Python 如何将列表对转换为元组对

如何通过使用简单编程（例如for loop）将包含对的列表转换为包含元组对的列表？x、 y=

我的代码：

def read_numbers():
    numbers = ['68,125', '113,69', '65,86', '108,149', '152,53', '78,90']
    numbers.split(',')
    x,y = tuple numbers
    return numbers

期望输出：

[(68,125), (113,69), (65,86), (108,149), (152,53), (78,90)]

---

尝试使用嵌套列表理解：

o = [tuple(int(y) for y in x.split(',')) for x in numbers]

[
    tuple(                              # Convert whatever is between these parentheses into a tuple
            int(y)                      # Make y an integer
            for y in                    # Where y is each element in
            x.split(",")                # x.split(","). Where x is a string and x.split(",") is a list
                                        # where the string is split into a list delimited by a comma.
    ) for x in numbers                  # x is each element in numbers
]

---

尝试使用嵌套列表理解：

o = [tuple(int(y) for y in x.split(',')) for x in numbers]

[
    tuple(                              # Convert whatever is between these parentheses into a tuple
            int(y)                      # Make y an integer
            for y in                    # Where y is each element in
            x.split(",")                # x.split(","). Where x is a string and x.split(",") is a list
                                        # where the string is split into a list delimited by a comma.
    ) for x in numbers                  # x is each element in numbers
]

---

只需使用列表理解。阅读更多关于它的信息

以下是列表理解的分解和解释（注释）：

o = [tuple(int(y) for y in x.split(',')) for x in numbers]

[
    tuple(                              # Convert whatever is between these parentheses into a tuple
            int(y)                      # Make y an integer
            for y in                    # Where y is each element in
            x.split(",")                # x.split(","). Where x is a string and x.split(",") is a list
                                        # where the string is split into a list delimited by a comma.
    ) for x in numbers                  # x is each element in numbers
]

---

但是，如果您只是为一个列表执行此操作，则无需创建函数。

只需使用列表理解即可。阅读更多关于它的信息

以下是列表理解的分解和解释（注释）：

o = [tuple(int(y) for y in x.split(',')) for x in numbers]

[
    tuple(                              # Convert whatever is between these parentheses into a tuple
            int(y)                      # Make y an integer
            for y in                    # Where y is each element in
            x.split(",")                # x.split(","). Where x is a string and x.split(",") is a list
                                        # where the string is split into a list delimited by a comma.
    ) for x in numbers                  # x is each element in numbers
]

但是，如果只针对一个列表执行此操作，则无需创建函数。

尝试以下操作：

def read_numbers():
    numbers = ['68,125', '113,69', '65,86', '108,149', '152,53', '78,90']
    final_list = []
    [final_list.append(tuple(int(test_str) for test_str in number.split(','))) for number in numbers]
    return final_list

试试这个：

def read_numbers():
    numbers = ['68,125', '113,69', '65,86', '108,149', '152,53', '78,90']
    final_list = []
    [final_list.append(tuple(int(test_str) for test_str in number.split(','))) for number in numbers]
    return final_list

---

请在网上搜索帐户：如果我在网上找到它，我不会在这里发布：）可能重复的请在网上搜索帐户：如果我在网上找到它，我不会在这里发布：）可能重复的谢谢！成功了！没有map（）方法，你怎么能做到这一点呢？谢谢！成功了！如果没有map（）方法，您如何做到这一点？您可以省略

元组

构造函数中的

[]

括号。无需构建

列表

，只需传递生成器表达式即可。您可以省略

元组

构造函数中的

[]

括号。无需构建

列表

，只需传递生成器表达式即可。您可以省略

元组

构造函数中的

[]

括号。无需构建

列表

，只需传递生成器表达式即可。您可以省略

元组

构造函数中的

[]

括号。无需构建

列表

，只需传递生成器表达式即可。