我想从另一个字符串中取消封装字符串(在Python中)。
例如,来自:
>>> string1
"u'abcde'"
我想得到:
>>> string2
'abcde'
答案 0 :(得分:1)
好像你想要函数eval
>>> stri = "u'abcde'"
>>> eval(stri)
'abcde'
>>> help(eval)
Help on built-in function eval in module builtins:
eval(...)
eval(source[, globals[, locals]]) -> value
Evaluate the source in the context of globals and locals.
The source may be a string representing a Python expression
or a code object as returned by compile().
The globals must be a dictionary and locals can be any mapping,
defaulting to the current globals and locals.
If only globals is given, locals defaults to it.
根据以下注释,您需要使用ast.literal_eval
函数而不是内置eval函数,因为eval将评估任意(和潜在危险)代码,而ast.literal_eval
将仅评估Python文字。
>>> import ast
>>> stri = "u'abcde'"
>>> ast.literal_eval(stri)
'abcde'