Python：如何在函数的本地范围内运行eval（）

Question

我尝试在函数的局部范围内使用eval（）。但是，它始终在全局范围内进行评估。

自包含的示例：

1-此代码有效：

var1 = 1
var2 = 2
var3 = 3    
myDict = dict((name, eval(name)) for name in ["var1",
                                              "var2",
                                              "var3"])
print(myDict["var1"])

2-抛出NameError为lvar1

def test1():
   lvar1 = 1
   lvar2 = 2
   lvar3 = 3
   myDict = dict((name, eval(name)) for name in ["lvar1",
                                                 "lvar2",
                                                 "lvar3"])
   print(myDict["lvar1"])

3-与2相同的结果。

def test2():
    lvar1 = 1
    lvar2 = 2
    lvar3 = 3
    myDict = dict((name, eval(name), locals()) for name in ["lvar1",
                                                            "lvar2",
                                                            "lvar3"])
    print(myDict["lvar1"])

Answer 1

保存locals()（或vars()）调用的结果以返回函数的本地范围。否则，locals()在生成器表达式内部将返回gen-expr的本地范围。

def test3():
    lvar1 = 1
    lvar2 = 2
    lvar3 = 3
    scope = locals()
    myDict = dict((name, eval(name, scope)) for name in [
                  "lvar1", "lvar2", "lvar3"])
    print(myDict["lvar1"])

顺便说一句，您不需要明确的理解即可构建该命令：

# copy() avoids quirky, unexpected updates if something else (like a debugger)
# accesses locals() or f_locals
myDict = locals().copy()  # or vars().copy()