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'}

另一个例子:

for key, igroup in itertools.groupby(xrange(12), lambda x: x // 5):
    print key, list(igroup)

结果

0 [0, 1, 2, 3, 4]
1 [5, 6, 7, 8, 9]
2 [10, 11]

注意，igroup是一个迭代器(文档称之为子迭代器)。

这对于分块生成器很有用:

def chunker(items, chunk_size):
    '''Group items in chunks of chunk_size'''
    for _key, group in itertools.groupby(enumerate(items), lambda x: x[0] // chunk_size):
        yield (g[1] for g in group)

with open('file.txt') as fobj:
    for chunk in chunker(fobj):
        process(chunk)

groupby的另一个例子-当键没有排序时。在以下示例中，xx中的项按yy中的值进行分组。在这种情况下，首先输出一组0，然后是一组1，然后又是一组0。

xx = range(10)
yy = [0, 0, 0, 1, 1, 1, 0, 0, 0, 0]
for group in itertools.groupby(iter(xx), lambda x: yy[x]):
    print group[0], list(group[1])

生产:

0 [0, 1, 2]
1 [3, 4, 5]
0 [6, 7, 8, 9]

@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!

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

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)]

我如何使用Python的itertools.groupby()?

您可以使用groupby来对迭代进行分组。你给groupby一个可迭代对象，和一个可选的键函数/可调用对象，用来检查从可迭代对象中取出的项，它返回一个迭代器，给出一个由可调用键的结果和另一个可迭代对象中的实际项组成的二元组。来自帮助:

groupby(iterable[, keyfunc]) -> create an iterator which returns
(key, sub-iterator) grouped by each value of key(value).

下面是groupby使用协程按计数分组的例子，它使用一个键可调用对象(在本例中是corroutine .send)来输出迭代次数的计数和元素的分组子迭代器:

import itertools


def grouper(iterable, n):
    def coroutine(n):
        yield # queue up coroutine
        for i in itertools.count():
            for j in range(n):
                yield i
    groups = coroutine(n)
    next(groups) # queue up coroutine

    for c, objs in itertools.groupby(iterable, groups.send):
        yield c, list(objs)
    # or instead of materializing a list of objs, just:
    # return itertools.groupby(iterable, groups.send)

list(grouper(range(10), 3))

打印

[(0, [0, 1, 2]), (1, [3, 4, 5]), (2, [6, 7, 8]), (3, [9])]