Python：2.6：尝试构建像 exec('if' + varname + '< 1.0:') 这样的语句

发布于 2024-12-21 09:23:56 字数 428 浏览 1 评论 0 原文

我的目标是在变量名称更改的情况下在 for 循环中具有相同的 if 语句这些语句位于不同文件的函数中：适当的导入方案已设置。

示例：

for i in range(0,10):

    myvar = ''
    myvar = myvar + str(i)
    exec('if' + myvar + '< 0:')
`do something`

它说的是“做某事”部分“意外的身份” 我删除了身份，它说；关键错误：'myvar1' 我尝试使用它，

if vars()[myvar] < 0:

    do something

它给了我相同的 KeyError: 'myvar1'

有没有办法使这样的变量语句起作用？

原文

My goal here is to have the same if statement in a for loop were the variable name changes
Those statements are in a function in a different file: the appropriate Import scheme
as been set.

Examples:

for i in range(0,10):

    myvar = ''
    myvar = myvar + str(i)
    exec('if' + myvar + '< 0:')
`do something`

It say about the do something part "unexpected ident "
I remove the ident and it say; KeyError: 'myvar1'
I tried using this instead

if vars()[myvar] < 0:

    do something

It gives me the same KeyError: 'myvar1'

Is there a way to make such a variable statement to work ?

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淑女气质 2024-12-28 09:23:56

您将字符串“if 0 < 0:”传递给 exec() 函数。这不是一个语法上有效的构造——它缺少受控套件。您可能会考虑：然而

if eval(myvar) < 0:
    do_something()

，作为更一般的观点，在生产程序中动态评估变量名几乎总是错误的。通常最好使用字典将键映射到值（这就是它们的用途，并且避免使用 exec() 和 eval()）。

You are passing the string "if 0 < 0:" to the exec() function. This is not a syntactically valid construct -- it's missing the controlled suite. You might consider instead:

if eval(myvar) < 0:
    do_something()

As a more general point, however, dynamic evaluation of variable names is almost invariably the wrong thing to do in a production program. You are usually better to use a dict to map keys to values (that's what they are for, and you avoid the use of exec() and eval()).

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情域 2024-12-28 09:23:56

因此，您实际上应该重构代码以将 myvarN 存储为列表。然后你可以这样做：

for element in myvars:
    if element < 0:
        doSomething(element)

So you should really just refactor your code to store myvarN as a list. Then you can just do:

for element in myvars:
    if element < 0:
        doSomething(element)

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浪推晚风 2024-12-28 09:23:56

如果您尝试执行以下操作：

var1 = 10
var2 = 12
var3 = 18

for i in range(10):
  if locals().get('var%d' % i, 0) > 10:
    # something here for the conditions that pass.

该代码应该可以工作。与您所做的最大区别是我在 locals() 上调用了 .get() ，这样如果您尝试测试一个不存在的变量，它就会返回与某事。老实说，这是一种奇怪的方法，但有时奇怪的方法是必要的。

If you're trying to do something like this:

var1 = 10
var2 = 12
var3 = 18

for i in range(10):
  if locals().get('var%d' % i, 0) > 10:
    # something here for the conditions that pass.

That code should work. The biggest difference from what you'd done is that I called .get() on locals() so that if you try to test a non-existent variable it comes back with something. This ... is an odd approach, tbh, but sometimes odd approaches are necessary.