逆轉（或簡化）笛卡爾產品？

Question

爲了使事情更容易，但也更復雜，我試圖實現「組合/簡潔標籤」的概念，這些概念進一步擴展爲多個基本標籤形式。

在這種情況下，標籤包括（一個或多個）「子（多個）標籤」，由分號分隔：

food:fruit:apple:sour/sweet 

drink:coffee/tea:hot/cold 

wall/bike:painted:red/blue 

斜線指示「子標籤」互換性。 因此，解釋它們轉換爲這樣的：

food:fruit:apple:sour 
food:fruit:apple:sweet 

drink:coffee:hot 
drink:coffee:cold 
drink:tea:hot 
drink:tea:cold 

wall:painted:red 
wall:painted:blue 
bike:painted:red 
bike:painted:blue 

代碼中使用（不完美，但工程）：

import itertools 

def slash_split_tag(tag): 
    if not '/' in tag: 
     return tag 
    subtags = tag.split(':') 
    pattern, v_pattern =(),() 
    for subtag in subtags: 
     if '/' in subtag: 
      pattern += (None,) 
      v_pattern += (tuple(subtag.split('/')),) 
     else: 
      pattern += (subtag,) 
    def merge_pattern_and_product(pattern, product): 
     ret = list(pattern) 
     for e in product: 
      ret[ret.index(None)] = e 
     return ret 
    CartesianProduct = tuple(itertools.product(*v_pattern)) # http://stackoverflow.com/a/170248 
    return [ ':'.join(merge_pattern_and_product(pattern, product)) for product in CartesianProduct ] 

#=============================================================================== 
# T E S T 
#=============================================================================== 

for tag in slash_split_tag('drink:coffee/tea:hot/cold'): 
    print tag 
print 
for tag in slash_split_tag('A1/A2:B1/B2/B3:C1/C2:D1/D2/D3/D4/EE'): 
    print tag 

問：我怎麼可能恢復這一進程？出於可讀性的原因，我需要這個。

Answer 1

這裏是在這樣一個功能簡單的，第一遍嘗試：

def compress_list(alist): 
    """Compress a list of colon-separated strings into a more compact 
    representation. 
    """ 
    components = [ss.split(':') for ss in alist] 

    # Check that every string in the supplied list has the same number of tags 
    tag_counts = [len(cc) for cc in components] 
    if len(set(tag_counts)) != 1: 
     raise ValueError("Not all of the strings have the same number of tags") 

    # For each component, gather a list of all the applicable tags. The set 
    # at index k of tag_possibilities is all the possibilities for the 
    # kth tag 
    tag_possibilities = list() 
    for tag_idx in range(tag_counts[0]): 
     tag_possibilities.append(set(cc[tag_idx] for cc in components)) 

    # Now take the list of tags, and turn them into slash-separated strings 
    tag_possibilities_strs = ['/'.join(tt) for tt in tag_possibilities] 

    # Finally, stitch this together with colons 
    return ':'.join(tag_possibilities_strs) 

希望這些意見在解釋它是如何工作充分。幾個注意事項，但是：

• 它沒有做任何事情，理智比如逃逸反斜槓如果發現他們的標籤列表。

• 這不識別是否存在更微妙的分割，或者它是否得到不完整的標籤列表。考慮下面這個例子：

  fish:cheese:red 
  chips:cheese:red 
  fish:chalk:red 
  

  它不會意識到，只有cheese既有fish和chips，而會崩潰這fish/chips:cheese/chalk:red。

• 成品字符串中標籤的順序是隨機的（或者至少，我不認爲它與給定列表中字符串的順序有關）。如果這很重要，您可以在加入斜線之前對tt進行排序。

測試在這個問題給出的三個列表似乎工作，但正如我所說，順序可能會有所不同初始字符串：

food:fruit:apple:sweet/sour 
drink:tea/coffee:hot/cold 
wall/bike:painted:blue/red