检查字符串是否在列表中，取决于最后两个字符 [英] Check if string in list, depending on last two characters

问题描述

设置

我正在使用Scrapy抓取房屋广告.每个广告我检索一个邮政编码，该邮政编码由四个数字和两个字母组成，例如1053ZM.

I am using Scrapy to scrape housing ads. Per ad I retrieve a postal code which consists of four numbers followed by 2 letters, e.g. 1053ZM.

我有一个Excel工作表，可通过以下方式将地区与邮政编码联系起来，

I have a excel sheet linking districts to postal codes in the following way,

district    postcode_min    postcode_max
   A           1011AB           1011BD
   A           1011BG           1011CE
   A           1011CH           1011CZ

因此，第二行指出范围为1011AB, 1011AC,..., 1011AZ, 1011BA,...,1011BD的邮政编码属于区A.

So, the second row states that postcodes ranging from 1011AB, 1011AC,..., 1011AZ, 1011BA,...,1011BD belong to district A.

实际列表包含1214行.

The actual list contains 1214 rows.

---

问题

我想使用其邮政编码和列表将每个广告与各自的地区进行匹配.

I'd like to match each ad with its respective district, using its postal code and the list.

我不确定什么是最好的方法以及如何做到这一点.

I am not sure what would be the best way to do this, and how to do this.

我想出了两种不同的方法:

I've come up with two different approaches:

1. 在postcode_min和postcode_max之间创建所有邮政编码，将所有邮政编码及其各自的地区分配给字典，以便随后使用循环进行匹配.

1. Create all postcodes between postcode_min and postcode_max, assign all postcodes and their respective districts to a dictionary to subsequently match using a loop.

即创建

 d = {'A': ['1011AB','1011AC',...,'1011BD',
            '1011BG','1011BH',...,'1011CE',
            '1011CH','1011CI',...,'1011CZ'],
      'B': [...],           
      }

然后

found = False
for distr in d.keys(): # loop over districts
     for code in d[distr]: # loop over district's postal codes
         if postal_code in code: # assign if ad's postal code in code                 
             district = distr
             found = True
             break
         else:
             district = 'unknown'
     if found:
         break

1. 让Python理解postcode_min和postcode_max之间存在一个范围，将范围及其各自的区域分配给字典，并使用循环进行匹配.

1. Make Python understand there is a range between the postcode_min and the postcode_max, assign ranges and their respective districts to a dictionary, and match using a loop.

即像

d = {'A': [range(1011AB,1011BD), range(1011BG,1011CE),range(1011CH,1011CZ)],
     'B': [...]
    }

然后

found = False
for distr in d.keys(): # loop over districts
     for range in d[distr]: # loop over district's ranges
         if postal_code in range: # assign if ad's postal code in range                 
             district = distr
             found = True
             break
         else:
             district = 'unknown'
     if found:
         break

---

问题

对于方法1:

• 如何创建所有邮政编码并将其分配给词典?

对于方法2:

我将range()用于说明目的，但是我知道range()不能像这样工作.

I used range() for explanatory purpose but I know range() does not work like this.

• 如上例所示，我需要什么才能有效地拥有range()?
• 如何正确遍历这些范围?

我认为我更喜欢方法2，但我很高兴与任何一个一起工作.或使用其他解决方案(如果有).

I think my preference lies with approach 2, but I am happy to work with either one. Or with another solution if you have one.

推荐答案

您可以像这样在excel中收集值

You can just collect the values in excel like this

d = {'A': ['1011AB', '1011BD', '1011BG', '1011CE',  '1011CH', '1011CZ'],
     'B': ['1061WB', '1061WB'],
     }

def is_in_postcode_range(current_postcode, min, max):
    return min <= current_postcode <= max

def get_district_by_post_code(postcode):
    for district, codes in d.items():
        first_code = codes[0]
        last_code = codes[-1]
        if is_in_postcode_range(postcode, first_code, last_code):
            if any(is_in_postcode_range(postcode, codes[i], codes[i+1]) for i in range(0, len(codes), 2)):
                return district
            else:
                return None

用法:

print get_district_by_post_code('1011AC'): A
print get_district_by_post_code('1011BE'): None
print get_district_by_post_code('1061WB'): B

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