Python 将字符串列表转换为元组

我有以下字符串列表：

['(39.2947500000, -76.6565600000)', '(39.3423900000, -76.5698300000)', '(39.3199500000, -76.6222000000)', '(39.2533200000, -76.6263600000)', '(39.3068100000, -76.6549700000)', '(39.2937500000, -76.6233700000)', '(39.3146700000, -76.6425300000)', '(39.3073300000, -76.6015900000)', '(39.2451900000, -76.6336400000)', '(39.3283000000, -76.5893200000)', '(39.3215400000, -76.6736800000)', '(39.3010000000, -76.5977400000)', '(39.3122600000, -76.6194200000)', '(39.3161400000, -76.5663900000)', '(39.3573500000, -76.6005300000)', '(39.3311200000, -76.6315100000)', '(39.3311200000, -76.6315100000)', '(39.2832900000, -76.5996300000)', '(39.2868200000, -76.6063900000)', '(39.3031200000, -76.6461100000)']

我需要将此字符串转换为元组，以便输出为：

[(39.2947500000, -76.6565600000),(39.3423900000, -76.5698300000)......]

我尝试使用float方法，但它给出了以下错误：

ValueError:无法将字符串转换为浮点：（39.2947500000，-76.6565600000）

提前谢谢

>>> L=['(39.2947500000, -76.6565600000)', '(39.3423900000, -76.5698300000)', '(39.3199500000, -76.6222000000)', '(39.2533200000, -76.6263600000)', '(39.3068100000, -76.6549700000)', '(39.2937500000, -76.6233700000)', '(39.3146700000, -76.6425300000)', '(39.3073300000, -76.6015900000)', '(39.2451900000, -76.6336400000)', '(39.3283000000, -76.5893200000)', '(39.3215400000, -76.6736800000)', '(39.3010000000, -76.5977400000)', '(39.3122600000, -76.6194200000)', '(39.3161400000, -76.5663900000)', '(39.3573500000, -76.6005300000)', '(39.3311200000, -76.6315100000)', '(39.3311200000, -76.6315100000)', '(39.2832900000, -76.5996300000)', '(39.2868200000, -76.6063900000)', '(39.3031200000, -76.6461100000)']
>>> import ast
>>> list(map(lambda x:ast.literal_eval(x), L))
[(39.29475, -76.65656), (39.34239, -76.56983), (39.31995, -76.6222), (39.25332, -76.62636), (39.30681, -76.65497), (39.29375, -76.62337), (39.31467, -76.64253), (39.30733, -76.60159), (39.24519, -76.63364), (39.3283, -76.58932), (39.32154, -76.67368), (39.301, -76.59774), (39.31226, -76.61942), (39.31614, -76.56639), (39.35735, -76.60053), (39.33112, -76.63151), (39.33112, -76.63151), (39.28329, -76.59963), (39.28682, -76.60639), (39.30312, -76.64611)]

对于Python2.x:

map（lambda x:ast.literal\u eval（x，L）

编辑：一些解释：

ast

代表抽象语法树

literal\u eval（）

比

eval（）

安全得多

引用官方文件：

ast.literal_eval（节点或字符串）安全地计算表达式节点或 一个Unicode或Latin-1编码的字符串，包含Python文本或 容器显示。提供的字符串或节点只能由 以下Python文本结构：字符串、数字、元组、列表、， 口授、布尔语和无语

这可以用于安全地评估包含Python的字符串 来自不受信任源的值，无需解析这些值 自己它不能计算任意复杂的参数 表达式，例如涉及运算符或索引的表达式

对于Python2.x:

map（lambda x:ast.literal\u eval（x，L）

编辑：一些解释：

ast

代表抽象语法树

literal\u eval（）

比

eval（）

安全得多

引用官方文件：

ast.literal_eval（节点或字符串）安全地计算表达式节点或 一个Unicode或Latin-1编码的字符串，包含Python文本或 容器显示。提供的字符串或节点只能由 以下Python文本结构：字符串、数字、元组、列表、， 口授、布尔语和无语

这可以用于安全地评估包含Python的字符串 来自不受信任源的值，无需解析这些值 自己它不能计算任意复杂的参数 表达式，例如涉及运算符或索引的表达式

---

您可以使用

eval

：

a = ['(39.2947500000, -76.6565600000)', '(39.3423900000, -76.5698300000)', '(39.3199500000, -76.6222000000)', '(39.2533200000, -76.6263600000)', '(39.3068100000, -76.6549700000)', '(39.2937500000, -76.6233700000)', '(39.3146700000, -76.6425300000)', '(39.3073300000, -76.6015900000)', '(39.2451900000, -76.6336400000)', '(39.3283000000, -76.5893200000)', '(39.3215400000, -76.6736800000)', '(39.3010000000, -76.5977400000)', '(39.3122600000, -76.6194200000)', '(39.3161400000, -76.5663900000)', '(39.3573500000, -76.6005300000)', '(39.3311200000, -76.6315100000)', '(39.3311200000, -76.6315100000)', '(39.2832900000, -76.5996300000)', '(39.2868200000, -76.6063900000)', '(39.3031200000, -76.6461100000)'] 
b = [ eval(x) for x in a ]

---

您可以使用

eval

：

a = ['(39.2947500000, -76.6565600000)', '(39.3423900000, -76.5698300000)', '(39.3199500000, -76.6222000000)', '(39.2533200000, -76.6263600000)', '(39.3068100000, -76.6549700000)', '(39.2937500000, -76.6233700000)', '(39.3146700000, -76.6425300000)', '(39.3073300000, -76.6015900000)', '(39.2451900000, -76.6336400000)', '(39.3283000000, -76.5893200000)', '(39.3215400000, -76.6736800000)', '(39.3010000000, -76.5977400000)', '(39.3122600000, -76.6194200000)', '(39.3161400000, -76.5663900000)', '(39.3573500000, -76.6005300000)', '(39.3311200000, -76.6315100000)', '(39.3311200000, -76.6315100000)', '(39.2832900000, -76.5996300000)', '(39.2868200000, -76.6063900000)', '(39.3031200000, -76.6461100000)'] 
b = [ eval(x) for x in a ]

此解决方案不使用eval（我不喜欢使用eval），它迭代字符串，在“，”上拆分它们，对它们进行子串以删除“（“，”和“）”，然后将它们强制转换为浮点

---

此解决方案不使用eval（我不喜欢使用eval），它迭代字符串，在“，”上拆分它们，对它们进行子串以删除“（“，”和“）”，然后将它们强制转换为浮点值（如果您希望避免

import

s，并且感觉
为了清晰起见，这可能是避免列表理解的地方*：

lst = ['(39.2947500000, -76.6565600000)', '(39.3423900000, -76.5698300000)', '(39.3199500000, -76.6222000000)', '(39.2533200000, -76.6263600000)', '(39.3068100000, -76.6549700000)', '(39.2937500000, -76.6233700000)', '(39.3146700000, -76.6425300000)', '(39.3073300000, -76.6015900000)', '(39.2451900000, -76.6336400000)', '(39.3283000000, -76.5893200000)', '(39.3215400000, -76.6736800000)', '(39.3010000000, -76.5977400000)', '(39.3122600000, -76.6194200000)', '(39.3161400000, -76.5663900000)', '(39.3573500000, -76.6005300000)', '(39.3311200000, -76.6315100000)', '(39.3311200000, -76.6315100000)', '(39.2832900000, -76.5996300000)', '(39.2868200000, -76.6063900000)', '(39.3031200000, -76.6461100000)']

out = list()
for i in lst:
    j,k = i.split(",")
    out.append(tuple((float(j[1:]), float(k[:-1]))))
print(out)

---

*如果您希望避免导入

并且您觉得

为了清晰起见，这可能是避免列表理解的地方*：

lst = ['(39.2947500000, -76.6565600000)', '(39.3423900000, -76.5698300000)', '(39.3199500000, -76.6222000000)', '(39.2533200000, -76.6263600000)', '(39.3068100000, -76.6549700000)', '(39.2937500000, -76.6233700000)', '(39.3146700000, -76.6425300000)', '(39.3073300000, -76.6015900000)', '(39.2451900000, -76.6336400000)', '(39.3283000000, -76.5893200000)', '(39.3215400000, -76.6736800000)', '(39.3010000000, -76.5977400000)', '(39.3122600000, -76.6194200000)', '(39.3161400000, -76.5663900000)', '(39.3573500000, -76.6005300000)', '(39.3311200000, -76.6315100000)', '(39.3311200000, -76.6315100000)', '(39.2832900000, -76.5996300000)', '(39.2868200000, -76.6063900000)', '(39.3031200000, -76.6461100000)']

out = list()
for i in lst:
    j,k = i.split(",")
    out.append(tuple((float(j[1:]), float(k[:-1]))))
print(out)



*请参见
即使它使用了危险的
eval
，这也是简短的。即使它使用了危险的
eval
，这也是简短的。