使用逗号分隔键访问jsonable嵌套对象 [英] Access jsonable nested object with comma separated key

问题描述

首先让我们创建一个可在python中jsonable的嵌套对象:

First lets create a nested object which is jsonable in python:

ExampleObject1 = [ {'a':0, 'b':1} ,  {'c':2, 'd':3} ]
ExampleObject2 = [ {'a':0, 'b':ExampleObject1}, {'c':2, 'd':3} ]
ExampleObject3 = [ {'a':0, 'b':ExampleObject1}, {'c':ExampleObject2, 'd':3} ]

我们可以像这样通过链接方括号轻松访问元素:

We can easily access an element with chaining square brackets like so:

print ( ExampleObject3[0]['b'][0]['b'] )

>>> 1

如何使用键列表访问同一元素，而不需要方括号?

How can I access the same element with a list of keys instead of needing the square brackets?

print ( ExampleObject3[ (0,'b',0,'b') ] )

>>> TypeError: list indices must be integers or slices, not tuple

注意:我可以这样访问numpy array. 一旦我尝试用逗号分隔的键访问字典，事情就会中断.

Note: I can access numpy arrays this way. As soon as I try to access a dictionary with comma separated key's then things break.

请参阅:将切片索引存储为对象.

原因:我只希望能够传递一个任意密钥，该密钥可用于以后从内存中的某个大对象获取数据.

Reason: I just want to be able to pass around an arbitrary key which can be used to go get data later from some large object sitting in memory.

使用键可以在原始对象中更改值也很好:

It would also be nice to be able to change values in the original object using the key:

ExampleObject3[ (0,'b',0,'b') ] = 'alpha'

推荐答案

您不能直接使用像这样的键列表建立索引，但是您可以创建一个简单的函数来为您完成此操作. functools.reduce() 对此很方便:

You can't index directly with a list of key like this, but you could make a simple function to do it for you. functools.reduce() is handy for this:

from functools import reduce

def fromKeyList(obj, key_list):
    return reduce(lambda o, k: o[k], key_list, obj)

ExampleObject1 = [ {'a':0, 'b':1} ,  {'c':2, 'd':3} ]
ExampleObject2 = [ {'a':0, 'b':ExampleObject1} ,  {'c':2, 'd':3} ]
ExampleObject3 = [ {'a':0, 'b':ExampleObject1} ,  {'c':ExampleObject2, 'd':3} ]

fromKeyList(ExampleObject3, (0,'b',0,'b'))
# 1
fromKeyList(ExampleObject3, (1,'c',0,'b',1, 'c'))
# 2

根据更多信息进行编辑

要设置一个项目，您可以从密钥列表中获取除最后一个项目以外的所有项目，然后使用最后一个密钥来设置该项目.看起来可能像这样:

To set an item, you can get all but the last item from the key list, then use the last key to set the item. That might look something like:

def setFromKeyList(obj, key_list, val):
    last = reduce(lambda o, k: o[k], key_list[:-1], obj)
    last[key_list[-1]] = val

fromKeyList(ExampleObject3, (0,'b',0,'b'))
# 1

setFromKeyList(ExampleObject3, (0,'b',0,'b'), 10)

fromKeyList(ExampleObject3, (0,'b',0,'b'))
#10

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