我尝试在函数的局部范围内使用 eval()。然而它总是在全局范围内评估。
独立的示例:
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- 为 lvar1
抛出 NameError
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"])
最佳答案
保存 locals()
(或 vars()
)调用的结果以返回函数的本地作用域。否则,生成器表达式内的 locals()
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来构建该字典:
# copy() avoids quirky, unexpected updates if something else (like a debugger)
# accesses locals() or f_locals
myDict = locals().copy() # or vars().copy()
关于Python:如何在函数的本地范围内运行 eval(),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36616739/