Python 迭代器工具包【推荐】
0x01 介绍了迭代器的概念,即定义了 __iter__() 和 __next__() 方法的对象,或者通过 yield 简化定义的“可迭代对象”,而在一些函数式编程语言(见 0x02 Python 中的函数式编程)中,类似的迭代器常被用于产生特定格式的列表(或序列),这时的迭代器更像是一种数据结构而非函数(当然在一些函数式编程语言中,这两者并无本质差异)。Python 借鉴了 APL, Haskell, and SML 中的某些迭代器的构造方法,并在 itertools 中实现(该模块是通过 C 实现,源代码:/Modules/itertoolsmodule.c)。
itertools 模块提供了如下三类迭代器构建工具:
无限迭代
整合两序列迭代
组合生成器
1. 无限迭代
所谓无限(infinite)是指如果你通过 for...in... 的语法对其进行迭代,将陷入无限循环,包括:
count(start, [step]) cycle(p) repeat(elem [,n])
从名字大概可以猜出它们的用法,既然说是无限迭代,我们自然不会想要将其所有元素依次迭代取出,而通常是结合 map/zip 等方法,将其作为一个取之不尽的数据仓库,与有限长度的可迭代对象进行组合操作:
from itertools import cycle, count, repeat print(count.__doc__) count(start=0, step=1) --> count object Return a count object whose .__next__() method returns consecutive values. Equivalent to: def count(firstval=0, step=1): x = firstval while 1: yield x x += step counter = count() print(next(counter)) print(next(counter)) print(list(map(lambda x, y: x+y, range(10), counter))) odd_counter = map(lambda x: 'Odd#{}'.format(x), count(1, 2)) print(next(odd_counter)) print(next(odd_counter)) 0 1 [2, 4, 6, 8, 10, 12, 14, 16, 18, 20] Odd#1 Odd#3 print(cycle.__doc__) cycle(iterable) --> cycle object Return elements from the iterable until it is exhausted. Then repeat the sequence indefinitely. cyc = cycle(range(5)) print(list(zip(range(6), cyc))) print(next(cyc)) print(next(cyc)) [(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 0)] 1 2 print(repeat.__doc__) repeat(object [,times]) -> create an iterator which returns the object for the specified number of times. If not specified, returns the object endlessly. print(list(repeat('Py', 3))) rep = repeat('p') print(list(zip(rep, 'y'*3))) ['Py', 'Py', 'Py'] [('p', 'y'), ('p', 'y'), ('p', 'y')]
2. 整合两序列迭代
所谓整合两序列,是指以两个有限序列为输入,将其整合操作之后返回为一个迭代器,最为常见的 zip 函数就属于这一类别,只不过 zip 是内置函数。这一类别完整的方法包括:
accumulate() chain()/chain.from_iterable() compress() dropwhile()/filterfalse()/takewhile() groupby() islice() starmap() tee() zip_longest()
这里就不对所有的方法一一举例说明了,如果想要知道某个方法的用法,基本通过 print(method.__doc__) 就可以了解,毕竟 itertools 模块只是提供了一种快捷方式,并没有隐含什么深奥的算法。这里只对下面几个我觉得比较有趣的方法进行举例说明。
from itertools import cycle, compress, islice, takewhile, count # 这三个方法(如果使用恰当)可以限定无限迭代 # print(compress.__doc__) print(list(compress(cycle('PY'), [1, 0, 1, 0]))) # 像操作列表 l[start:stop:step] 一样操作其它序列 # print(islice.__doc__) print(list(islice(cycle('PY'), 0, 2))) # 限制版的 filter # print(takewhile.__doc__) print(list(takewhile(lambda x: x < 5, count()))) ['P', 'P'] ['P', 'Y'] [0, 1, 2, 3, 4] from itertools import groupby from operator import itemgetter print(groupby.__doc__) for k, g in groupby('AABBC'): print(k, list(g)) db = [dict(name='python', script=True), dict(name='c', script=False), dict(name='c++', script=False), dict(name='ruby', script=True)] keyfunc = itemgetter('script') db2 = sorted(db, key=keyfunc) # sorted by `script' for isScript, langs in groupby(db2, keyfunc): print(', '.join(map(itemgetter('name'), langs))) groupby(iterable[, keyfunc]) -> create an iterator which returns (key, sub-iterator) grouped by each value of key(value). A ['A', 'A'] B ['B', 'B'] C ['C'] c, c++ python, ruby from itertools import zip_longest # 内置函数 zip 以较短序列为基准进行合并, # zip_longest 则以最长序列为基准,并提供补足参数 fillvalue # Python 2.7 中名为 izip_longest print(list(zip_longest('ABCD', '123', fillvalue=0))) [('A', '1'), ('B', '2'), ('C', '3'), ('D', 0)]
3. 组合生成器
关于生成器的排列组合:
product(*iterables, repeat=1):两输入序列的笛卡尔乘积 permutations(iterable, r=None):对输入序列的完全排列组合 combinations(iterable, r):有序版的排列组合 combinations_with_replacement(iterable, r):有序版的笛卡尔乘积 from itertools import product, permutations, combinations, combinations_with_replacement print(list(product(range(2), range(2)))) print(list(product('AB', repeat=2))) [(0, 0), (0, 1), (1, 0), (1, 1)] [('A', 'A'), ('A', 'B'), ('B', 'A'), ('B', 'B')] print(list(combinations_with_replacement('AB', 2))) [('A', 'A'), ('A', 'B'), ('B', 'B')] # 赛马问题:4匹马前2名的排列组合(A^4_2) print(list(permutations('ABCDE', 2))) [('A', 'B'), ('A', 'C'), ('A', 'D'), ('A', 'E'), ('B', 'A'), ('B', 'C'), ('B', 'D'), ('B', 'E'), ('C', 'A'), ('C', 'B'), ('C', 'D'), ('C', 'E'), ('D', 'A'), ('D', 'B'), ('D', 'C'), ('D', 'E'), ('E', 'A'), ('E', 'B'), ('E', 'C'), ('E', 'D')] # 彩球问题:4种颜色的球任意抽出2个的颜色组合(C^4_2) print(list(combinations('ABCD', 2))) [('A', 'B'), ('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D'), ('C', 'D')]

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