将字符串列表转换为元组

Question

我有这个字符串列表:

['(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)......]

我试图使用float方法,但它给出了这个错误:

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

提前致谢

Answer 1

>>> 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python 2.x: map(lambda x:ast.literal_eval(x), L)

编辑:一些解释:

ast代表抽象语法树.literal_eval()比安全得多eval().

引自官方文件:

ast.literal_eval(node_or_string)安全地评估表达式节点或包含Python文字或容器显示的Unicode或Latin-1编码字符串.提供的字符串或节点可能只包含以下Python文字结构:字符串,数字,元组,列表,dicts,布尔值和None.

这可用于安全地评估包含来自不受信任来源的Python值的字符串,而无需自己解析值.它无法评估任意复杂的表达式,例如涉及运算符或索引.