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itertools 模块提供了非常多的方法。

Infinite iterators

Iterator	Arguments	Results	Example
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

import itertools as it

# count
count_iter = it.count(10)
print([next(count_iter) for i in range(5)])

# cycle
cycle_iter = it.cycle('ABCD')
print([next(cycle_iter) for i in range(8)])

# repeat
repeat_iter = it.repeat(10, 3)
print([next(repeat_iter) for i in range(3)])

Iterators terminating on the shortest input sequence

Iterator	Arguments	Results	Example
accumulate()	p [,func]	p0, p0+p1, p0+p1+p2, …	accumulate([1,2,3,4,5]) --> 1 3 6 10 15
chain()	p, q, …	p0, p1, … plast, q0, q1, …	chain(‘ABC’, ‘DEF’) --> A B C D E F
chain.from_iterable()	iterable	p0, p1, … plast, q0, q1, …	chain.from_iterable([‘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
filterfalse()	pred, seq	elements of seq where pred(elem) is false	filterfalse(lambda x: x%2, range(10)) --> 0 2 4 6 8
groupby()	iterable[, key]	sub-iterators grouped by value of key(v)
islice()	seq, [start,] stop [, step]	elements from seq[start:stop:step]	islice(‘ABCDEFG’, 2, None) --> C D E F G
starmap()	func, seq	func(*seq[0]), func(*seq[1]), …	starmap(pow, [(2,5), (3,2), (10,3)]) --> 32 9 1000
takewhile()	pred, seq	seq[0], seq[1], until pred fails	takewhile(lambda x: x<5, [1,4,6,4,1]) --> 1 4
tee()	it, n	it1, it2, … itn splits one iterator into n
zip_longest()	p, q, …	(p[0], q[0]), (p[1], q[1]), …	zip_longest(‘ABCD’, ‘xy’, fillvalue=’-’) --> Ax By C- D-

import itertools as it

# accumulate
print(list(it.accumulate([1, 2, 3, 4, 5])))

# chain
print(list(it.chain('ABC', 'DEF')))

# chain.from_iterable
print(list(it.chain.from_iterable(['ABC', 'DEF'])))

# compress
print(list(it.compress('ABCDEF', [1, 0, 1, 0, 1, 1])))

# dropwhile
print(list(it.dropwhile(lambda x: x < 5, [1, 4, 6, 4, 1])))

# filterfalse
print(list(it.filterfalse(lambda x: x % 2, range(10))))

# groupby
print([list(g) for k, g in it.groupby('AAAABBBCCD')])

# islice
print(list(it.islice('ABCDEFG', 2, None)))

# starmap
print(list(it.starmap(pow, [(2,5), (3,2), (10,3)])))

# takewhile
print(list(it.takewhile(lambda x: x < 5, [1, 4, 6, 4, 1])))

# takewhile
print(list(it.takewhile(lambda x: x < 5, [1, 4, 6, 4, 1])))

# tee
my_iter = iter(range(10))
it1, it2 = it.tee(my_iter, 2)
print(list(it1))
print(list(it2))

# zip_longest
print(list(it.zip_longest('ABCD', 'xy', fillvalue='-')))

Combinatoric iterators

Iterator	Arguments	Results
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

import itertools as it

# product
print(list(it.product('ABCD', repeat=2)))

# permutations
# n! / (n-r)!
print(list(it.permutations('ABCD', 2)))

# combinations
# n! / r! / (n-r)!
print(list(it.combinations('ABCD', 2)))

# combinations_with_replacement
# (n+r-1)! / r! / (n-1)!
print(list(it.combinations_with_replacement('ABCD', 2)))

自定义函数

尽管 Python 提供了大量的函数为函数式编程服务，有时候我们还是需要自定义函数，一是为了实现特殊功能，二是为了封装已有函数使可读性更好。如，我们想实现类似 SQL 取前 10 条数据的功能，我们可以这样写。

list(islice(iterable, n))

为了可读性更好，我们可以定义一个函数来封装它。

def top(n, iterable):
    return list(islice(iterable, n))

自定义函数要遵循 currying 原则，一个函数只实现一个功能，每次只处理一个元素，避免消耗内存，返回一个可迭代对象。

– 更多参见：Python 精萃
– 声 明：转载请注明出处
– Last Updated on 2018-10-05
– Written by ShangBo on 2018-10-04
– End