groupby的一个新技巧是在一行中运行长度编码:

[(c,len(list(cgen))) for c,cgen in groupby(some_string)]

会给你一个二元组列表，其中第一个元素是char，第二个元素是重复的次数。

编辑:注意这是itertools的区别。来自SQL GROUP BY语义的groupby: itertools不会(通常也不能)提前对迭代器排序，因此具有相同“key”的组不会合并。

Python文档中的示例非常简单:

groups = []
uniquekeys = []
for k, g in groupby(data, keyfunc):
    groups.append(list(g))      # Store group iterator as a list
    uniquekeys.append(k)

在你的例子中，data是一个节点列表，keyfunc是criteria函数的逻辑所在，然后groupby()对数据进行分组。

在调用groupby之前，必须小心地按照条件对数据进行排序，否则它将不起作用。Groupby方法实际上只是遍历一个列表，每当键改变时，它就创建一个新组。

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)]

我想再举一个例子，说明没有排序的groupby是行不通的。改编自James Sulak的例子

from itertools import groupby

things = [("vehicle", "bear"), ("animal", "duck"), ("animal", "cactus"), ("vehicle", "speed boat"), ("vehicle", "school bus")]

for key, group in groupby(things, lambda x: x[0]):
    for thing in group:
        print "A %s is a %s." % (thing[1], key)
    print " "

输出是

A bear is a vehicle.

A duck is a animal.
A cactus is a animal.

A speed boat is a vehicle.
A school bus is a vehicle.

有两组有车辆，而我们只能期待一组

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]}

我遇到的一个有用的例子可能会有帮助:

from itertools import groupby

#user input

myinput = input()

#creating empty list to store output

myoutput = []

for k,g in groupby(myinput):

    myoutput.append((len(list(g)),int(k)))

print(*myoutput)

示例输入:14445221

样本输出:(1,1)(3,4)(1,5)(2,2)(1,1)