@CaptSolo，我试过你的例子，但没用。

from itertools import groupby 
[(c,len(list(cs))) for c,cs in groupby('Pedro Manoel')]

输出:

[('P', 1), ('e', 1), ('d', 1), ('r', 1), ('o', 1), (' ', 1), ('M', 1), ('a', 1), ('n', 1), ('o', 1), ('e', 1), ('l', 1)]

如你所见，有两个o和两个e，但它们被分成了不同的组。这时我意识到需要对传递给groupby函数的列表进行排序。所以，正确的用法是:

name = list('Pedro Manoel')
name.sort()
[(c,len(list(cs))) for c,cs in groupby(name)]

输出:

[(' ', 1), ('M', 1), ('P', 1), ('a', 1), ('d', 1), ('e', 2), ('l', 1), ('n', 1), ('o', 2), ('r', 1)]

记住，如果列表没有排序，groupby函数将不起作用!

使用itertools的关键是要认识到。Groupby是指只有在可迭代对象中是顺序的项才会被分组在一起。这就是排序工作的原因，因为基本上你在重新排列集合，以便所有满足callback(item)的项现在都按顺序出现在排序的集合中。

也就是说，您不需要对列表进行排序，只需要一个键-值对的集合，其中的值可以根据groupby生成的每个group iterable增长。例如，列表字典。

>>> things = [("vehicle", "bear"), ("animal", "duck"), ("animal", "cactus"), ("vehicle", "speed boat"), ("vehicle", "school bus")]
>>> coll = {}
>>> for k, g in itertools.groupby(things, lambda x: x[0]):
...     coll.setdefault(k, []).extend(i for _, i in g)
...
{'vehicle': ['bear', 'speed boat', 'school bus'], 'animal': ['duck', 'cactus']}

排序和分组

from itertools import groupby

val = [{'name': 'satyajit', 'address': 'btm', 'pin': 560076}, 
       {'name': 'Mukul', 'address': 'Silk board', 'pin': 560078},
       {'name': 'Preetam', 'address': 'btm', 'pin': 560076}]


for pin, list_data in groupby(sorted(val, key=lambda k: k['pin']),lambda x: x['pin']):
...     print pin
...     for rec in list_data:
...             print rec
... 
o/p:

560076
{'name': 'satyajit', 'pin': 560076, 'address': 'btm'}
{'name': 'Preetam', 'pin': 560076, 'address': 'btm'}
560078
{'name': 'Mukul', 'pin': 560078, 'address': 'Silk board'}

from random import randint
from itertools import groupby

 l = [randint(1, 3) for _ in range(20)]

 d = {}
 for k, g in groupby(l, lambda x: x):
     if not d.get(k, None):
         d[k] = list(g)
     else:
         d[k] = d[k] + list(g)

上面的代码展示了如何使用groupby根据提供的lambda函数/键对列表进行分组。唯一的问题是输出没有合并，这可以使用字典轻松解决。

例子:

l = [2, 1, 2, 3, 1, 3, 2, 1, 3, 3, 1, 3, 2, 3, 1, 2, 1, 3, 2, 3]

应用groupby后，结果将是:

for k, g in groupby(l, lambda x:x):
    print(k, list(g))

2 [2]
1 [1]
2 [2]
3 [3]
1 [1]
3 [3]
2 [2]
1 [1]
3 [3, 3]
1 [1]
3 [3]
2 [2]
3 [3]
1 [1]
2 [2]
1 [1]
3 [3]
2 [2]
3 [3]

一旦字典被使用如下所示的结果可以很容易地迭代:

{2: [2, 2, 2, 2, 2, 2], 1: [1, 1, 1, 1, 1, 1], 3: [3, 3, 3, 3, 3, 3, 3, 3]}

itertools。Groupby是一个对项目进行分组的工具。

从文档中，我们进一步收集了它可能做的事情:

# [k for k, g in groupby('AAAABBBCCDAABBB')]——> AB CDA B # [list(g) for k, g in groupby('AAAABBBCCD')]——> AAAABBBCC

Groupby对象产生键-组对，其中组是一个生成器。

特性

A.将连续的项目组合在一起 B.给定一个已排序的可迭代对象，对一个项目的所有出现进行分组 C.指定如何使用键功能*对项目进行分组

比较

# Define a printer for comparing outputs
>>> def print_groupby(iterable, keyfunc=None):
...    for k, g in it.groupby(iterable, keyfunc):
...        print("key: '{}'--> group: {}".format(k, list(g)))

# Feature A: group consecutive occurrences
>>> print_groupby("BCAACACAADBBB")
key: 'B'--> group: ['B']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A', 'A']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A']
key: 'C'--> group: ['C']
key: 'A'--> group: ['A', 'A']
key: 'D'--> group: ['D']
key: 'B'--> group: ['B', 'B', 'B']

# Feature B: group all occurrences
>>> print_groupby(sorted("BCAACACAADBBB"))
key: 'A'--> group: ['A', 'A', 'A', 'A', 'A']
key: 'B'--> group: ['B', 'B', 'B', 'B']
key: 'C'--> group: ['C', 'C', 'C']
key: 'D'--> group: ['D']

# Feature C: group by a key function
>>> # islower = lambda s: s.islower()                      # equivalent
>>> def islower(s):
...     """Return True if a string is lowercase, else False."""   
...     return s.islower()
>>> print_groupby(sorted("bCAaCacAADBbB"), keyfunc=islower)
key: 'False'--> group: ['A', 'A', 'A', 'B', 'B', 'C', 'C', 'D']
key: 'True'--> group: ['a', 'a', 'b', 'b', 'c']

Uses

Anagrams (see notebook) Binning Group odd and even numbers Group a list by values Remove duplicate elements Find indices of repeated elements in an array Split an array into n-sized chunks Find corresponding elements between two lists Compression algorithm (see notebook)/Run Length Encoding Grouping letters by length, key function (see notebook) Consecutive values over a threshold (see notebook) Find ranges of numbers in a list or continuous items (see docs) Find all related longest sequences Take consecutive sequences that meet a condition (see related post)

注意:后面的几个例子来自Víctor Terrón的PyCon (talk)(西班牙语)，“Kung Fu at Dawn with Itertools”。请参见用C语言编写的groupby源代码。

*一个函数，其中所有项都被传递和比较，影响结果。其他具有key函数的对象包括sorted()， max()和min()。

响应

# OP: Yes, you can use `groupby`, e.g. 
[do_something(list(g)) for _, g in groupby(lxml_elements, criteria_func)]

遗憾的是，我不认为使用itertools.groupby()是明智的。要安全使用它太难了，而且只需要几行代码就可以写出一些可以按照预期工作的东西。

def my_group_by(iterable, keyfunc):
    """Because itertools.groupby is tricky to use

    The stdlib method requires sorting in advance, and returns iterators not
    lists, and those iterators get consumed as you try to use them, throwing
    everything off if you try to look at something more than once.
    """
    ret = defaultdict(list)
    for k in iterable:
        ret[keyfunc(k)].append(k)
    return dict(ret)

像这样使用它:

def first_letter(x):
    return x[0]

my_group_by('four score and seven years ago'.split(), first_letter)

得到

{'f': ['four'], 's': ['score', 'seven'], 'a': ['and', 'ago'], 'y': ['years']}