開箱元組

Question

列表我怎樣才能解開下面的列表

[('1', 'GENERAL', '1'), ('1.1', 'RELATED DOCUMENTS', '1'), ('1.2', 'SUMMARY', '1'), ('1.3', 'DEFINITIONS', '1'), ('1.4', 'INFORMATIONAL SUBMITTALS', '2'), ('1.5', 'GENERAL COORDINATION PROCEDURES', '2'), ('1.6', 'COORDINATION DRAWINGS', '3'), ('1.7', 'REQUESTS FOR INFORMATION (RFIs)', '4'), ('1.8', 'PROJECT MEETINGS', '6')] 
[[('2', 'PRODUCTS – NOT APPLICABLE', '10')]] 

從其他文章中，我試圖解決方案。

Part, Title, Page = zip(*text_good[0]) 

但得到的錯誤

too many values to unpack (expected 3) 

而且我也試過

Part1[a].append(Part for Part, Title, Page in text_good[0]) 
Part2[a].append(Part for Part, Title, Page in text_good[1]) 
Part3[a].append(Part for Part, Title, Page in text_good[2]) 

但是，這似乎是在內存中返回一個點，因爲我收到一個錯誤我無法打開陣列說明這是不可取的。

感謝

更新： 分配text_good的

for i in range(0, len(text_between_parts)): 
    text_good[i].append(re.findall(r'\s*(\b\d+(?:[.]\d+)?)\W+\s*(.*?)\s*(\b\d+\b)', text_between_parts[i])) 

更新2：當我做text_good [0]我得到

[[('1', 'GENERAL', '1'), ('1.1', 'RELATED DOCUMENTS', '1'), ('1.2', 'SUMMARY', '1'), ('1.3', 'DEFINITIONS', '1'), ('1.4', 'INFORMATIONAL SUBMITTALS', '2'), ('1.5', 'GENERAL COORDINATION PROCEDURES', '2'), ('1.6', 'COORDINATION DRAWINGS', '3'), ('1.7', 'REQUESTS FOR INFORMATION (RFIs)', '4'), ('1.8', 'PROJECT MEETINGS', '6')]] 

當我做text_good [0] [ 0]我得到

[('1', 'GENERAL', '1'), ('1.1', 'RELATED DOCUMENTS', '1'), ('1.2', 'SUMMARY', '1'), ('1.3', 'DEFINITIONS', '1'), ('1.4', 'INFORMATIONAL SUBMITTALS', '2'), ('1.5', 'GENERAL COORDINATION PROCEDURES', '2'), ('1.6', 'COORDINATION DRAWINGS', '3'), ('1.7', 'REQUESTS FOR INFORMATION (RFIs)', '4'), ('1.8', 'PROJECT MEETINGS', '6')] 

當我做text_good [0]時注意額外的括號。

Answer 1

好吧，我想我們首先需要在這裏做一點澄清。我在完全名單是什麼，所以我會做如下假設（如果這些假設都錯了，請讓我知道這樣我就可以解決這些問題）有點糊塗：

text_good = [[('1', 'GENERAL', '1'), ('1.1', 'RELATED DOCUMENTS', '1'), ('1.2', 'SUMMARY', '1'), ('1.3', 'DEFINITIONS', '1'), ('1.4', 'INFORMATIONAL SUBMITTALS', '2'), ('1.5', 'GENERAL COORDINATION PROCEDURES', '2'), ('1.6', 'COORDINATION DRAWINGS', '3'), ('1.7', 'REQUESTS FOR INFORMATION (RFIs)', '4'), ('1.8', 'PROJECT MEETINGS', '6')], [('2', 'PRODUCTS - NOT APPLICABLE', '10')]] 

現在在哪裏，如果我做text_good[0]我得到：

[('1', 'GENERAL', '1'), 
('1.1', 'RELATED DOCUMENTS', '1'), 
('1.2', 'SUMMARY', '1'), 
('1.3', 'DEFINITIONS', '1'), 
('1.4', 'INFORMATIONAL SUBMITTALS', '2'), 
('1.5', 'GENERAL COORDINATION PROCEDURES', '2'), 
('1.6', 'COORDINATION DRAWINGS', '3'), 
('1.7', 'REQUESTS FOR INFORMATION (RFIs)', '4'), 
('1.8', 'PROJECT MEETINGS', '6')] 

和text_good[1]是：

[('2', 'PRODUCTS - NOT APPLICABLE', '10')] 

對我來說，這似乎是你有一個元組的列表，其中('1', 'GENERAL', '1')將按照該順序對應於Part, Title, Page。

那麼，如果這是你需要可以做這樣的事情的情況下：

Parts, Title, Page = zip(*[t for l in text_good for t in l]) 

凡在這種情況下，您可以：

print Parts # ('1', '1.1', '1.2', '1.3', '1.4', '1.5', '1.6', '1.7', '1.8', '2') 
print Title # ('GENERAL', 
      # 'RELATED DOCUMENTS', 
      # 'SUMMARY', 
      # 'DEFINITIONS', 
      # 'INFORMATIONAL SUBMITTALS', 
      # 'GENERAL COORDINATION PROCEDURES', 
      # 'COORDINATION DRAWINGS', 
      # 'REQUESTS FOR INFORMATION (RFIs)', 
      # 'PROJECT MEETINGS', 
      # 'PRODUCTS - NOT APPLICABLE') 

print Page # ('1', '1', '1', '1', '2', '2', '3', '4', '6', '10') 

最後編輯： 因爲@JStuff有list of lists of lists of tuples，我們技術上需要3個循環才能提取他想要的定義。

Parts, Title, Page = [t for l in text_good for ll in l for t in ll] # Yay for list comprehension?