从字符串更改 int 和 string

【问题标题】：Changing the int and string from a string从字符串更改 int 和 string
【发布时间】：2020-10-02 17:30:42
【问题描述】：

format_address 函数将地址字符串的一部分分离为新字符串：house_number 和 street_name，并返回："house number X on street named Y"。

输入字符串的格式为：数字门牌号，后跟街道名称，可以包含数字，但不能单独包含数字，可以是几个单词。例如，"123 Main Street"、"1001 1st Ave" 或 "55 North Center Drive"。

填补空白以完成此功能。

def format_address(address_string):
  # Declare variables
  house_number=' '
  street_name=" "

  # Separate the address string into parts
  x=address_string.split(" ")
  # Traverse through the address parts
  for y in x:
    if(y.isdigit()):
      house_number=y
    else:
      street_name+=y
      street_name+=' '
    # Determine if the address part is the
    # house number or part of the street name
 # Does anything else need to be done 
  # before returning the result?
  
  # Return the formatted string  
    return "house number {} on street named {}".format(house_number,street_name)

print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"

但它显示输出为：

house number 123 on street named  
house number 1001 on street named  
house number 55 on street named

【问题讨论】：

  • @TomServo 我认为使用 {} 语法它只会从左到右填充它

标签： python string function formatting integer


【解决方案1】：

当您使用str.split() 时，您可以选择要将字符串拆分为多少段（只要不超过最大值）。在您的情况下，字符串应该只拆分一次，以将门牌号与街道名称分开。此外，您可以格式化字符串：

def format_address(address_string):
    num, st = address_string.split(' ',1)
    return f"house number {num} on street named {st}"

print(format_address("123 Main Street"))

输出：

house number 123 on street named Main Street

【讨论】：

  • 非常好的答案，不仅仅是直接修复明显的错误。
【解决方案2】：

在我看来，您的 return 语句发生在 for 循环内，这意味着您在第一项之后就退出了循环！但到那时你还没有确定街道名称。

只需将return ... 移回一个标签，它应该可以正常工作。

【讨论】：

  • 好地方！这是一个很好的例子，说明为什么让您的 cmets 与他们评论的代码保持一致是一种很好的做法；很难在视觉上跟随压痕的垂直线，因为 cmets 到处都是。
【解决方案3】： 
def format_address(address_string):
  # Declare variables
  house_number = 0
  street_name = []
  # Separate the address string into parts
  address = address_string.split()
  # Traverse through the address parts
  for item in address:
    if item.isnumeric():
      house_number = item
    else:
      street_name.append(item)
    # Determine if the address part is the
    # house number or part of the street name

  # Does anything else need to be done 
  # before returning the result?
  
  # Return the formatted string  
  return "house number {} on street named {}".format(house_number, " ".join(street_name))

print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"

print(format_address("1001 1st Ave"))
# Should print: "house number 1001 on street named 1st Ave"

print(format_address("55 North Center Drive"))
# Should print "house number 55 on street named North Center Drive"

【讨论】：

    【解决方案4】： 
    def format_address(address_string):
  # Declare variables
  number = ''
  street = ''
  # Separate the address string into parts
  address_string = address_string.split()
  # Traverse through the address parts
  for add in address_string:
      # Determine if the address part is the
      # house number or part of the street name
      if add.isdigit():
          number += add 
      # Does anything else need to be done 
      # before returning the result?
      else :
          street += " " + add
      # Return the formatted string  
      return "house number {} on street named {}".format(number, street)

print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"

print(format_address("1001 1st Ave"))
# Should print: "house number 1001 on street named 1st Ave"

print(format_address("55 North Center Drive"))
# Should print "house number 55 on street named North Center Drive"

    【讨论】：

      【解决方案5】： 
      def format_address(address_string):
# Declare variables
house_number=''
street_name=''
# Separate the address string into parts
address = address_string.split()
# Traverse through the address parts
for add in address:
    # Determine if the address part is the
    # house number or part of the street name
    if add.isnumeric():
        house_number = int(add)
    else:
        street_name += add + " "
        # Does anything else need to be done 
        # before returning the result?
    # Return the formatted string
return "house number {} on street named {}".format(house_number,street_name.strip())

      【讨论】：

      • 你好 Ghyxen。欢迎来到堆栈溢出。虽然此代码可能会回答问题，但提供有关此代码为何和/或如何回答问题的额外上下文可提高其长期价值。 How to Answer。此外，这个问题有很多答案。您能否解释一下为什么我应该使用这个而不是现有答案之一。亲切的问候。
      【解决方案6】：

      那是因为你的返回不是在好的压痕中。它在第一个循环中被调用，因此您的 boucle for 只进行了一次迭代。

      def format_address(address_string):
  # Declare variables
  house_number=' '
  street_name=" "

  # Separate the address string into parts
  x=address_string.split(" ")
  # Traverse through the address parts
  for y in x:
    if(y.isdigit()):
      house_number=y
    else:
      street_name+=y
      street_name+=' '
    # Determine if the address part is the
    # house number or part of the street name
 # Does anything else need to be done 
  # before returning the result?

  # Return the formatted string  
  return "house number {} on street named {}".format(house_number,street_name)

print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"```

      【讨论】：

        【解决方案7】：

        或者你可以这样做！

        def format_address(address_string):
  # Declare variables
  space = address_string.find(" ")
  house_number = address_string[:space]
  street_name = address_string[space:]
  return "house number {} on street named{}".format(house_number, street_name)

print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"

print(format_address("1001 1st Ave"))
# Should print: "house number 1001 on street named 1st Ave"

print(format_address("55 North Center Drive"))
# Should print "house number 55 on street named North Center Drive"

        【讨论】：

          【解决方案8】： 
          def format_address(address_string):
  house_number =''
  street_name =''

  # Declare variables

  # Separate the address string into parts
  spi = address_string.split()
  # Traverse through the address parts
  for ele in spi:
    if ele.isdigit():
      house_number = ele
    else:
      street_name += ele
      street_name += ' '
`

`# Return the formatted string`
  `return "house number {} on street named {}".format(house_number, street_name)`


print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"

print(format_address("1001 1st Ave"))
# Should print: "house number 1001 on street named 1st Ave"

print(format_address("55 North Center Drive"))
# Should print "house number 55 on street named North Center Drive"`

          【讨论】：

            【解决方案9】：

            这对我有用

            def format_address(address_string):
    # Declare variables
    hn =""
    ha =""
  # Separate the address string into parts
    address_string = address_string.split()
  # Traverse through the address parts
    for add in address_string:
        if add.isnumeric():
            hn += add
        else:
            ha += add + " "
    # Determine if the address part is the
    # house number or part of the street name
  # Does anything else need to be done
  # before returning the result?

  # Return the formatted string
    return "house number {} on street named {}".format(hn,ha)

print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"

print(format_address("1001 1st Ave"))
# Should print: "house number 1001 on street named 1st Ave"

print(format_address("55 North Center Drive"))
# Should print "house number 55 on street named North Center Drive"

            【讨论】：

              【解决方案10】： 
              def format_address(address_string):
# Declare variables
  house_number=[]
  street_name=[]
  addresses=[]
# Separate the address string into parts
  addresses = address_string.split()
# Traverse through the address parts
  for address in addresses:
# Determine if the address part is the house number or part of the street name
    
    if address.isnumeric() == True:
       house_number.append(address)
    else:
       street_name.append(address)
  
  house=" ".join(house_number)
  street=" ".join(street_name)
# Return the formatted string  
  return "house number {} on street named {}".format(house,street)

print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"

print(format_address("1001 1st Ave"))
# Should print: "house number 1001 on street named 1st Ave"

print(format_address("55 North Center Drive"))
# Should print "house number 55 on street named North Center Drive"

              【讨论】：

                【解决方案11】： 
                def format_address(address_string):
  words = address_string.split()
  return "house number {} on street named {}".format(words[0],' '.join(words[1:]))

                【讨论】：

                  【解决方案12】： 
                  def format_address(address_string):
  # Declare variables
  addres_list = []
  name = " "
  # Separate the address string into parts
  addres_list = address_string.split(" ")
  # Traverse through the address parts
  number = addres_list[0]
  name = name.join(addres_list[1:])
  # Return the formatted string  
  return "house number {} on street named {}".format(number,name)

                  【讨论】：

                    【解决方案13】： 
                    def format_address(address_string):

  address = address_string.split(" ",1)
  house=address[0]
  street=address[1]
  return "house number {} on street named {}".format(house,street)

print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"

print(format_address("1001 1st Ave"))
# Should print: "house number 1001 on street named 1st Ave"

print(format_address("55 North Center Drive"))
# Should print "house number 55 on street named North Center Drive"

                    【讨论】：

                      【解决方案14】：

                      我在不使用数字的 if 语句的情况下解决它，我将列表分成 2 部分并将每个部分添加到适当的变量中。

                      def format_address(address_string):
      house_number = ''
      street_name = ''
      parts = address_string.split(' ',1)
      house_number += parts[0]
      street_name += parts[1]
    
    
      return "house number {} on street named {}".format(house_number,street_name)

                      【讨论】：

                        【解决方案15】： 
                        def format_address(address_string):
     # Declare variables
     house_number =' '
     street_name =" "

     # Separate the address string into parts
     x= address_string.split(" ")
      # Traverse through the address parts
      for y in x:
        if(y.isdigit()):
          house_number=y
        else:
          street_name+=y
          street_name+=' '
    # Determine if the address part is the
    # house number or part of the street name
    # Does anything else need to be done
    # before returning the result?

    # Return the formatted string
    return "house number {} on street named {}".format(house_number, street_name)

    print(format_address("123 Main Street"))
    # Should print: "house number 123 on street named Main Street"

    print(format_address("1001 1st Ave"))
    # Should print: "house number 1001 on stre
    print(format_address("55 North Center Drive"))
    # Should print "house number 55 on street named North Center Drive"

                        【讨论】：

                        • 欢迎来到 StackOverflow。虽然此代码可能会回答问题，但提供有关 如何 和/或 为什么 解决问题的附加上下文将提高​​答案的长期价值。
                        【解决方案16】： 
                        def format_address(address_string):
  street_name = ""
  houseandstreet = address_string.split()
  for i in houseandstreet:
    if i.isdigit():
      house_number = i
    else:
      street_name = street_name + i +" "   
  return "house number {} on street named {}".format(house_number,street_name)

                        【讨论】：

                        • 这里的 house_number 永远是个位数
                        • @arunna house_number 并不总是一个数字，因为我在第 3 行使用了 split() 方法。
                        【解决方案17】：

                        我也会参与进来。 有时数字不在开头，所以这里的最短答案在这种情况下不起作用。

                        这是我的版本：

                        def format_address(address_string):
  # Declare variables
    house_nr = ""
    street = ""
  # Separate the address string into parts
    addy_parts = address_string.split()
  # Traverse through the address parts
    for p in addy_parts:
    # Determine if the address part is the
    # house number or part of the street name
        if p.isnumeric() and house_nr == "":
            house_nr = p
        else:
            street += p + " "
            
  # Does anything else need to be done 
  # before returning the result?
  
  # Return the formatted string  
    return "house number {} on street named {street}".format(house_nr, street = street.rstrip())

print(format_address("Main Street 123"))
# Should print: "house number 123 on street named Main Street"

print(format_address("1001 1st Ave Apt 20"))
# Should print: "house number 1001 on street named 1st Ave"

print(format_address("55 North Center Drive"))
# Should print "house number 55 on street named North Center Drive"

                        它还将考虑地址中超过 1 个数字字，例如当给出一个公寓号码，或者一个人给出的任何愚蠢的条目时。

                        我知道这不是这个 Q 来自 Google 课程的原始问题的一部分，但它有助于理解它是如何工作的。

                        如果您只想要简短的答案，请使用@Ann Zen 的答案，很好。

                        【讨论】：

                          【解决方案18】： 
                          def format_address(address_string):
  # Declare variables
  z = address_string.split(" ")
  house_number = z[0]
  street_name = z[1:]
  # Return the formatted string  
  return "house number {} on street named {}".format(house_number, " ".join(street_name))

print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"

print(format_address("1001 1st Ave"))
# Should print: "house number 1001 on street named 1st Ave"

print(format_address("55 North Center Drive"))
# Should print "house number 55 on street named North Center Drive"

                          【讨论】：

                            【解决方案19】：

                            x[0] 将是门牌号，而 " ".join(x[1:]) 将是街道名称。这是可行的，因为数字总是在字符串中排在第一位。

                            def format_address(address_string):
    x = address_string.split()
    return "house number {} on street named {}".format(x[0], " ".join(x[1:]))

print(format_address("123 Main Street"))
print(format_address("1001 1st Ave"))
print(format_address("55 North Center Drive"))

                            【讨论】：

                            • 正如目前所写，您的答案尚不清楚。请edit 添加其他详细信息，以帮助其他人了解这如何解决所提出的问题。你可以找到更多关于如何写好答案的信息in the help center。
                            【解决方案20】：

                            定义格式地址（地址字符串）：

                            return "house number {} on street named {}".format(address_string.split()[0], " ".join(address_string.split()[1:]))

print(format_address("123 Main Street"))
# Should print: "house number 123 on street named Main Street"

print(format_address("1001 1st Ave"))
# Should print: "house number 1001 on street named 1st Ave"

print(format_address("55 North Center Drive"))
# Should print "house number 55 on street named North Center Drive"

                            【讨论】：

                            猜你喜欢
                            • 1970-01-01
                            • 1970-01-01
                            • 1970-01-01
                            • 1970-01-01
                            • 1970-01-01
                            • 1970-01-01
                            • 2017-06-15
                            • 1970-01-01
                            • 2011-03-05
                            相关资源
                            最近更新 更多
                            热门标签
                            Java Python linux javascript Mysql C# Docker 算法 前端 SpringBoot Redis Vue spring 设计模式 .net core .net kubernetes c++ 数据库 数据结构 大数据 js 机器学习 微服务 Android Go 程序员 面试 JVM ASP.net core 云原生 人工智能 后端 PHP git CSS golang k8s Nginx Django mybatis 深度学习 多线程 React 架构 devops 爬虫 云计算 Spring Boot LeetCode