如果内存服务于我，在R中有一个名为factor的数据类型，当在DataFrame中使用时，它可以自动解压缩到回归设计矩阵的必要列中。例如，包含True / False / Maybe值的因子将转换为：
1 0 0
0 1 0
or
0 0 1
为了使用较低级别的回归代码。有没有办法用pandas库实现类似的东西？我看到Pandas中有一些回归支持，但由于我有自己的定制回归例程，我真的很感兴趣的是从异构数据构建设计矩阵（一个2d numpy数组或矩阵），支持映射和堡垒之间numpy对象的列和派生它的Pandas DataFrame。
更新：这是一个数据矩阵的示例，其中包含我正在考虑的异类数据（该示例来自Pandas手册）：
>>> df2 = DataFrame({'a' : ['one', 'one', 'two', 'three', 'two', 'one', 'six'],'b' : ['x', 'y', 'y', 'x', 'y', 'x', 'x'],'c' : np.random.randn(7)})
>>> df2
a b c
0 one x 0.000343
1 one y -0.055651
2 two y 0.249194
3 three x -1.486462
4 two y -0.406930
5 one x -0.223973
6 six x -0.189001
>>>
'a'列应转换为4个浮点列（尽管有意义，只有四个唯一原子），'b'列可以转换为单个浮点列，'c'列应该是设计矩阵中未经修改的最后一列。
谢谢，
那么setjmp
有一个名为patsy的新模块可以解决这个问题。下面链接的快速入门在几行代码中完全解决了上述问题。
以下是一个示例用法：
import pandas
import patsy
dataFrame = pandas.io.parsers.read_csv("salary2.txt")
#salary2.txt is a re-formatted data set from the textbook
#Introductory Econometrics: A Modern Approach
#by Jeffrey Wooldridge
y,X = patsy.dmatrices("sl ~ 1+sx+rk+yr+dg+yd",dataFrame)
#X.design_info provides the meta data behind the X columns
print X.design_info
产生：
> DesignInfo(['Intercept',
> 'sx[T.male]',
> 'rk[T.associate]',
> 'rk[T.full]',
> 'dg[T.masters]',
> 'yr',
> 'yd'],
> term_slices=OrderedDict([(Term([]), slice(0, 1, None)), (Term([EvalFactor('sx')]), slice(1, 2, None)),
> (Term([EvalFactor('rk')]), slice(2, 4, None)),
> (Term([EvalFactor('dg')]), slice(4, 5, None)),
> (Term([EvalFactor('yr')]), slice(5, 6, None)),
> (Term([EvalFactor('yd')]), slice(6, 7, None))]),
> builder=<patsy.build.DesignMatrixBuilder at 0x10f169510>)
import pandas
import numpy as np
num_rows = 7;
df2 = pandas.DataFrame(
{
'a' : ['one', 'one', 'two', 'three', 'two', 'one', 'six'],
'b' : ['x', 'y', 'y', 'x', 'y', 'x', 'x'],
'c' : np.random.randn(num_rows)
}
)
a_attribute_list = ['one', 'two', 'three', 'six']; #Or use list(set(df2['a'].values)), but that doesn't guarantee ordering.
b_attribute_list = ['x','y']
a_membership = [ np.reshape(np.array(df2['a'].values == elem).astype(np.float64), (num_rows,1)) for elem in a_attribute_list ]
b_membership = [ np.reshape((df2['b'].values == elem).astype(np.float64), (num_rows,1)) for elem in b_attribute_list ]
c_column = np.reshape(df2['c'].values, (num_rows,1))
design_matrix_a = np.hstack(tuple(a_membership))
design_matrix_b = np.hstack(tuple(b_membership))
design_matrix = np.hstack(( design_matrix_a, design_matrix_b, c_column ))
# Print out the design matrix to see that it's what you want.
for row in design_matrix:
print row
我得到这个输出：
[ 1. 0. 0. 0. 1. 0. 0.36444463]
[ 1. 0. 0. 0. 0. 1. -0.63610264]
[ 0. 1. 0. 0. 0. 1. 1.27876991]
[ 0. 0. 1. 0. 1. 0. 0.69048607]
[ 0. 1. 0. 0. 0. 1. 0.34243241]
[ 1. 0. 0. 0. 1. 0. -1.17370649]
[ 0. 0. 0. 1. 1. 0. -0.52271636]
因此，第一列是DataFrame位置为“one”的指示符，第二列是DataFrame位置为“two”的指示符，依此类推。第4列和第5列分别是DataFrame位置的指示符，分别为“x”和“y”，最后一列只是随机数据。
从2014年2月3日起，熊猫0.13.1有一个方法：
>>> pd.Series(['one', 'one', 'two', 'three', 'two', 'one', 'six']).str.get_dummies()
one six three two
0 1 0 0 0
1 1 0 0 0
2 0 0 0 1
3 0 0 1 0
4 0 0 0 1
5 1 0 0 0
6 0 1 0 0
在许多情况下，patsy.dmatrices可能运作良好。如果你只有一个矢量 - 一个pandas.Series - 那么下面的代码可能会产生退化设计矩阵而没有拦截列。
def factor(series):
"""Convert a pandas.Series to pandas.DataFrame design matrix.
Parameters
----------
series : pandas.Series
Vector with categorical values
Returns
-------
pandas.DataFrame
Design matrix with ones and zeroes.
See Also
--------
patsy.dmatrices : Converts categorical columns to numerical
Examples
--------
>>> import pandas as pd
>>> design = factor(pd.Series(['a', 'b', 'a']))
>>> design.ix[0,'[a]']
1.0
>>> list(design.columns)
['[a]', '[b]']
"""
levels = list(set(series))
design_matrix = np.zeros((len(series), len(levels)))
for row_index, elem in enumerate(series):
design_matrix[row_index, levels.index(elem)] = 1
name = series.name or ""
columns = map(lambda level: "%s[%s]" % (name, level), levels)
df = pd.DataFrame(design_matrix, index=series.index,
columns=columns)
return df
import pandas as pd
import numpy as np
def get_design_matrix(data_in,columns_index,ref):
columns_index_temp = columns_index.copy( )
design_matrix = pd.DataFrame(np.zeros(shape = [len(data_in),len(columns_index)-1]))
columns_index_temp.remove(ref)
design_matrix.columns = columns_index_temp
for ii in columns_index_temp:
loci = list(map(lambda x:x == ii,data_in))
design_matrix.loc[loci,ii] = 1
return(design_matrix)
get_design_matrix(data_in = ['one','two','three','six','one','two'],
columns_index = ['one','two','three','six'],
ref = 'one')
Out[3]:
two three six
0 0.0 0.0 0.0
1 1.0 0.0 0.0
2 0.0 1.0 0.0
3 0.0 0.0 1.0
4 0.0 0.0 0.0
5 1.0 0.0 0.0