13

Python 的简洁语法通过它的电池允许冗长的代码行以可读的单行表示。考虑以下示例

====================================================|
for a in range(3):                                  |
    for b in range(3):                              |
        for c in range(3):                          |
            print (a,b,c),                          |
-  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -|
for e in product(range(3), repeat=3):               |
    print e,                                        |
====================================================|
for a in range(3):                                  |
    for b in range(a , 3):                          |
        for c in range(b , 3):                      |
            print (a,b,c),                          |
-  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -|
for e in combinations_with_replacement(range(3), 3):|
    print e,                                        |
====================================================|
for a in range(3):                                  |
    for b in range(a + 1, 3):                       |
        for c in range(b + 1, 3):                   |
            print (a,b,c),                          |
-  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -|
for e in combinations(range(3), 3):                 |
    print e,                                        |
====================================================|
for a in range(3):                                  |
    for b in range(3):                              |
        for c in range(3):                          |
            if len(set([a,b,c])) == 3:              |
                print (a,b,c),                      |
-  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -  -|
for e in permutations(range(3)):                    |
    print e,                                        |
====================================================|

最近我得到了一个深度嵌套的依赖循环，我试图简洁地表达但失败了

循环的结构如下

for a in A():
    for b in B(a):
        for c in C(b):
            foo(a,b,c)

这种结构可以用等效的 itertools 符号表示吗？

4

2 回答 2

5

没有确切的itertools解决方案，但功能的简单组合itertools就足够了：

def chain_imap_accumulate(seq, f):
    def acc_f(x):
        for n in f(x[-1]):
            yield x + (n,)
    return chain.from_iterable(imap(acc_f, seq))

def accumulative_product(*generators):
    head, tail = generators[0], generators[1:]
    head = imap(tuple, head())
    return reduce(chain_imap_accumulate, tail, head)

快速测试。定义：

from itertools import chain, imap, izip
chain_ = chain.from_iterable

def A():
    yield 'A'
    yield 'B'

def B(x):
    yield int(x, 16)
    yield int(x, 16) + 1

def C(x):
    yield str(x) + 'Z'
    yield str(x) + 'Y'

结果：

>>> list(accumulative_product(A, B, C))
[('A', 10, '10Z'), ('A', 10, '10Y'), 
 ('A', 11, '11Z'), ('A', 11, '11Y'), 
 ('B', 11, '11Z'), ('B', 11, '11Y'), 
 ('B', 12, '12Z'), ('B', 12, '12Y')]

正如上面代码的快速“推导”所示，几乎所有的复杂性都来自输入的累积。最终的( c) 值可以仅使用几个嵌套itertools结构生成：

>>> list(chain_(imap(C, chain_(imap(B, (A()))))))
['10Z', '10Y', '11Z', '11Y', '11Z', '11Y', '12Z', '12Y']

这可以概括为reduce。要使用reduce，chain_imap不能使用标准imap参数顺序。它必须被交换：

def chain_imap(seq, f):
    return chain.from_iterable(imap(f, seq))

这给出了相同的结果：

>>> list(reduce(chain_imap, [B, C], A()))
['10Z', '10Y', '11Z', '11Y', '11Z', '11Y', '12Z', '12Y']

最后的任务是累积初始值，以便您可以访问a、b和c。这需要一些思考才能正确，但实现相当简单——我们只需转换f为忽略除最后一个输入值之外的所有输入值的函数，并将新值附加到完整输入：

def chain_imap_accumulate(seq, f):
    def acc_f(x):
        for n in f(x[-1]):
            yield x + (n,)
    return chain.from_iterable(imap(acc_f, seq))

这要求将第一个输入包装在元组中，因此我们映射A为tuple：

>>> list(reduce(chain_imap_accumulate, [B, C], imap(tuple, A())))
[('A', 10, '10Z'), ('A', 10, '10Y'), 
 ('A', 11, '11Z'), ('A', 11, '11Y'), 
 ('B', 11, '11Z'), ('B', 11, '11Y'), 
 ('B', 12, '12Z'), ('B', 12, '12Y')]

为了清楚起见，重写上面的代码，结果是这个答案顶部的代码。

顺便说一句，chain_imap_accumulate可以使用genex更简洁地重写。这可以与一个较短版本的accumulative_product非常紧凑的定义相结合（如果你对那种东西感兴趣的话）。这也恰好完全消除了 itertools 依赖：

def chain_map_accumulate(seq, f):
    return (x + (n,) for x in seq for n in f(x[-1]))

def accumulative_product2(*gens):
    return reduce(chain_map_accumulate, gens[1:], (tuple(x) for x in gens[0]()))
于 2013-02-23T16:22:13.430 回答
4

没有，但你可以做一个：

def chainGang(steps, currentVars=None):
    thisOne = steps[0]
    if currentVars is None:
        for item in thisOne():
            for gang in chainGang(steps[1:], [item]):
                yield gang
    elif len(steps) == 1:       
        for item in thisOne(currentVars[-1]):
            yield currentVars + [item]
    else:
        for item in thisOne(currentVars[-1]):
            for gang in chainGang(steps[1:], currentVars + [item]):
                yield gang

接着：

>>> outer = lambda: ["A", "B", "C", "D"]
>>> middle = lambda letter: [letter, letter*2, letter*3]
>>> inner = lambda s: range(len(s)+1)
>>> for a in chainGang([outer, middle, inner]):
...     print a
[u'A', u'A', 0]
[u'A', u'A', 1]
[u'A', u'AA', 0]
[u'A', u'AA', 1]
[u'A', u'AA', 2]
[u'A', u'AAA', 0]
[u'A', u'AAA', 1]
[u'A', u'AAA', 2]
[u'A', u'AAA', 3]
[u'B', u'B', 0]
[u'B', u'B', 1]
[u'B', u'BB', 0]
[u'B', u'BB', 1]
[u'B', u'BB', 2]
[u'B', u'BBB', 0]
[u'B', u'BBB', 1]
[u'B', u'BBB', 2]
[u'B', u'BBB', 3]
[u'C', u'C', 0]
[u'C', u'C', 1]
[u'C', u'CC', 0]
[u'C', u'CC', 1]
[u'C', u'CC', 2]
[u'C', u'CCC', 0]
[u'C', u'CCC', 1]
[u'C', u'CCC', 2]
[u'C', u'CCC', 3]
[u'D', u'D', 0]
[u'D', u'D', 1]
[u'D', u'DD', 0]
[u'D', u'DD', 1]
[u'D', u'DD', 2]
[u'D', u'DDD', 0]
[u'D', u'DDD', 1]
[u'D', u'DDD', 2]
[u'D', u'DDD', 3]
于 2013-02-23T05:44:28.867 回答