Python itertools模塊詳解

更新時間：2015年05月09日 12:19:48 投稿：junjie

這篇文章主要介紹了Python itertools模塊詳解,本文基本是基于文檔的翻譯和補充,相當于翻譯了,需要的朋友可以參考下

這貨很強大, 必須掌握

文檔鏈接 http://docs.python.org/2/library/itertools.html

pymotw 鏈接 http://pymotw.com/2/itertools/

基本是基于文檔的翻譯和補充，相當于翻譯了

itertools用于高效循環(huán)的迭代函數(shù)集合

組成

總體，整體了解

無限迭代器

復(fù)制代碼代碼如下:

迭代器         參數(shù)         結(jié)果                                                例子

count()     start, [step]   start, start+step, start+2*step, ...                count(10) --> 10 11 12 13 14 ...

cycle()     p               p0, p1, ... plast, p0, p1, ...                      cycle('ABCD') --> A B C D A B C D ...

repeat()    elem [,n]       elem, elem, elem, ... endlessly or up to n times    repeat(10, 3) --> 10 10 10

處理輸入序列迭代器

復(fù)制代碼代碼如下:

迭代器          參數(shù)            結(jié)果                                        例子

chain()     p, q, ...           p0, p1, ... plast, q0, q1, ...              chain('ABC', 'DEF') --> A B C D E F

compress()  data, selectors     (d[0] if s[0]), (d[1] if s[1]), ...         compress('ABCDEF', [1,0,1,0,1,1]) --> A C E F

dropwhile() pred, seq           seq[n], seq[n+1], starting when pred fails  dropwhile(lambda x: x<5, [1,4,6,4,1]) --> 6 4 1

groupby()   iterable[, keyfunc] sub-iterators grouped by value of keyfunc(v)

ifilter()   pred, seq           elements of seq where pred(elem) is True    ifilter(lambda x: x%2, range(10)) --> 1 3 5 7 9

ifilterfalse()  pred, seq       elements of seq where pred(elem) is False   ifilterfalse(lambda x: x%2, range(10)) --> 0 2 4 6 8

islice()    seq, [start,] stop [, step] elements from seq[start:stop:step]  islice('ABCDEFG', 2, None) --> C D E F G

imap()      func, p, q, ...     func(p0, q0), func(p1, q1), ...             imap(pow, (2,3,10), (5,2,3)) --> 32 9 1000

starmap()   func, seq           func(*seq[0]), func(*seq[1]), ...           starmap(pow, [(2,5), (3,2), (10,3)]) --> 32 9 1000

tee()       it, n               it1, it2 , ... itn splits one iterator into n

takewhile() pred, seq           seq[0], seq[1], until pred fails            takewhile(lambda x: x<5, [1,4,6,4,1]) --> 1 4

izip()      p, q, ...           (p[0], q[0]), (p[1], q[1]), ...             izip('ABCD', 'xy') --> Ax By

izip_longest()  p, q, ...       (p[0], q[0]), (p[1], q[1]), ...             izip_longest('ABCD', 'xy', fillvalue='-') --> Ax By C- D-

組合生成器

復(fù)制代碼代碼如下:

迭代器          參數(shù)                        結(jié)果

product()       p, q, ... [repeat=1]        cartesian product, equivalent to a nested for-loop

permutations()  p[, r]                      r-length tuples, all possible orderings, no repeated elements

combinations()  p, r                        r-length tuples, in sorted order, no repeated elements

combinations_with_replacement() p, r        r-length tuples, in sorted order, with repeated elements

product('ABCD', repeat=2)                   AA AB AC AD BA BB BC BD CA CB CC CD DA DB DC DD

permutations('ABCD', 2)                     AB AC AD BA BC BD CA CB CD DA DB DC

combinations('ABCD', 2)                     AB AC AD BC BD CD

combinations_with_replacement('ABCD', 2)    AA AB AC AD BB BC BD CC CD DD

第一部分

itertools.count(start=0, step=1)

創(chuàng)建一個迭代器，生成從n開始的連續(xù)整數(shù)，如果忽略n，則從0開始計算（注意：此迭代器不支持長整數(shù)）

如果超出了sys.maxint，計數(shù)器將溢出并繼續(xù)從-sys.maxint-1開始計算。

定義

復(fù)制代碼代碼如下:

def count(start=0, step=1):

    # count(10) --> 10 11 12 13 14 ...

    # count(2.5, 0.5) -> 2.5 3.0 3.5 ...

    n = start

    while True:

        yield n

        n += step

等同于(start + step * i for i in count())

使用

復(fù)制代碼代碼如下:

from itertools import *

for i in izip(count(1), ['a', 'b', 'c']):
print i

(1, 'a')
(2, 'b')
(3, 'c')

itertools.repeat(object[, times])

創(chuàng)建一個迭代器，重復(fù)生成object，times（如果已提供）指定重復(fù)計數(shù)，如果未提供times，將無止盡返回該對象。

定義

復(fù)制代碼代碼如下:

def repeat(object, times=None):

    # repeat(10, 3) --> 10 10 10

    if times is None:

        while True:

            yield object

    else:

        for i in xrange(times):

            yield object

使用

復(fù)制代碼代碼如下:

from itertools import *

for i in repeat('over-and-over', 5):
print i

over-and-over
over-and-over
over-and-over
over-and-over
over-and-over

第二部分
itertools.chain(*iterables)

將多個迭代器作為參數(shù), 但只返回單個迭代器, 它產(chǎn)生所有參數(shù)迭代器的內(nèi)容, 就好像他們是來自于一個單一的序列.

復(fù)制代碼代碼如下:

def chain(*iterables):

    # chain('ABC', 'DEF') --> A B C D E F

    for it in iterables:

        for element in it:

            yield element

使用

復(fù)制代碼代碼如下:

from itertools import *

for i in chain([1, 2, 3], ['a', 'b', 'c']):
print i
1
2
3
a
b
c

from itertools import chain, imap
def flatmap(f, items):
return chain.from_iterable(imap(f, items))
>>> list(flatmap(os.listdir, dirs))
>>> ['settings.py', 'wsgi.py', 'templates', 'app.py',
'templates', 'index.html, 'config.json']

itertools.compress(data, selectors)

提供一個選擇列表，對原始數(shù)據(jù)進行篩選

復(fù)制代碼代碼如下:

def compress(data, selectors):

    # compress('ABCDEF', [1,0,1,0,1,1]) --> A C E F

    return (d for d, s in izip(data, selectors) if s)

itertools.dropwhile(predicate, iterable)

創(chuàng)建一個迭代器，只要函數(shù)predicate(item)為True，就丟棄iterable中的項，如果predicate返回False，就會生成iterable中的項和所有后續(xù)項。

即：在條件為false之后的第一次, 返回迭代器中剩下來的項.

復(fù)制代碼代碼如下:

def dropwhile(predicate, iterable):

    # dropwhile(lambda x: x<5, [1,4,6,4,1]) --> 6 4 1

    iterable = iter(iterable)

    for x in iterable:

        if not predicate(x):

            yield x

            break

    for x in iterable:

        yield x

使用

復(fù)制代碼代碼如下:

from itertools import *

def should_drop(x):
print 'Testing:', x
return (x<1)

for i in dropwhile(should_drop, [ -1, 0, 1, 2, 3, 4, 1, -2 ]):
print 'Yielding:', i

Testing: -1
Testing: 0
Testing: 1
Yielding: 1
Yielding: 2
Yielding: 3
Yielding: 4
Yielding: 1
Yielding: -2

itertools.groupby(iterable[, key])

返回一個產(chǎn)生按照key進行分組后的值集合的迭代器.

如果iterable在多次連續(xù)迭代中生成了同一項，則會定義一個組，如果將此函數(shù)應(yīng)用一個分類列表，那么分組將定義該列表中的所有唯一項，key（如果已提供）是一個函數(shù)，應(yīng)用于每一項，如果此函數(shù)存在返回值，該值將用于后續(xù)項而不是該項本身進行比較，此函數(shù)返回的迭代器生成元素(key, group)，其中key是分組的鍵值，group是迭代器，生成組成該組的所有項。

即：按照keyfunc函數(shù)對序列每個元素執(zhí)行后的結(jié)果分組(每個分組是一個迭代器), 返回這些分組的迭代器

等價于

復(fù)制代碼代碼如下:

class groupby(object):

    # [k for k, g in groupby('AAAABBBCCDAABBB')] --> A B C D A B

    # [list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D

    def __init__(self, iterable, key=None):

        if key is None:

            key = lambda x: x

        self.keyfunc = key

        self.it = iter(iterable)

        self.tgtkey = self.currkey = self.currvalue = object()

    def __iter__(self):

        return self

    def next(self):

        while self.currkey == self.tgtkey:

            self.currvalue = next(self.it)    # Exit on StopIteration

            self.currkey = self.keyfunc(self.currvalue)

        self.tgtkey = self.currkey

        return (self.currkey, self._grouper(self.tgtkey))

    def _grouper(self, tgtkey):

        while self.currkey == tgtkey:

            yield self.currvalue

            self.currvalue = next(self.it)    # Exit on StopIteration

            self.currkey = self.keyfunc(self.currvalue)

應(yīng)用

復(fù)制代碼代碼如下:

from itertools import groupby
qs = [{'date' : 1},{'date' : 2}]
[(name, list(group)) for name, group in itertools.groupby(qs, lambda p:p['date'])]

Out[77]: [(1, [{'date': 1}]), (2, [{'date': 2}])]

>>> from itertools import *
>>> a = ['aa', 'ab', 'abc', 'bcd', 'abcde']
>>> for i, k in groupby(a, len):
... print i, list(k)
...
2 ['aa', 'ab']
3 ['abc', 'bcd']
5 ['abcde']

另一個例子

復(fù)制代碼代碼如下:

from itertools import *
from operator import itemgetter

d = dict(a=1, b=2, c=1, d=2, e=1, f=2, g=3)
di = sorted(d.iteritems(), key=itemgetter(1))
for k, g in groupby(di, key=itemgetter(1)):
print k, map(itemgetter(0), g)

1 ['a', 'c', 'e']
2 ['b', 'd', 'f']
3 ['g']

itertools.ifilter(predicate, iterable)

返回的是迭代器類似于針對列表的內(nèi)置函數(shù) filter() , 它只包括當測試函數(shù)返回true時的項. 它不同于 dropwhile()

創(chuàng)建一個迭代器，僅生成iterable中predicate(item)為True的項，如果predicate為None，將返回iterable中所有計算為True的項

對函數(shù)func執(zhí)行返回真的元素的迭代器

復(fù)制代碼代碼如下:

def ifilter(predicate, iterable):

    # ifilter(lambda x: x%2, range(10)) --> 1 3 5 7 9

    if predicate is None:

        predicate = bool

    for x in iterable:

        if predicate(x):

            yield x

使用

復(fù)制代碼代碼如下:

from itertools import *

def check_item(x):
print 'Testing:', x
return (x<1)

for i in ifilter(check_item, [ -1, 0, 1, 2, 3, 4, 1, -2 ]):
print 'Yielding:', i

Testing: -1
Yielding: -1
Testing: 0
Yielding: 0
Testing: 1
Testing: 2
Testing: 3
Testing: 4
Testing: 1
Testing: -2
Yielding: -2

itertools.ifilterfalse(predicate, iterable)

和ifilter(函數(shù)相反，返回一個包含那些測試函數(shù)返回false的項的迭代器)

創(chuàng)建一個迭代器，僅生成iterable中predicate(item)為False的項，如果predicate為None，則返回iterable中所有計算為False的項對函數(shù)func執(zhí)行返回假的元素的迭代器

復(fù)制代碼代碼如下:

def ifilterfalse(predicate, iterable):

    # ifilterfalse(lambda x: x%2, range(10)) --> 0 2 4 6 8

    if predicate is None:

        predicate = bool

    for x in iterable:

        if not predicate(x):

            yield x

使用

復(fù)制代碼代碼如下:

from itertools import *

def check_item(x):
print 'Testing:', x
return (x<1)

for i in ifilterfalse(check_item, [ -1, 0, 1, 2, 3, 4, 1, -2 ]):
print 'Yielding:', i

Testing: -1
Testing: 0
Testing: 1
Yielding: 1
Testing: 2
Yielding: 2
Testing: 3
Yielding: 3
Testing: 4
Yielding: 4
Testing: 1
Yielding: 1
Testing: -2

itertools.islice(iterable, stop)

itertools.islice(iterable, start, stop[, step])

返回的迭代器是返回了輸入迭代器根據(jù)索引來選取的項

創(chuàng)建一個迭代器，生成項的方式類似于切片返回值： iterable[start : stop : step]，將跳過前start個項，迭代在stop所指定的位置停止，step指定用于跳過項的步幅。與切片不同，負值不會用于任何start，stop和step，如果省略了start，迭代將從0開始，如果省略了step，步幅將采用1.

返回序列seq的從start開始到stop結(jié)束的步長為step的元素的迭代器

復(fù)制代碼代碼如下:

def islice(iterable, *args):

    # islice('ABCDEFG', 2) --> A B

    # islice('ABCDEFG', 2, 4) --> C D

    # islice('ABCDEFG', 2, None) --> C D E F G

    # islice('ABCDEFG', 0, None, 2) --> A C E G

    s = slice(*args)

    it = iter(xrange(s.start or 0, s.stop or sys.maxint, s.step or 1))

    nexti = next(it)

    for i, element in enumerate(iterable):

        if i == nexti:

            yield element

            nexti = next(it)

使用

復(fù)制代碼代碼如下:

from itertools import *

print 'Stop at 5:'
for i in islice(count(), 5):
print i

print 'Start at 5, Stop at 10:'
for i in islice(count(), 5, 10):
print i

print 'By tens to 100:'
for i in islice(count(), 0, 100, 10):
print i

Stop at 5:
0
1
2
3
4
Start at 5, Stop at 10:
5
6
7
8
9
By tens to 100:
0
10
20
30
40
50
60
70
80
90

itertools.imap(function, *iterables)

創(chuàng)建一個迭代器，生成項function(i1, i2, ..., iN)，其中i1，i2...iN分別來自迭代器iter1，iter2 ... iterN，如果function為None，則返回(i1, i2, ..., iN)形式的元組，只要提供的一個迭代器不再生成值，迭代就會停止。

即：返回一個迭代器, 它是調(diào)用了一個其值在輸入迭代器上的函數(shù), 返回結(jié)果. 它類似于內(nèi)置函數(shù) map() , 只是前者在任意輸入迭代器結(jié)束后就停止(而不是插入None值來補全所有的輸入).

返回序列每個元素被func執(zhí)行后返回值的序列的迭代器

復(fù)制代碼代碼如下:

def imap(function, *iterables):

    # imap(pow, (2,3,10), (5,2,3)) --> 32 9 1000

    iterables = map(iter, iterables)

    while True:

        args = [next(it) for it in iterables]

        if function is None:

            yield tuple(args)

        else:

            yield function(*args)

使用

復(fù)制代碼代碼如下:

from itertools import *

print 'Doubles:'
for i in imap(lambda x:2*x, xrange(5)):
print i

print 'Multiples:'
for i in imap(lambda x,y:(x, y, x*y), xrange(5), xrange(5,10)):
print '%d * %d = %d' % i

Doubles:
0
2
4
6
8
Multiples:
0 * 5 = 0
1 * 6 = 6
2 * 7 = 14
3 * 8 = 24
4 * 9 = 36

itertools.starmap(function, iterable)

創(chuàng)建一個迭代器，生成值func(*item),其中item來自iterable，只有當iterable生成的項適用于這種調(diào)用函數(shù)的方式時，此函數(shù)才有效。

對序列seq的每個元素作為func的參數(shù)列表執(zhí)行, 返回執(zhí)行結(jié)果的迭代器

復(fù)制代碼代碼如下:

def starmap(function, iterable):

    # starmap(pow, [(2,5), (3,2), (10,3)]) --> 32 9 1000

    for args in iterable:

        yield function(*args)

使用

復(fù)制代碼代碼如下:

from itertools import *

values = [(0, 5), (1, 6), (2, 7), (3, 8), (4, 9)]
for i in starmap(lambda x,y:(x, y, x*y), values):
print '%d * %d = %d' % i

0 * 5 = 0
1 * 6 = 6
2 * 7 = 14
3 * 8 = 24
4 * 9 = 36

itertools.tee(iterable[, n=2])

返回一些基于單個原始輸入的獨立迭代器(默認為2). 它和Unix上的tee工具有點語義相似, 也就是說它們都重復(fù)讀取輸入設(shè)備中的值并將值寫入到一個命名文件和標準輸出中

從iterable創(chuàng)建n個獨立的迭代器，創(chuàng)建的迭代器以n元組的形式返回，n的默認值為2，此函數(shù)適用于任何可迭代的對象，但是，為了克隆原始迭代器，生成的項會被緩存，并在所有新創(chuàng)建的迭代器中使用，一定要注意，不要在調(diào)用tee()之后使用原始迭代器iterable，否則緩存機制可能無法正確工作。

把一個迭代器分為n個迭代器, 返回一個元組.默認是兩個

復(fù)制代碼代碼如下:

def tee(iterable, n=2):

    it = iter(iterable)

    deques = [collections.deque() for i in range(n)]

    def gen(mydeque):

        while True:

            if not mydeque:             # when the local deque is empty

                newval = next(it)       # fetch a new value and

                for d in deques:        # load it to all the deques

                    d.append(newval)

            yield mydeque.popleft()

    return tuple(gen(d) for d in deques)

使用

復(fù)制代碼代碼如下:

from itertools import *

r = islice(count(), 5)
i1, i2 = tee(r)

for i in i1:
print 'i1:', i
for i in i2:
print 'i2:', i

i1: 0
i1: 1
i1: 2
i1: 3
i1: 4
i2: 0
i2: 1
i2: 2
i2: 3
i2: 4

itertools.takewhile(predicate, iterable)

和dropwhile相反

創(chuàng)建一個迭代器，生成iterable中predicate(item)為True的項，只要predicate計算為False，迭代就會立即停止。

即：從序列的頭開始, 直到執(zhí)行函數(shù)func失敗.

復(fù)制代碼代碼如下:

def takewhile(predicate, iterable):

    # takewhile(lambda x: x<5, [1,4,6,4,1]) --> 1 4

    for x in iterable:

        if predicate(x):

            yield x

        else:

            break

使用

復(fù)制代碼代碼如下:

from itertools import *

def should_take(x):
print 'Testing:', x
return (x<2)

for i in takewhile(should_take, [ -1, 0, 1, 2, 3, 4, 1, -2 ]):
print 'Yielding:', i

Testing: -1
Yielding: -1
Testing: 0
Yielding: 0
Testing: 1
Yielding: 1
Testing: 2

itertools.izip(*iterables)

返回一個合并了多個迭代器為一個元組的迭代器. 它類似于內(nèi)置函數(shù)zip(), 只是它返回的是一個迭代器而不是一個列表

創(chuàng)建一個迭代器，生成元組(i1, i2, ... iN)，其中i1，i2 ... iN 分別來自迭代器iter1，iter2 ... iterN，只要提供的某個迭代器不再生成值，迭代就會停止，此函數(shù)生成的值與內(nèi)置的zip()函數(shù)相同。

復(fù)制代碼代碼如下:

izip(iter1, iter2, ... iterN):
返回:(it1[0],it2 [0], it3[0], ..), (it1[1], it2[1], it3[1], ..)...

def izip(*iterables):
    # izip('ABCD', 'xy') --> Ax By
    iterators = map(iter, iterables)
    while iterators:
        yield tuple(map(next, iterators))

使用

復(fù)制代碼代碼如下:

from itertools import *

for i in izip([1, 2, 3], ['a', 'b', 'c']):
print i
(1, 'a')
(2, 'b')
(3, 'c')

itertools.izip_longest(*iterables[, fillvalue])

與izip()相同，但是迭代過程會持續(xù)到所有輸入迭代變量iter1,iter2等都耗盡為止，如果沒有使用fillvalue關(guān)鍵字參數(shù)指定不同的值，則使用None來填充已經(jīng)使用的迭代變量的值。

復(fù)制代碼代碼如下:

class ZipExhausted(Exception):
pass

def izip_longest(*args, **kwds):
    # izip_longest('ABCD', 'xy', fillvalue='-') --> Ax By C- D-
    fillvalue = kwds.get('fillvalue')
    counter = [len(args) - 1]
    def sentinel():
        if not counter[0]:
            raise ZipExhausted
        counter[0] -= 1
        yield fillvalue
    fillers = repeat(fillvalue)
    iterators = [chain(it, sentinel(), fillers) for it in args]
    try:
        while iterators:
            yield tuple(map(next, iterators))
    except ZipExhausted:
        pass

第三部分

itertools.product(*iterables[, repeat])

笛卡爾積

創(chuàng)建一個迭代器，生成表示item1，item2等中的項目的笛卡爾積的元組，repeat是一個關(guān)鍵字參數(shù)，指定重復(fù)生成序列的次數(shù)。

復(fù)制代碼代碼如下:

def product(*args, **kwds):

    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy

    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111

    pools = map(tuple, args) * kwds.get('repeat', 1)

    result = [[]]

    for pool in pools:

        result = [x+[y] for x in result for y in pool]

    for prod in result:

        yield tuple(prod)

例子

復(fù)制代碼代碼如下:

import itertools
a = (1, 2, 3)
b = ('A', 'B', 'C')
c = itertools.product(a,b)
for elem in c:
print elem

(1, 'A')
(1, 'B')
(1, 'C')
(2, 'A')
(2, 'B')
(2, 'C')
(3, 'A')
(3, 'B')
(3, 'C')

itertools.permutations(iterable[, r])

排列

創(chuàng)建一個迭代器，返回iterable中所有長度為r的項目序列，如果省略了r，那么序列的長度與iterable中的項目數(shù)量相同：返回p中任意取r個元素做排列的元組的迭代器

復(fù)制代碼代碼如下:

def permutations(iterable, r=None):
    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = range(n)
    cycles = range(n, n-r, -1)
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return
也可以用product實現(xiàn)

def permutations(iterable, r=None):
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    for indices in product(range(n), repeat=r):
        if len(set(indices)) == r:
            yield tuple(pool[i] for i in indices)

itertools.combinations(iterable, r)

創(chuàng)建一個迭代器，返回iterable中所有長度為r的子序列，返回的子序列中的項按輸入iterable中的順序排序 (不帶重復(fù))

復(fù)制代碼代碼如下:

def combinations(iterable, r):
    # combinations('ABCD', 2) --> AB AC AD BC BD CD
    # combinations(range(4), 3) --> 012 013 023 123
    pool = tuple(iterable)
    n = len(pool)
    if r > n:
        return
    indices = range(r)
    yield tuple(pool[i] for i in indices)
    while True:
        for i in reversed(range(r)):
            if indices[i] != i + n - r:
                break
        else:
            return
        indices[i] += 1
        for j in range(i+1, r):
            indices[j] = indices[j-1] + 1
        yield tuple(pool[i] for i in indices)

#或者
def combinations(iterable, r):
    pool = tuple(iterable)
    n = len(pool)
    for indices in permutations(range(n), r):
        if sorted(indices) == list(indices):
            yield tuple(pool[i] for i in indices)

itertools.combinations_with_replacement(iterable, r)

創(chuàng)建一個迭代器，返回iterable中所有長度為r的子序列，返回的子序列中的項按輸入iterable中的順序排序 (帶重復(fù))

復(fù)制代碼代碼如下:

def combinations_with_replacement(iterable, r):

    # combinations_with_replacement('ABC', 2) --> AA AB AC BB BC CC

    pool = tuple(iterable)

    n = len(pool)

    if not n and r:

        return

    indices = [0] * r

    yield tuple(pool[i] for i in indices)

    while True:

        for i in reversed(range(r)):

            if indices[i] != n - 1:

                break

        else:

            return

        indices[i:] = [indices[i] + 1] * (r - i)

        yield tuple(pool[i] for i in indices)

或者

def combinations_with_replacement(iterable, r):

    pool = tuple(iterable)

    n = len(pool)

    for indices in product(range(n), repeat=r):

        if sorted(indices) == list(indices):

            yield tuple(pool[i] for i in indices)

第四部分

擴展

使用現(xiàn)有擴展功能

復(fù)制代碼代碼如下:

def take(n, iterable):
"Return first n items of the iterable as a list"
return list(islice(iterable, n))

def tabulate(function, start=0):
"Return function(0), function(1), ..."
return imap(function, count(start))

def consume(iterator, n):
    "Advance the iterator n-steps ahead. If n is none, consume entirely."
    # Use functions that consume iterators at C speed.
    if n is None:
        # feed the entire iterator into a zero-length deque
        collections.deque(iterator, maxlen=0)
    else:
        # advance to the empty slice starting at position n
        next(islice(iterator, n, n), None)

def nth(iterable, n, default=None):
"Returns the nth item or a default value"
return next(islice(iterable, n, None), default)

def quantify(iterable, pred=bool):
"Count how many times the predicate is true"
return sum(imap(pred, iterable))

def padnone(iterable):
"""Returns the sequence elements and then returns None indefinitely.

    Useful for emulating the behavior of the built-in map() function.
    """
    return chain(iterable, repeat(None))

def ncycles(iterable, n):
"Returns the sequence elements n times"
return chain.from_iterable(repeat(tuple(iterable), n))

def dotproduct(vec1, vec2):
return sum(imap(operator.mul, vec1, vec2))

def flatten(listOfLists):
"Flatten one level of nesting"
return chain.from_iterable(listOfLists)

def repeatfunc(func, times=None, *args):
"""Repeat calls to func with specified arguments.

    Example: repeatfunc(random.random)
    """
    if times is None:
        return starmap(func, repeat(args))
    return starmap(func, repeat(args, times))

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    pending = len(iterables)
    nexts = cycle(iter(it).next for it in iterables)
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))

def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    # unique_everseen('AAAABBBCCDAABBB') --> A B C D
    # unique_everseen('ABBCcAD', str.lower) --> A B C D
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in ifilterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element

def unique_justseen(iterable, key=None):
    "List unique elements, preserving order. Remember only the element just seen."
    # unique_justseen('AAAABBBCCDAABBB') --> A B C D A B
    # unique_justseen('ABBCcAD', str.lower) --> A B C A D
    return imap(next, imap(itemgetter(1), groupby(iterable, key)))

def iter_except(func, exception, first=None):
""" Call a function repeatedly until an exception is raised.

    Converts a call-until-exception interface to an iterator interface.
    Like __builtin__.iter(func, sentinel) but uses an exception instead
    of a sentinel to end the loop.

    Examples:
        bsddbiter = iter_except(db.next, bsddb.error, db.first)
        heapiter = iter_except(functools.partial(heappop, h), IndexError)
        dictiter = iter_except(d.popitem, KeyError)
        dequeiter = iter_except(d.popleft, IndexError)
        queueiter = iter_except(q.get_nowait, Queue.Empty)
        setiter = iter_except(s.pop, KeyError)

    """
    try:
        if first is not None:
            yield first()
        while 1:
            yield func()
    except exception:
        pass

def random_product(*args, **kwds):
    "Random selection from itertools.product(*args, **kwds)"
    pools = map(tuple, args) * kwds.get('repeat', 1)
    return tuple(random.choice(pool) for pool in pools)

def random_permutation(iterable, r=None):
    "Random selection from itertools.permutations(iterable, r)"
    pool = tuple(iterable)
    r = len(pool) if r is None else r
    return tuple(random.sample(pool, r))

def random_combination(iterable, r):
    "Random selection from itertools.combinations(iterable, r)"
    pool = tuple(iterable)
    n = len(pool)
    indices = sorted(random.sample(xrange(n), r))
    return tuple(pool[i] for i in indices)

def random_combination_with_replacement(iterable, r):
    "Random selection from itertools.combinations_with_replacement(iterable, r)"
    pool = tuple(iterable)
    n = len(pool)
    indices = sorted(random.randrange(n) for i in xrange(r))
    return tuple(pool[i] for i in indices)

def tee_lookahead(t, i):
"""Inspect the i-th upcomping value from a tee object
while leaving the tee object at its current position.

Raise an IndexError if the underlying iterator doesn't
have enough values.

    """
    for value in islice(t.__copy__(), i, None):
        return value
    raise IndexError(i)

自定義擴展

將序列按大小切分,更好的性能

復(fù)制代碼代碼如下:

from itertools import chain, islice
def chunks(iterable, size, format=iter):
    it = iter(iterable)
    while True:
        yield format(chain((it.next(),), islice(it, size - 1)))

>>> l = ["a", "b", "c", "d", "e", "f", "g"]
>>> for chunk in chunks(l, 3, tuple):...
print chunk...
("a", "b", "c")
("d", "e", "f")
("g",)

補充

迭代工具，你最好的朋友

迭代工具模塊包含了操做指定的函數(shù)用于操作迭代器。想復(fù)制一個迭代器出來？鏈接兩個迭代器？以one liner（這里的one-liner只需一行代碼能搞定的任務(wù))用內(nèi)嵌的列表組合一組值？不使用list創(chuàng)建Map/Zip？···，你要做的就是 import itertools，舉個例子吧：

四匹馬賽跑到達終點排名的所有可能性：

復(fù)制代碼代碼如下:

>>> horses = [1, 2, 3, 4]

>>> races = itertools.permutations(horses)

>>> print(races)

<itertools.permutations object at 0xb754f1dc]]>

>>> print(list(itertools.permutations(horses)))

[(1, 2, 3, 4),

 (1, 2, 4, 3),

 (1, 3, 2, 4),

 (1, 3, 4, 2),

 (1, 4, 2, 3),

 (1, 4, 3, 2),

 (2, 1, 3, 4),

 (2, 1, 4, 3),

 (2, 3, 1, 4),

 (2, 3, 4, 1),

 (2, 4, 1, 3),

 (2, 4, 3, 1),

 (3, 1, 2, 4),

 (3, 1, 4, 2),

 (3, 2, 1, 4),

 (3, 2, 4, 1),

 (3, 4, 1, 2),

 (3, 4, 2, 1),

 (4, 1, 2, 3),

 (4, 1, 3, 2),

 (4, 2, 1, 3),

 (4, 2, 3, 1),

 (4, 3, 1, 2),

 (4, 3, 2, 1)]

理解迭代的內(nèi)部機制：迭代(iteration）就是對可迭代對象（iterables，實現(xiàn)了__iter__()方法）和迭代器（iterators，實現(xiàn)了__next__()方法）的一個操作過程?？傻鷮ο笫侨魏慰煞祷匾粋€迭代器的對象，迭代器是應(yīng)用在迭代對象中迭代的對象，換一種方式說的話就是：iterable對象的__iter__()方法可以返回iterator對象，iterator通過調(diào)用next()方法獲取其中的每一個值(譯者注)，讀者可以結(jié)合Java API中的 Iterable接口和Iterator接口進行類比。

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