Pandas:我使用的应用函数给出了错误的结果

Pandas: The apply function I am using is giving me wrong results

我有一个看起来像这样的数据集

     a_id b_received brand_id c_consumed type_received       date  output  \
0    sam       soap     bill        oil       edibles 2011-01-01       1   
1    sam        oil    chris        NaN       utility 2011-01-02       1   
2    sam      brush      dan       soap       grocery 2011-01-03       0   
3  harry        oil      sam      shoes      clothing 2011-01-04       1   
4  harry      shoes     bill        oil       edibles 2011-01-05       1   
5  alice       beer      sam       eggs     breakfast 2011-01-06       0   
6  alice      brush    chris      brush      cleaning 2011-01-07       1   
7  alice       eggs      NaN        NaN       edibles 2011-01-08       1   

我正在使用以下代码

 def probability(x):
    y=[]
    for i in range(len(x)):
        y.append(float(x[i])/float(len(x)))
    return y

 df2['prob']= (df2.groupby('a_id')
           .apply(probability(['output']))
           .reset_index(level='a_id', drop=True))

理想的结果应该是一个具有以下值的新列

    prob  
 0  0.333334  
 1  0.333334  
 2  0.0  
 3  0.5  
 4  0.5  
 5  0     
 6  0.333334     
 7  0.333334     

但我遇到了错误

y.append(float(x[i])/float(len(x)))
ValueError: could not convert string to float: output

列输出为int格式。我不明白为什么会出现此错误。

我正在尝试计算每个人消费列输出给出的产品的输出概率。例如,如果 sam 收到 soap,并且 soap 也出现在 'c_consumed' 列中,则结果为 1,否则结果为 0。

现在,由于 sam 收到了 3 件产品,他消费了 2 件,因此每件产品消费的概率是 1/3。因此,输出为 1 的概率应为 0.333334,输出为 0 的概率应为 0。

如何达到预期的效果?

我认为您可以简单地将 output 列连同已计算的分组 .groupby('a_id')['output'] 传递给 GroupBy 对象,然后使用函数 probability,其中 return 仅将列 output 与其 len:

相除
def probability(x):
    #print x
    return x / len(x)

df2['prob']= (df2.groupby('a_id')['output']
           .apply(probability)
           .reset_index(level='a_id', drop=True))

lambda:

df2['prob']= (df2.groupby('a_id')['output']
           .apply(lambda x: x / len(x) )
           .reset_index(level='a_id', drop=True))

更简单、更快速的解决方案是 transform:

df2['prob']= df2['output'] / df2.groupby('a_id')['output'].transform('count')
print df2
    a_id b_received brand_id c_consumed type_received        date  output  \
0    sam       soap     bill        oil       edibles  2011-01-01       1   
1    sam        oil    chris        NaN       utility  2011-01-02       1   
2    sam      brush      dan       soap       grocery  2011-01-03       0   
3  harry        oil      sam      shoes      clothing  2011-01-04       1   
4  harry      shoes     bill        oil       edibles  2011-01-05       1   
5  alice       beer      sam       eggs     breakfast  2011-01-06       0   
6  alice      brush    chris      brush      cleaning  2011-01-07       1   
7  alice       eggs      NaN        NaN       edibles  2011-01-08       1   

       prob  
0  0.333333  
1  0.333333  
2  0.000000  
3  0.500000  
4  0.500000  
5  0.000000  
6  0.333333  
7  0.333333  

时间:

In [505]: %timeit (df2.groupby('a_id')['output'].apply(lambda x: x / len(x) ).reset_index(level='a_id', drop=True))
The slowest run took 10.99 times longer than the fastest. This could mean that an intermediate result is being cached 
100 loops, best of 3: 1.73 ms per loop

In [506]: %timeit df2['output'] / df2.groupby('a_id')['output'].transform('count')
The slowest run took 5.03 times longer than the fastest. This could mean that an intermediate result is being cached 
1000 loops, best of 3: 449 µs per loop