访问python中的字典列表并使用值对其进行排序（嵌套字典）

Question

I have a list of 50 dictionaries and want to be sorted by a value of 'Key2' of that dictionary.

list1= [{'outside_key1': [ { 'key1': 'one', 'key2': 'xyz','key3':'three'}]} ,
        {'outside_key2': [ { 'key1': 'one', 'key2': 'abc','key3':'three'}]}]

Key2 can statically be accessed by:

>>>print list1[0]['outside_key1'][0]['key2']
   xyz

Now sorting based on 'key2' Like:

sorted_list = sorted(list1, key=lambda k: k[???])

Final sorted by value becomes:

[{'outside_key2': [ { 'key1': 'one', 'key2': 'abc','key3':'three'}]} ,
 {'outside_key1': [ { 'key1': 'one', 'key2': 'xyz','key3':'three'}]}]

So my questions:
1. How can i dynamically access the value of 'Key2'?
2. How can i sort list of dictionary based on the value of 'Key2'?

Answer 1

list1= [{'outside_key1': [ { 'key1': 'one', 'key2': 'xyz','key3':'three'}]} ,
        {'outside_key2': [ { 'key1': 'one', 'key2': 'abc','key3':'three'}]}]
sort_on = "key2"
decorated = [(dict_.values()[0][0][sort_on], dict_) for dict_ in list1]
decorated.sort()
result = [dict_ for (key, dict_) in decorated]
print result

dict_.values() get the outside dic values, the first [0] get the value of outside_key1 , the second [0] get the first value of inner list.

Here's the fastest way to do it, as it avoids using a custom comparison function, instead using builtin comparisons. You can get more information from Sorting Lists of Dictionaries

Answer 2

Assuming you have only a single key for each of the dictionaries, you could do something like:

sorted(list1, key=lambda d: d[next(iter(d))][0]['key2'])

Here: d[next(iter(d))] will give the value of the first key associated with d . Of course, dictionaries are unordered, so the "first" key only has meaning when you have only a single key-value pair in the dict...

---

FWIW, it seems like your data-structure is really getting in the way of a clean solution here. It seems like a flat list of dict would be a better way to store the data:

list1 = [
    {'key1': 'one', 'key2': 'abc', 'key3':'three'},
    { 'key1': 'one', 'key2': 'xyz','key3':'three'},
]

As having dicts with only one value and lists with only one value tend to make the container a bit overkill.