列的MultiLevel索引：将value_counts作为pandas中的列 [英] MultiLevel index to columns : getting value_counts as columns in pandas

问题描述

从一般意义上讲，我要解决的问题是将多级索引的一个组件更改为列。也就是说，我有一个 Series ，它包含一个多级索引，我希望将最低级别的索引更改为 dataframe 。以下是我试图解决的实际示例问题，

In a very general sense, the problem I am looking to solve is changing one component of a multi-level index into columns. That is, I have a Series that contains a multilevel index and I want the lowest level of the index changed into columns in a dataframe. Here is the actual example problem I'm trying to solve,

这里我们可以生成一些示例数据：

Here we can generate some sample data:

foo_choices = ["saul", "walter", "jessee"]
bar_choices = ["alpha", "beta", "foxtrot", "gamma", "hotel", "yankee"]

df = DataFrame([{"foo":random.choice(foo_choices), 
                 "bar":random.choice(bar_choices)} for _ in range(20)])
df.head()

这给了我们，

     bar     foo
0    beta    jessee
1    gamma   jessee
2    hotel   saul
3    yankee  walter
4    yankee  jessee
...

现在，我可以分组 bar 并获取 foo 字段的value_counts，

Now, I can groupby bar and get value_counts of the foo field,

dfgb = df.groupby('foo')
dfgb['bar'].value_counts()

并输出，

foo            
jessee  hotel      4
        gamma      2
        yankee     1
saul    foxtrot    3
        hotel      2
        gamma      1
        alpha      1
walter  hotel      2
        gamma      2
        foxtrot    1
        beta       1

但我想要的是，

          hotel    beta    foxtrot    alpha    gamma    yankee
foo                        
jessee     1       1       5          4        1        1
saul       0       3       0          0        1        0
walter     1       0       0          1        1        0

我的解决方案是编写以下内容：

My solution was to write the following bit:

for v in df['bar'].unique():
    if v is np.nan: continue
    df[v] = np.nan
    df.ix[df['bar'] == v, v] = 1

dfgb = df.groupby('foo')
dfgb.count()[df['bar'].unique()]

推荐答案

我想你想要：

dfgb['bar'].value_counts().unstack().fillna(0.)

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