使用列表/元组元素作为键创建字典 [英] Creating dictionary using list/tuple elements as key

问题描述

我需要生成这样的字典:

<代码>{'新环境':{'新项目':{'新比较':{实例":[]，'n_thing': '新事物'}}}}

来自一个元组，像这样:('newEnv','newProj','newComp','newThing') 但前提是它不存在.所以，我尝试了这个:

myDict = {}(env,proj,comp,thing) = ('newEnv','newProj','newComp','newThing')如果 env 不在 myDict 中:我的字典 [env] = {}如果 proj 不在 myDict[env] 中:myDict[env][proj] = {}如果 comp 不在 myDict[env][proj] 中:myDict[env][proj][comp] = {'n_thing': 东西，'instances': []}

这非常有效，但不确定它的效率如何，或者我是否应该这样做.有什么建议)??

解决方案

可以使用循环(只有前 3 个键，newThing 不是链中的键):

myDict = {}路径 = ('newEnv','newProj','newComp')当前 = 我的字典对于路径中的键:当前 = current.setdefault(key, {})

其中 current 最终成为最里面的字典，让您在其上设置 'n_thing' 和 'instances' 键.p>

您可以使用 reduce() 将其折叠成单行:

myDict = {}路径 = ('newEnv','newProj','newComp')减少(lambda d，k:d.setdefault(k，{})，路径，myDict)

reduce 调用返回最里面的字典，因此您可以使用它来分配最终值:

myDict = {}路径 = ('newEnv','newProj','newComp')inner = reduce(lambda d, k: d.setdefault(k, {}), path, myDict)inner.update({'n_thing': 'newThing', 'instances': []})

I need to generate a dictionary like this:

{
  'newEnv': {
     'newProj': {
        'newComp': {
           'instances': [],
           'n_thing': 'newThing'
        }
     }
  }
}

from a tuple, like this: ('newEnv','newProj','newComp','newThing') but only if that doesn't already exists. So, I tried this:

myDict = {}
(env,proj,comp,thing) = ('newEnv','newProj','newComp','newThing')

if env not in myDict:
    myDict[env] = {}
if proj not in myDict[env]:
    myDict[env][proj] = {}
if comp not in myDict[env][proj]:
    myDict[env][proj][comp] = {'n_thing': thing, 'instances': []}

which is pretty much working but not sure how efficient is that or if I should be doing this way at all. Any suggestion(s)??

解决方案

You can use a loop (with just the first 3 keys, newThing is not a key in the chain):

myDict = {}
path = ('newEnv','newProj','newComp')
current = myDict
for key in path:
    current = current.setdefault(key, {})

where current ends up as the innermost dictionary, letting you set the 'n_thing' and 'instances' keys on that.

You could use reduce() to collapse that into a one-liner:

myDict = {}
path = ('newEnv','newProj','newComp')
reduce(lambda d, k: d.setdefault(k, {}), path, myDict)

The reduce call returns the innermost dictionary, so you can use that to assign your final value:

myDict = {}
path = ('newEnv','newProj','newComp')
inner = reduce(lambda d, k: d.setdefault(k, {}), path, myDict)
inner.update({'n_thing': 'newThing', 'instances': []})

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