如何离散化 Pandas DataFrame 中的值并转换为二进制矩阵？
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How do I discretize values in a pandas DataFrame and convert to a binary matrix?
提问by Uri Laserson
I mean something like this:
我的意思是这样的：
I have a DataFramewith columns that may be categorical or nominal. For each observation (row), I want to generate a new row where every possible value for the variables is now its own binary variable. For example, this matrix (first row is column labels)
我有一个DataFrame可能是分类或名义的列。对于每个观察（行），我想生成一个新行，其中变量的每个可能值现在都是它自己的二进制变量。例如，这个矩阵（第一行是列标签）
'a' 'b' 'c'
one 0.2 0
two 0.4 1
two 0.9 0
three 0.1 2
one 0.0 4
two 0.2 5
would be converted into something like this:
将被转换成这样的：
'a' 'b' 'c'
one two three [0.0,0.2) [0.2,0.4) [0.4,0.6) [0.6,0.8) [0.8,1.0] 0 1 2 3 4 5
1 0 0 0 1 0 0 0 1 0 0 0 0 0
0 1 0 0 0 0 0 1 0 1 0 0 0 0
0 1 0 0 0 0 0 1 1 0 0 0 0 0
0 0 1 1 0 0 0 0 0 0 1 0 0 0
1 0 0 1 0 0 0 0 0 0 0 0 1 0
0 1 0 0 1 0 0 0 0 0 0 0 0 1
Each variable (column) in the initial matrix get binned into all the possible values. If it's categorical, then each possible value becomes a new column. If it's a float, then the values are binned some way (say, always splitting into 10 bins). If it's an int, then it can be every possibel int value, or perhaps also binning.
初始矩阵中的每个变量（列）都被分箱为所有可能的值。如果它是分类的，那么每个可能的值都会成为一个新列。如果它是一个浮点数，那么这些值会以某种方式分箱（例如，总是分成 10 个分箱）。如果它是一个 int，那么它可以是每个可能的 int 值，或者也可以是分箱。
FYI: in my real application, the table has up to 2 million rows, and the full "expanded" matrix may have hundreds of columns.
仅供参考：在我的实际应用中，该表最多有 200 万行，完整的“扩展”矩阵可能有数百列。
Is there an easy way to perform this operation?
有没有简单的方法来执行这个操作？
Separately, I would also be willing to skip this step, as I am really trying to compute a Burt table (which is a symmetric matrix of the cross-tabulations). Is there an easy way to do something similar with the crosstabfunction? Otherwise, computing the cross tabulation is just a simple matrix multiplication.
另外，我也愿意跳过这一步，因为我真的想计算一个 Burt 表（它是交叉表的对称矩阵）。有没有一种简单的方法可以对crosstab函数做类似的事情？否则，计算交叉表只是一个简单的矩阵乘法。
采纳答案by lbolla
You can use some kind of broadcasting:
您可以使用某种广播：
In [58]: df
Out[58]:
a b c
0 one 0.2 0
1 two 0.4 1
2 two 0.9 0
3 three 0.1 2
4 one 0.0 4
5 two 0.2 5
In [41]: (df.a.values[:,numpy.newaxis] == df.a.unique()).astype(int)
Out[41]:
array([[1, 0, 0],
[0, 1, 0],
[0, 1, 0],
[0, 0, 1],
[1, 0, 0],
[0, 1, 0]])
In [54]: ((0 <= df.b.values[:,numpy.newaxis]) & (df.b.values[:,numpy.newaxis] < 0.2)).astype(int)
Out[54]:
array([[0],
[0],
[0],
[1],
[1],
[0]])
In [59]: (df.c.values[:,numpy.newaxis] == df.c.unique()).astype(int)
Out[59]:
array([[1, 0, 0, 0, 0],
[0, 1, 0, 0, 0],
[1, 0, 0, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 0, 1, 0],
[0, 0, 0, 0, 1]])
And then join all the pieces together with pandas.concator similar.
然后将所有部件连接在一起pandas.concat或类似。
回答by Wes McKinney
Note that I have implemented new cutand qcutfunctions for discretizing continuous data:
请注意，我已经实现新的cut和qcut功能离散连续的数据：
http://pandas-docs.github.io/pandas-docs-travis/basics.html#discretization-and-quantiling
http://pandas-docs.github.io/pandas-docs-travis/basics.html#discretization-and-quantiling
回答by wonderkid2
For labeled columns like the aand ccolumn in your example you can use the pandas build-in method get_dummies().
对于像示例中的a和列这样的标记列c，您可以使用熊猫内置方法get_dummies()。
Ex.:
前任。：
import pandas as pd
s1 = ['a', 'b', np.nan]
pd.get_dummies(s1)
a b
0 1 0
1 0 1
2 0 0
回答by elyase
I doubt you will beat patsy's simplicity. It was designed precisely for this task:
我怀疑你会打败patsy的简单性。它专为此任务而设计：
>>> from patsy import dmatrix
>>> dmatrix('C(a) + C(b) + C(c) - 1', df, return_type='dataframe')
C(a)[one] C(a)[three] C(a)[two] C(b)[T.0.1] C(b)[T.0.2] C(b)[T.0.4] C(b)[T.0.9] C(c)[T.1] C(c)[T.2] C(c)[T.4] C(c)[T.5]
0 1 0 0 0 1 0 0 0 0 0 0
1 0 0 1 0 0 1 0 1 0 0 0
2 0 0 1 0 0 0 1 0 0 0 0
3 0 1 0 1 0 0 0 0 1 0 0
4 1 0 0 0 0 0 0 0 0 1 0
5 0 0 1 0 1 0 0 0 0 0 1
Here the C(a)means convert the variable to categorical and the -1is to avoid outputting an intercept column.
这里的C(a)意思是将变量转换为分类变量-1，避免输出截距列。
回答by Tim
Putting together a couple of other comments into a single response answering OPs questions.
将一些其他评论放在一个回答 OP 问题的单一回复中。
d = {'a' : pd.Series(['one', 'two', 'two', 'three', 'one', 'two']),
'b' : pd.Series([0.2, 0.4, 0.9, 0.1, 0.0, 0.2]),
'c' : pd.Series([0, 1, 0, 2, 4, 5]) }
data = pd.DataFrame(d)
a_cols = pd.crosstab(data.index, [data.a])
b_bins = pd.cut(data.b, [0.0, 0.2, 0.4, 0.6, 0.8, 1.0], right=False)
b_cols = pd.crosstab(data.index, b_bins)
c_cols = pd.crosstab(data.index, [data.c], )
new_data = a_cols.join(b_cols).join(c_cols)
new_data.index.names = ['']
print new_data.to_string()
"""
one three two [0, 0.2) [0.2, 0.4) [0.4, 0.6) [0.8, 1) 0 1 2 4 5
0 1 0 0 0 1 0 0 1 0 0 0 0
1 0 0 1 0 0 1 0 0 1 0 0 0
2 0 0 1 0 0 0 1 1 0 0 0 0
3 0 1 0 1 0 0 0 0 0 1 0 0
4 1 0 0 1 0 0 0 0 0 0 1 0
5 0 0 1 0 1 0 0 0 0 0 0 1
"""
