Question

我如何制作：

("A","b"),("b","C"),("A",),("b",),("C",)

这

("A","b","C")

或许可以使用itertool，我似乎无法找到适合的功能。

更新

请注意（“A”，“C”）不在预期的输出中，因为我希望子集包含彼此连续的成员。

另一个例子：

subset(("A","b","C","D"))

应该产生：

("A","b","C"),
("b","C","D"),
("A","b"),
("b","C"),
("C","D"),
("A",),
("b",),
("C",),
("D",)

Answer 1

您可以使用滚动窗口配方：

def window(seq, n=2):
    """
    Returns a sliding window (of width n) over data from the sequence
    s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...
    """
    for i in range(len(seq)-n+1):
        yield tuple(seq[i:i+n])

def shrinking_window(seq):
    for i in range(len(seq)-1, 0, -1):
        yield from window(seq, i)

print(list(shrinking_window('AbC')))
# [('A', 'b'), ('b', 'C'), ('A',), ('b',), ('C',)]
print(list(shrinking_window('AbCD')))
# [('A', 'b', 'C'), ('b', 'C', 'D'), ('A', 'b'), ('b', 'C'), ('C', 'D'), ('A',), ('b',), ('C',), ('D',)]

Answer 2

这是一种方式：

input_tuple = ("A", "b", "C", "d")
output_tuples = []
for subtuple_length in reversed(xrange(1, len(input_tuple))):
    for start_index in xrange(0, (len(input_tuple) + 1 - subtuple_length)):
        output_tuples.append(input_tuple[start_index:start_index + subtuple_length])

构建连续子元素的列表 - 您也可以print，或者yield将它们作为生成器或其他任何元素。它是输入元组长度的二次方，但是你的预期结果集的大小也是如此，所以我不确定是否有办法解决这个问题。

Answer 3

subset(("A","b","C","D"))

应该产生：

("A","b","C"), 
("b","C","D"),
("A","b"),
("b","C"),
("C","D"),
("A",),
("b",),
("C",),
("D",)

滑动窗户很难。迭代地缩小或增长窗口是双倍的。

首先列出要解决的步骤，然后创建一个遵循这些步骤的函数：

从最大窗口大小开始（比示例代码中的总长度少一个）。
然后计算覆盖数据集所需的窗口数。
然后对于每个窗口，您可以重复使用该数字作为起始索引，并且需要将起始索引添加到窗口大小以确定每个窗口停止的位置：

结果函数：

def subset(data):
    total_length = len(data)
    for window_length in range(total_length - 1, 0, -1): # biggest first
        n_windows = total_length - window_length + 1
        for each_window in range(n_windows):
            start = each_window
            stop = start + window_length
            yield data[start:stop]

示例数据：

data = ("A","b","C","D")

现在，在subset上调用data会返回一个生成器，如果我们传递给list，则会生成结果：

>>> subset(data)
<generator object subset at 0x7fbc3d7f3570>
>>> list(subset(data))
[('A', 'b', 'C'), ('b', 'C', 'D'), ('A', 'b'), ('b', 'C'), ('C', 'D'), ('A',), ('b',), ('C',), ('D',)]

Deque解决方案：

我对使用deque（来自集合模块）滚动窗口的想法很着迷，并决定证明这一点：

import collections
import pprint

def shrinking_windows(iterable):
    '''
    Given an ordered iterable (meaningless for unordered ones)
    return a list of tuples representing each possible set
    of consecutive items from the original list. e.g.
    shrinking_windows(['A', 'b', 'c']) returns 
    [('A', 'b', 'c'), ('A', 'b'), ('b', 'c') ...] but not ('A', 'c')
    '''
    window_generator = range(len(iterable), 0, -1)
    results = []
    for window in window_generator:
        d = collections.deque((), maxlen=window)
        for i in iterable:
            d.append(i)
            if len(d) == window:
                results.append(tuple(d))
    return results

pprint.pprint(shrinking_windows('AbCd'))

很好地返回：

[('A', 'b', 'C', 'd'),
 ('A', 'b', 'C'),
 ('b', 'C', 'd'),
 ('A', 'b'),
 ('b', 'C'),
 ('C', 'd'),
 ('A',),
 ('b',),
 ('C',),
 ('d',)]

元组的子集

3 个答案:

Deque解决方案：