如何用空格分割字符串的字符，然后用特殊字符和数字分割列表的结果元素，然后再次加入它们？

Question

所以，我想做的是将字符串中的一些单词转换为字典中的相应单词，然后按原样 is.For 示例输入为：

standarisationn("well-2-34 2   @$%23beach bend com")

我希望输出为：

"well-2-34 2 @$%23bch bnd com"

我使用的代码是：

def standarisationn(addr):
a=re.sub(',', ' ', addr)
lookp_dict = {"allee":"ale","alley":"ale","ally":"ale","aly":"ale",
              "arcade":"arc",
               "apartment":"apt","aprtmnt":"apt","aptmnt":"apt",
               "av":"ave","aven":"ave","avenu":"ave","avenue":"ave","avn":"ave","avnue":"ave",
              "beach":"bch",
              "bend":"bnd",
              "blfs":"blf","bluf":"blf","bluff":"blf","bluffs":"blf",
              "boul":"blvd","boulevard":"blvd","boulv":"blvd",
              "bottm":"bot","bottom":"bot",
              "branch":"br","brnch":"br",
              "brdge":"brg","bridge":"brg",
              "bypa":"byp","bypas":"byp","bypass":"byp","byps":"byp",
              "camp":"cmp",
              "canyn":"cny","canyon":"cny","cnyn":"cny",
              "southwest":"sw" ,"northwest":"nw"}

temp=re.findall(r"[A-Za-z0-9]+|\S", a)
print(temp)
res = []
for wrd in temp:
     res.append(lookp_dict.get(wrd,wrd))
res = ' '.join(res)
return str(res)

但它给出了错误的输出：

'well - 2 - 34 2 @ $ % 23beach bnd com'

有太多 space，甚至没有将“beach”转换为“bch”。所以，这就是我认为的 issue.What 首先将字符串拆分为 spaces 然后用特殊字符和数字拆分结果元素并使用字典，然后首先用没有 space 的特殊字符连接分隔的字符串，然后用 space.Can 连接所有列表任何人都建议如何去关于这个或任何更好的方法？

Answer 1

您可以使用字典的键构建正则表达式，确保它们不包含在另一个词中（即不直接在字母之前或之后）：

import re
def standarisationn(addr):
    addr = re.sub(r'(,|\s+)', " ", addr)
    lookp_dict = {"allee":"ale","alley":"ale","ally":"ale","aly":"ale",
                "arcade":"arc",
                "apartment":"apt","aprtmnt":"apt","aptmnt":"apt",
                "av":"ave","aven":"ave","avenu":"ave","avenue":"ave","avn":"ave","avnue":"ave",
                "beach":"bch",
                "bend":"bnd",
                "blfs":"blf","bluf":"blf","bluff":"blf","bluffs":"blf",
                "boul":"blvd","boulevard":"blvd","boulv":"blvd",
                "bottm":"bot","bottom":"bot",
                "branch":"br","brnch":"br",
                "brdge":"brg","bridge":"brg",
                "bypa":"byp","bypas":"byp","bypass":"byp","byps":"byp",
                "camp":"cmp",
                "canyn":"cny","canyon":"cny","cnyn":"cny",
                "southwest":"sw" ,"northwest":"nw"}

    for wrd in lookp_dict:
        addr = re.sub(rf'(?:^|(?<=[^a-zA-Z])){wrd}(?=[^a-zA-Z]|$)', lookp_dict[wrd], addr)
    return addr

print(standarisationn("well-2-34 2   @$%23beach bend com"))

表达式由三部分组成：

^匹配字符串的开头
(?<=[^a-zA-Z]) 是后视（即非捕获表达式），检查前面的字符是否为字母
{wrd} 是你字典的关键字
(?=[^a-zA-Z]|$) 是先行（即非捕获表达式），检查后面的字符是字母还是字符串的结尾

输出：

well-2-34 2 @$%23bch bnd com

编辑：如果将循环替换为：

，则可以编译整个表达式并仅使用 re.sub 一次

repl_pattern = re.compile(rf"(?:^|(?<=[^a-zA-Z]))({'|'.join(lookp_dict.keys())})(?=([^a-zA-Z]|$))")
addr = re.sub(repl_pattern, lambda x: lookp_dict[x.group(1)], addr)

如果你的字典变大，这应该会快得多，因为我们用你所有的字典键构建了一个表达式：

({'|'.join(lookp_dict.keys())}) 被解释为 (allee|alley|...
re.sub 中的 lambda 函数用 lookp_dict 中的相应值替换匹配元素（有关此的更多详细信息，请参见示例 this link）

如何用空格分割字符串的字符，然后用特殊字符和数字分割列表的结果元素，然后再次加入它们？

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