熊猫数据框分组：双分组和应用功能

Question

I have a question regarding pandas dataframes:

I have a dataframe like the following,

df = pd.DataFrame([[1,1,10],[1,1,30],[1,2,40],[2,3,50],[2,3,150],[2,4,100]],columns=["a","b","c"])   

   a  b    c
0  1  1   10
1  1  1   30
2  1  2   40
3  2  3   50
4  2  3  150
5  2  4  100

And i want to produce the following output,

  a "new col"
0 1 30
1 2 100

where the first line is calculated as the following:

1. Group df by the first column "a",
2. Then group each grouped object the "b"
3. calculate the mean of "c" for this b-group
4. calculate the means of all b-groupbs for one "a"
5. this is the final value stored in "new col" for one "a"

I can imagine that this is somehow confusing, but I hope this is understandable, nevertheless.

I achieved the desired result, but as i need it for a huge dataframe, my solution is probably much to slow,

pd.DataFrame([ [a, adata.groupby("b").agg({"c": lambda x:x.mean()}).mean()[0]] for a,adata in df.groupby("a") ],columns=["a","new col"])
   a  new col
0  1     30.0
1  2    100.0

Therefore, what I would need is something like (?) df.groupby("a").groupby("b")["c"].mean()

Thank you very much in advance!

Answer 1

Here's one way

In [101]: (df.groupby(['a', 'b'], as_index=False)['c'].mean()
             .groupby('a', as_index=False)['c'].mean()
             .rename(columns={'c': 'new col'}))
Out[101]:
   a  new col
0  1       30
1  2      100

Answer 2

In [57]: df.groupby(['a','b'])['c'].mean().mean(level=0).reset_index()
Out[57]:
   a    c
0  1   30
1  2  100

Answer 3

df.groupby(['a','b']).mean().reset_index().groupby('a').mean()
Out[117]: 
     b      c
a            
1  1.5   30.0
2  3.5  100.0