Question

设置

在伦敦搜索住房广告，我在每个广告中获得一个单元素列表中的地址，例如

address=['Brockham Drive, Brixton SW2']

我有一本字典链接伦敦自治市和他们的地区，例如

boroughs={ 
'Barking_Dagenham':['Barking', ..., 'Rush Green'],
'Barnet':['Arkley', ..., 'Woodside Park'],
    ⋮
'Westminster':['Bayswater', ..., 'Westminster'],
}

<小时/> 的问题

我想检查地区名称是否在address中。如果该区位于address，那么我想创建变量district和borough，表明该区及其相应的行政区。

<小时/> 代码尝试

（1）

for bor in boroughs.keys(): # loop over boroughs
   for distr in boroughs[bor]: # loop over borough's districts
      if distr in address[0]: # assign if district in address
         district = distr
         borough = bor
         break
      else:
         district = 'unknown'
         borough = 'unknown'

（1）不起作用。也就是说，所有内容都标记为'unknown'。

我不确定我是否正确执行break，也不确定if distr in address[0]:是否是在迭代时检查匹配的正确方法。

（2）

for bor in boroughs.keys(): # loop over boroughs
   for distr in boroughs[bor]: # loop over borough's districts
      district = re.search(r'\b'distr'\b', address[0]):    
      borough = ?
      break
   else:
      district = 'unknown'
      borough = 'unknown'

使用（2），我不确定在使用'\ b'时如何正确迭代'bor'。当迭代产生正确的区域匹配时，不确定如何分配相应的区域。另外，不确定我是否应该使用（2）而不是（1）。

我应该使用哪种方法，以及如何让其中至少一种方法起作用？

Answer 1

Your code try #1 is correct, but missing one key element. You are only breaking out of the inner for loop, but then your code continues to loop through the outer for loop. Add a variable to check if it is found to break out of the outer for loop.

found = False

for bor in boroughs.keys(): # loop over boroughs
  for distr in boroughs[bor]: # loop over borough's districts
    if distr in address[0]: # assign if district in address
      district = distr
      borough = bor
      found = True
      break
    else:
      district = 'unknown'
      borough = 'unknown'
  if found:
    break

Python：正则表达式迭代以匹配列表元素

1 个答案: