python itertools groubpy 实例

groupby(iterable [,key]):

创建一个迭代器，对iterable生成的连续项进行分组，在分组过程中会查找重复项。

如果iterable在多次连续迭代中生成了同一项，则会定义一个组，如果将此函数应用一个分类列表，那么分组将定义该列表中的所有唯一项，key（如果已提供）是一个函数，应用于每一项，如果此函数存在返回值，该值将用于后续项而不是该项本身进行比较，此函数返回的迭代器生成元素(key, group)，其中key是分组的键值，group是迭代器，生成组成该组的所有项。

an example:

 

from itertools import groupby
from operator import itemgetter

things =[('2009-09-02', 'mars', 1, 11),
          ('2009-09-02', 'mars', 1, 3),
          ('2009-09-03', 'ma', 2, 10),
          ('2009-09-03', 'ma', 2, 4),
          ('2009-09-03', 'ma', 2, 22),
          ('2009-09-06', 'm', 3, 33)]

for key, items in groupby(things, itemgetter(0)):
    print key
    for subitem in items:
        print subitem
    print '-' * 20

 The output looks like:

 

2009-09-02
('2009-09-02', 'mars, 11)
('2009-09-02', 'mars', 3)
--------------------
2009-09-03
('2009-09-03', 'ma', 10)
('2009-09-03', 'ma', 4)
('2009-09-03', 'ma', 22)
--------------------
2009-09-06
('2009-09-06', 'm', 33)
-------------------

another example:

key_func = lambda t: (t[0], t[1])
for key, items in groupby(things, key_func) :
    print key
    for subitem in items:
        print subitem
    print '-' * 20

 output:

('2009-09-02', 'mars')
('2009-09-02', 'mars, 11)
('2009-09-02', 'mars', 3)
--------------------
('2009-09-03', 'ma')
('2009-09-03', 'ma', 10)
('2009-09-03', 'ma', 4)
('2009-09-03', 'ma', 22)
--------------------
('2009-09-06', 'm')
('2009-09-06', 'm', 33)
-------------------