python - Pandas 中的分层组大小

Question

假设我在 Pandas 中有一个多级索引数据框，如下所示:

                     A         B         C
X      Y     Z                          
bar   one    a   -0.007381 -0.365315 -0.024817
             b   -1.219794  0.370955 -0.795125
baz   three  a    0.145578  1.428502 -0.408384
             b   -0.249321 -0.292967 -1.849202
      two    a   -0.249321 -0.292967 -1.849202
      four   a    0.21     -0.967123  1.202234
foo   one    b   -1.046479 -1.250595  0.781722
             a    1.314373  0.333150  0.133331
qux   one    c    0.716789  0.616471 -0.298493
      two    b    0.385795 -0.915417 -1.367644

我想知道:

1. 叶子大小每个级别的每个值。在上面的示例中，这将是:

   bar: 2
   bar & one: 2
   bar & one & a: 1
   bar & one & b: 1
   baz: 4
   baz & three: 2
   baz & three & a: 1
   baz & three & b: 1 
   etc.
   

2. 连续级别之间的相对大小。在上面的示例中，这将是:

   # First level -> Second level :
   bar: 1 (i.e. grouping ["one"])
   baz: 3 (i.e. grouping ["three", two", "four"])
   foo: 1 (i.e. grouping ["one"])
   qux: 2 (i.e. grouping ["one", "two"])
   
   # Second level -> Third level
   ... 
   
   # Third level -> Fourth level (if we had one)
   etc.
   

有没有什么方法可以在 Pandas 中执行此操作，并且(最好)也可以在数据框中获得结果？

Answer 1

好吧，既然你添加了另一部分，我会充实我的答案。要执行第 1 部分，我将使用列表理解来循环不同的分组级别并获取所有组的大小。然后 concat 将来自每个 groupby 的结果数据帧组合在一起:

print pd.concat([df.groupby(level=x).size() for x in [0,[0,1],[0,1,2]]])

bar                2
baz                4
foo                2
qux                2
(bar, one)         2
(baz, four)        1
(baz, three)       2
(baz, two)         1
(foo, one)         2
(qux, one)         1
(qux, two)         1
(bar, one, a)      1
(bar, one, b)      1
(baz, four, a)     1
(baz, three, a)    1
(baz, three, b)    1
(baz, two, a)      1
(foo, one, a)      1
(foo, one, b)      1
(qux, one, c)      1
(qux, two, b)      1

第 2 部分更复杂，但我认为我们可以使用相同的结构。可能有很多方法可以做到这一点，但我将在相同的基本列表理解中使用 ngroups 方法:

def group_count(df,x):
    by = df['A'].groupby(level=x[0])
    return by.agg(lambda g: g.groupby(level=x[1]).ngroups)

lvl = [0,[0,1],[0,1,2]]
print pd.concat([group_count(df,x) for x in zip(lvl[:-1],lvl[1:])])

bar             1
baz             3
foo             1
qux             2
(bar, one)      2
(baz, four)     1
(baz, three)    2
(baz, two)      1
(foo, one)      2
(qux, one)      1
(qux, two)      1

当然你可能不喜欢索引作为元组；如果您愿意，您可以在列表理解中重置索引以获得以下内容(例如，对于第 1 部分，此 if):

lvl = [0,[0,1],[0,1,2]]
print pd.concat([df.groupby(level=x).size().reset_index() for x in lvl])

   0    X      Y    Z
0  2  bar    NaN  NaN
1  4  baz    NaN  NaN
2  2  foo    NaN  NaN
3  2  qux    NaN  NaN
0  2  bar    one  NaN
1  1  baz   four  NaN
2  2  baz  three  NaN
3  1  baz    two  NaN
4  2  foo    one  NaN
5  1  qux    one  NaN
6  1  qux    two  NaN
0  1  bar    one    a
1  1  bar    one    b
2  1  baz   four    a
3  1  baz  three    a
4  1  baz  three    b
5  1  baz    two    a
6  1  foo    one    a
7  1  foo    one    b
8  1  qux    one    c
9  1  qux    two    b