如何在Python的pandas中对groupby结果执行函数？

Question

我用这段代码来计算每个用户在每个组里的不同质量指标的值。

>>> for name, group in df.groupby(["Cluster_id", "User"]):
...     print 'group name:', name
...     print 'group rows:'
...     print group
...     print 'counts of Quality values:'
...     print group["Quality"].value_counts()
...     raw_input()
...     

但是现在我得到的输出是：

group rows:
                tag                       user                    quality  cluster
676    black fabric  http://steve.nl/user_1002          usefulness-useful        1
708      blond wood  http://steve.nl/user_1002          usefulness-useful        1
709      blond wood  http://steve.nl/user_1002    problematic-misspelling        1
1410         eames?  http://steve.nl/user_1002      usefulness-not_useful        1
1411         eames?  http://steve.nl/user_1002  problematic-misperception        1
3649  rocking chair  http://steve.nl/user_1002          usefulness-useful        1
3650  rocking chair  http://steve.nl/user_1002  problematic-misperception        1
counts of Quality Values:
usefulness-useful            3
problematic-misperception    2
usefulness-not_useful        1
problematic-misspelling      1

我现在想做的是添加一个检查条件，也就是：

if quality==usefulness-useful:
 good = good + 1
else:
 bad = bad + 1

我尝试把输出：

counts of Quality Values:
usefulness-useful            3
problematic-misperception    2
usefulness-not_useful        1
problematic-misspelling      1

放到一个变量里，然后逐行遍历这个变量，但这并没有成功。有没有人能给我一些建议，告诉我怎么对某些行进行计算。

Answer 1

一旦你有了一个数据组，你可以用 .iterrows() 方法逐行遍历。这个方法会给你每一行的索引和对应的行数据：

In [33]: for row_number, row in group.iterrows():
   ....:     print row_number
   ....:     print row
   ....:     
676
Tag                        black fabric
User          http://steve.nl/user_1002
Quality               usefulness-useful
Cluster_id                            1
Name: 676
708
Tag                          blond wood
User          http://steve.nl/user_1002
Quality               usefulness-useful
Cluster_id                            1
Name: 708
[etc]

而且这些行数据可以像字典一样进行索引，比如：

In [48]: row
Out[48]: 
Tag                       rocking chair
User          http://steve.nl/user_1002
Quality       problematic-misperception
Cluster_id                            1
Name: 3650

In [49]: row["User"]
Out[49]: 'http://steve.nl/user_1002'

In [50]: row["Tag"]
Out[50]: 'rocking chair'

所以你可以这样写你的循环：

good = 0
bad = 0
for row_number, row in group.iterrows():
    if row['Quality'] == 'usefulness-useful':
        good += 1
    else:
        bad += 1
print 'good', good, 'bad', bad

这样就能得到：

good 3 bad 4

如果这个方法对你来说很清楚，那就没问题。另一种方法是直接从 Quality 列的计数入手：

In [54]: counts = group["Quality"].value_counts()

In [55]: counts
Out[55]: 
usefulness-useful            3
problematic-misperception    2
usefulness-not_useful        1
problematic-misspelling      1

In [56]: counts['usefulness-useful']
Out[56]: 3

而且因为坏的数量 = 总数 - 好的数量，所以我们有：

In [57]: counts.sum() - counts['usefulness-useful']
Out[57]: 4