    (0.68, 0.71], (0.71, 0.75], (0.75, 0.78],
                    (0.78, 0.82], (0.82, 0.86], (0.86, 0.9], (0.9, 0.95],
                    (0.95, 1.0065], (1.0065, 1.0513], (1.0513, 1.1],
                    (1.1, 1.15], (1.15, 1.2], (1.2, 1.25], (1.25, 1.3],
                    (1.3, 1.36], (1.36, 1.42], (1.42, 1.49],
                    (1.49, 1.57], (1.57, 1.66], (1.66, 1.76],
                    (1.76, 1.858], (1.858, 1.953], (1.953, 2.07],
                    (2.07, 2.2], (2.2, 2.327], (2.327, 2.47],
                    (2.47, 2.61], (2.61, 2.81], (2.81, 2.98],
                    (2.98, 3.21], (3.21, 3.42], (3.42, 3.791],
                    (3.791, 4.475], (4.475, 9.65]], dtype=object)

In [17]:
resid_df['bins'] = qcut(df['arsenic'], 40)
resid_df[:20]
Out [17]:
       resid   var            bins
1   0.842596  2.36   (2.327, 2.47]
2   1.281417  0.71    (0.68, 0.71]
3  -1.613751  2.07   (1.953, 2.07]
4   0.996195  1.15     (1.1, 1.15]
5   1.005102  1.10   (1.0513, 1.1]
6   0.592056  3.90  (3.791, 4.475]
7   0.941372  2.97    (2.81, 2.98]
8   0.640139  3.24    (3.21, 3.42]

找到“var”的正确bin，分配没有注意“resid”。

Answer 1

我想我明白了..在qcut分类（其结果）对象上有一个“标签”属性;标签带有一个数字，例如每个点的1,2,3，基于该点落入的四分位数。然后，如果将qcut结果分配到DataFrame上的新列Pandas将此“标签”与DataFrame的“索引”匹配。

Answer 2

我发现做问题的标题的唯一一般方法是：

quartiles = pd.qcut(df['ValToRank'], 4, labels=range(1,5))
df = df.assign(Quartile=quartiles.values)

这会将四分位数值指定为新的DataFrame列df['Quartile']。

A solution for a more generalized case, in which one wants to partition the cut by multiple columns, is given here

将qcut分配为新列

2 个答案: