合并基于模板

Question

列表的多个列表我有3个DB调用返回同一个名称，代码元组的元组和计数，像这样：

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))

我需要将这些全部合并在一起，以便他们拥有姓名，代码并计算每年，每月和每周的计数。

我的问题是，如果没有记录，我需要插入名称和代码和计数的0值。最终产品应该是这样的：

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)) 

我试图嵌套循环来检查，如果名字出现在DB调用和模板。如果如果将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 

这是追加DB值但不是零。我相信必须有更好的方式来做到这一点，并让它实际工作。

编辑

这是根据接收到的答案更新的代码。对我来说，它仍然是一个0

数据替换每个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)) 

代码：

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) 

结果：

(('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

这里有一个方式使用defaultdict以避免关心缺少的条目来处理您的问题：

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) 

输出：

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

Answer 2

不知道我理解作为您的示例输入，您试图实现的目标是n OT真正符合你的输出，但我认为你可以使用列表理解来构造的结果，检查这些项目是否在years，months和weeks名单，并增加了1或0分别为：

>>> 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)] 

如果这些罪名实际上可以从1是不同的，你应该先创建一些字典映射城市各自的年/月/周数，然后使用类似的列表理解如上：

>>> 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)]