pandas规范化groupby.sum

Question

我有一个像这样的pandas数据框：

      **I     SI     weights**
        1     3      0.3  
        2     4      0.2
        1     3      0.5
        1     5      0.5

我需要这样做：给定值I，考虑SI的每个值并添加总重量。最后，对于每个实现，我应该有类似的东西：

             I = 1     SI = 3      weight = 0.8
                       SI = 5      weight = 0.5

             I = 2     SI = 4      weight = 0.2

通过调用groupby和sum可以轻松实现这一点：

       name = ['I', 'SI','weight']
       Location = 'Simulationsdata/prova.csv'
       df = pd.read_csv(Location, names = name,sep='\t',encoding='latin1') 

       results = df.groupby(['I', 'real', 'SI']).weight.sum()

现在我想将权重标准化为1，因此它应该是这样的：

             I = 1     SI = 3      weight = 0.615
                       SI = 5      weight = 0.385

             I = 2     SI = 4      weight = 1

我试过这个：

        for idx2, j in enumerate(results.index.get_level_values(1).unique()):
            norm = [float(i)/sum(results.loc[j]) for i in results.loc[j]]

但是当我试图为每个我绘制SI的分布时，我发现SI也被标准化了，我不希望这种情况发生。

附：这个问题与this one有关，但由于它涉及问题的另一个方面，我认为最好单独问一下

Answer 1

你应该能够将weight列除以它自己的总和：

# example data
df
   I  SI   weight
0  1   3      0.3
1  2   4      0.2
2  1   3      0.5
3  1   5      0.5

# two-level groupby, with the result as a DataFrame instead of Series:
# df['col'] gives a Series, df[['col']] gives a DF
res = df.groupby(['I', 'SI'])[['weight']].sum()
res
       weight
I SI         
1 3       0.8
  5       0.5
2 4       0.2

# Get the sum of weights for each value of I,
# which will serve as denominators in normalization
denom = res.groupby('I')['weight'].sum()
denom
I
1    1.3
2    0.2
Name: weight, dtype: float64

# Divide each result value by its index-matched
# denominator value
res.weight = res.weight / denom
res
        weight
I SI          
1 3   0.615385
  5   0.384615
2 4   1.000000