合併基於模板

Question

列表的多個列表我有3個DB調用返回同一個名稱，代碼元組的元組和計數，像這樣：

year = (('Fanklin Grand Isle', '5560', 1), ('Windham', '3457', 1))

month = (('Fanklin Grand Isle', '5560', 1), ('Chittendon', '3367', 1))

week = (('Fanklin Grand Isle', '5560', 1), ('Chittendon', '3367', 1))

我需要將這些全部合併在一起，以便他們擁有姓名，代碼並計算每年，每月和每週的計數。

我的問題是，如果沒有記錄，我需要插入名稱和代碼和計數的0值。最終產品應該是這樣的：

result = (('Windham', '0905', 1, 1, 0), ('Windsor Windham', '7852', 0, 0, 0), ('Washington', '3292', 0, 0, 0), 
      ('Orleans Essex', '44072', 1, 1, 0), ('Addison', '2853', 0), ('Bennington', '3778', 0), 
      ('Fanklin Grand Isle', '5560', 0, 0 0), ('Caledonia', '1992', 0, 0, 0), 
      ('Rutland', '2395', 1, 0, 0), ('Chittendon', '3367', 1, 1, 0), ('Lamoille', '5229',0, 0 0)) 

我試圖嵌套循環來檢查，如果名字出現在DB調用和模板。如果如果將DB值附加到列表中，如果不附加0

i = 0 
for p in newlist: 
    try: 
     if p[0] == mlist[i][0]: 
      print("HERE: {} {}".format(p[0], mlist[i][0])) 
      p.append(mlist[i][-1]) 
      i += 1 
     else: 
      p.append(0) 
    except IndexError: 
     continue 

這是追加DB值但不是零。我相信必須有更好的方式來做到這一點，並讓它實際工作。

編輯

這是根據接收到的答案更新的代碼。對我來說，它仍然是一個0

數據替換每個year值：

year = (('Fanklin Grand Isle', '5560', 1), ('Windham', '0905', 0), ('Windsor Windham', '7852', 0), ('Washington', '3292', 0), ('Orleans Essex', '44072', 0), ('Chittendon', '18028633367', 1), ('Addison', '12853', 0), ('Bennington', '3778', 0), ('Caledonia', '11992', 0), ('Rutland', '1895', 0), ('Chittendon', '18367', 0), ('Lamoille', '1809', 0), ('Windham', '180905', 0), ('Windsor Windham', '180852', 0), ('Waston', '18022623292', 0), ('Orleans Essex', '18072', 0), ('Addison', '1853', 0), ('Bennington', '1778', 0), ('Fanklin Grand Isle', '18560', 0), ('Caledonia', '180292', 0), ('Rutland', '195', 0), ('Lamoille', '18229', 0)) 

month = (('Fanklin Grand Isle', '5560', 1), ('Chittendon', '18028633367', 1)) 

week = (('Fanklin Grand Isle', '5560', 1), ('Chittendon', '18367', 1)) 

代碼：

from collections import defaultdict 
joined_data = defaultdict([0, 0, 0].copy) 

for entry in year: 
    # we create the default entry by calling the defaultdict with a key 
    # and immediately grab the newly created list 
    count = joined_data[(entry[0],entry[1])] 
    # we swap *inplace* the value given by the DB query 
    count[0] = entry[2] 

# rinse and repeat with the rest of the data 
for entry in month: 
    count = joined_data[(entry[0], entry[1])] 
    count[1] = entry[2] 

for entry in week: 
    count = joined_data[(entry[0], entry[1])] 
    count[2] = entry[2] 

# Finally we format the data to the required format 
result = tuple(key+tuple(value) for key,value in joined_data.items()) 
print(result) 

結果：

(('Fanklin Grand Isle', '5560', 0, 1, 1), ('Windham', '0905', 0, 0, 0), ('Windsor Windham', '7852', 0, 0, 0), ('Washington', '3292', 0, 0, 0), ('Orleans Essex', '1072', 0, 0, 0), ('Chittendon', '13367', 0, 1, 1), ('Addison', '2853', 0, 0, 0), ('Bennington', '1878', 0, 0, 0), ('Caledonia', '1992', 0, 0, 0), ('Rutland', '2395', 0, 0, 0), ('Lamoille', '5229', 0, 0, 0))

Answer 1

這裏有一個方式使用defaultdict以避免關心缺少的條目來處理您的問題：

year = (('Fanklin Grand Isle', '5560', 1), ('Windham', '3457', 1)) 
month = (('Fanklin Grand Isle', '5560', 1), ('Chittendon', '3367', 1)) 
week = (('Fanklin Grand Isle', '5560', 1), ('Chittendon', '3367', 1)) 

#I'm using a defaultdict to deal with the missing entries 
from collections import defaultdict 
joined_data = defaultdict([0,0,0].copy) 

for entry in year: 
    #we create the default entry by calling the defaultdict with a key 
    #and immediately grab the newly created list 
    count = joined_data[(entry[0],entry[1])] 
    #we swap *inplace* the value given by the DB query 
    count[0] = entry[2] 

#rinse and repeat with the rest of the data 
for entry in month: 
    count = joined_data[(entry[0],entry[1])] 
    count[1] = entry[2] 

for entry in week: 
    count = joined_data[(entry[0],entry[1])] 
    count[2] = entry[2] 

#Finally we format the data to the required format 
result = tuple(key+tuple(value) for key,value in joined_data.items()) 
print(result) 

輸出：

>>>(('Chittendon', '3367', 0, 1, 1), ('Fanklin Grand Isle', '5560', 1, 1, 1), ('Windham', '3457', 1, 0, 0)) 

Answer 2

不知道我理解作爲您的示例輸入，您試圖實現的目標是n OT真正符合你的輸出，但我認爲你可以使用列表理解來構造的結果，檢查這些項目是否在years，months和weeks名單，並增加了1或0分別爲：

>>> year = (('Fanklin Grand Isle', '5560', 1), ('Windham', '3457', 1)) 
>>> month = (('Fanklin Grand Isle', '5560', 1), ('Chittendon', '3367', 1)) 
>>> week = (('Fanklin Grand Isle', '5560', 1), ('Chittendon', '3367', 1)) 
>>> [(x[0], x[1], int(x in year), int(x in month), int(x in week)) for x in set(year + week + month)] 
[('Chittendon', '3367', 0, 1, 1), 
('Windham', '3457', 1, 0, 0), 
('Fanklin Grand Isle', '5560', 1, 1, 1)] 

如果這些罪名實際上可以從1是不同的，你應該先創建一些字典映射城市各自的年/月/週數，然後使用類似的列表理解如上：

>>> year_counts = {(name, code): count for (name, code, count) in year} 
>>> month_counts = {(name, code): count for (name, code, count) in month} 
>>> week_counts = {(name, code): count for (name, code, count) in week} 
>>> all_cities = [(name, code) for (name, code, count) in set(year + month + week)] 
>>> [(x[0], x[1], year_counts.get(x, 0), month_counts.get(x, 0), week_counts.get(x, 0)) for x in all_cities] 
[('Chittendon', '3367', 0, 1, 1), 
('Windham', '3457', 1, 0, 0), 
('Fanklin Grand Isle', '5560', 1, 1, 1)]