遍历命名空间 [英] Iterate through a Namespace

问题描述

我有一个看起来像这样的命名空间:

I have got a namespace looking like this :

命名空间(aTQ =无，bE =无，bEQ =无，b =无，bQ =无，c =无， c = None，cJ = None，d = None，g = None，jR = ['xx'，'015']，lC = None，l = None)

Namespace(aTQ=None, bE=None, bEQ=None, b=None, bQ=None, c=None, c=None, cJ=None, d=None, g=None, jR=['xx', '015'], lC=None, l=None)

如何遍历它，以便可以找到并替换"jR"键的"xx"值?

How can I iterate through it so I can find and replace the 'xx' value for the "jR" key in place ?

推荐答案

我不确定您事先知道什么(名称jR，或者只有一个名称具有not None值)，但是您可以尝试使用vars()，就像__dict__一样，这是一个以名称空间中的名称为其键的字典.因此，如果字符串'jR'在某处，

I'm not sure what you know in advance, (the name jR, or only that one name has a not None value) but you can try using vars() which, like __dict__ should be a dict that has the names in your namespace as its keys. So if you have the string 'jR' somewhere,

vars()['jR'] = vars()['jR'][0]

同样，您可以使用namespace.__dict__而不是vars()

likewise, you can get the same dict using namespace.__dict__ instead of vars()

会将['xx','015']中的第一个值保留为jR的唯一值.

will leave you with the first value in ['xx','015'] as the only value for jR.

为清楚起见，由于vars()和__dict__都返回dict，因此您可以按以下方式遍历它们:

To be clear, since vars() and __dict__ both return dicts, you can iterate through them as:

for k in namespace.__dict__:
    if namespace.__dict__[k] is not None:
        <<do something>>

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