相关疑难解决方法(0)

我有一个复杂的字典结构,我想通过一个键列表访问,以解决正确的项目.

dataDict = {
    "a":{
        "r": 1,
        "s": 2,
        "t": 3
        },
    "b":{
        "u": 1,
        "v": {
            "x": 1,
            "y": 2,
            "z": 3
        },
        "w": 3
        }
}    

maplist = ["a", "r"]

要么

maplist = ["b", "v", "y"]

我已经制作了以下代码,但是我确信如果有人有想法,有更好更有效的方法.

# Get a given data from a dictionary with position provided as a list
def getFromDict(dataDict, mapList):    
    for k in mapList: dataDict = dataDict[k]
    return dataDict

# Set a given data in a dictionary with position provided as a list
def …

我希望能够做到这样的事情:

from dotDict import dotdictify

life = {'bigBang':
           {'stars':
               {'planets': []}
           }
       }

dotdictify(life)

# This would be the regular way:
life['bigBang']['stars']['planets'] = {'earth': {'singleCellLife': {}}}
# But how can we make this work?
life.bigBang.stars.planets.earth = {'singleCellLife': {}}

#Also creating new child objects if none exist, using the following syntax:
life.bigBang.stars.planets.earth.multiCellLife = {'reptiles':{},'mammals':{}}

我的动机是改进代码的简洁性,如果可能的话,使用与Javascript类似的语法来访问JSON对象,以实现高效的跨平台开发.(我也使用Py2JS和类似的.)

说我有在Python几个变量或对象a,b,c,...

如何轻松地将这些变量转储到Python中的命名空间中并在以后恢复它们？(例如,以相同的方式argparse将各种变量包装到命名空间中).

以下是我希望如何在命名空间之间转储内容的两个示例:

将局部变量转储到命名空间中

function (bar):
   # We start with a, b and c
   a = 10
   b = 20
   c = "hello world"

   # We can dump anything we want into e, by just passing things as arguments:
   e = dump_into_namespace(a, b, c) 
   del a, b, c

   print (e.a + e.b) # Prints 30
   return e  # We can return e if we want. This is just …

全新的python,让我说我有一个字典:

kidshair = {'allkids':{'child1':{'hair':'blonde'},
                      'child2':{'hair':'black'},
                      'child3':{'hair':'red'},
                      'child4':{'hair':'brown'}}}

如果child3定期更改头发颜色,我可能想编写一个应用程序来加速数据维护.在这个例子中,我使用:

kidshair['allkids']['child3']['hair'] = ...

有没有办法将此路径存储为变量,以便我可以在我的乐趣中访问它？明显

mypath = kidshair['allkids']['child3']['hair']

导致mypath ='red'.是否有任何可能的方法来硬编码路径本身,所以我可以使用:

mypath = 'blue' 

拒绝

kidshair['allkids']['child3']['hair'] = 'blue'

非常感谢,ATfPT

有没有一种很好的方法来处理不存在的字典键？

我有一个基于字典的数据库,其中的键看起来像这样

ID
   ver
      sub_ver
           A
           B
           C

我想比较每个ID/ver/sub_ver的A,B,C键的值,以确保:

 (A==B and C==None) or (A==C and B==None) or (A==B==C)

然而,并非所有"ID"都具有A和B以及C键/值

我不是很好的代码:

**loops outside of this for ID/ver/sub_ver** 
try:
    A = data_structure[ID][ver][sub_ver]['A']
    B = data_structure[ID][ver][sub_ver]['B']
    C = data_structure[ID][ver][sub_ver]['C']
except KeyError:
    try:
        A = data_structure[ID][ver][sub_ver]['A']
        B = data_structure[ID][ver][sub_ver]['B']
        C = None

    except KeyError: 
        try:
            A = data_structure[ID][ver][sub_ver]['A']
            B = None
            C = data_structure[ID][ver][sub_ver]['C']                    

接下来我检查所有值是否匹配

我使用set()以防万一A/B/C列表不合规

if not any((set(A)==set(B) and C==None, \
            set(A)==set(C) and B==None, \
            set(A)==set(B)==set(C))):
    set_of_problems.append([ID, ver, sub_ver, [A, B, C])

有没有更好的方法来执行嵌套try …