列表中多个字典按键值排序

方法一:

解析: 使用sorted 方法, 排序后的结果为一个元组. 可以字符串排序(那数字肯定更没问题了!)   

1:  按照键值(value)排序

a = {'a': 'China', 'c': 'USA', 'b': 'Russia', 'd': 'Canada'}
b = sorted(a.items(), key=lambda x: x[1], reverse=True)
结果:
[('c', 'USA'), ('b', 'Russia'), ('a', 'China'), ('d', 'Canada')]
2: 按照键名(key)排序
a = {'a': 'China', 'c': 'USA', 'b': 'Russia', 'd': 'Canada'}
b = sorted(a.items(), key=lambda x: x[0], reverse=True)
结果:
[('d', 'Canada'), ('c', 'USA'), ('b', 'Russia'), ('a', 'China')]
3: 嵌套字典, 按照字典键名(key)排序
a = {'a': {'b': 'China'}, 'c': {'d': 'USA'}, 'b': {'c': 'Russia'}, 'd': {'a': 'Canada'}}
b = sorted(a.items(), key=lambda x: x[1], reverse=True)
结果:
[('c', {'d': 'USA'}), ('b', {'c': 'Russia'}), ('a', {'b': 'China'}), ('d', {'a': 'Canada'})]
4: 嵌套列表, 针对列表第一个元素排序( 其实直接写 x: x[1] 就是按照第一个值排序. )
a = {'a': [1, 3], 'c': [3, 4], 'b': [0, 2], 'd': [2, 1]}
b = sorted(a.items(), key=lambda x: x[1][0], reverse=True)
结果:
[('c', [3, 4]), ('d', [2, 1]), ('a', [1, 3]), ('b', [0, 2])]
4-2: 嵌套列表, 按照列表其他元素排序  只需要修改列表对应的下标
a = {'a': [1, 3], 'c': [3, 4], 'b': [0, 2], 'd': [2, 1]}
b = sorted(a.items(), key=lambda x: x[1][1], reverse=True)
结果:
[('c', [3, 4]), ('a', [1, 3]), ('b', [0, 2]), ('d', [2, 1])]
总结:  此处使用lambda方法, x: x[1][1] 就可以看做是在访问字典的值, 想要按照哪个数值排序, 用相应的坐标对应即可, 但当字典过于复杂后, 应该选择用元组存储, 简化排序过程. 

方法二:
from operator import itemgetter
dict_list = [{"ming": 87}, {"mei": 93}, {"hua": 68}, {"jon": 75}, {"ston": 100}, {"jack": 56}]
mid_dict = {key: value for x in dict_list for key, value in x.items()}
mid_list = sorted(mid_dict.items(), key=itemgetter(1))
fin_list = [{x[0]: x[1]} for x in mid_list]
例子：T=[{'xgei-0/0/1/1': '9'}, {'xgei-0/0/1/2': '20'},{'xgei-0/0/1/15': '12'}]
def Sorted_listdict(dict_list ):
    New_List=[]
    New_D={}
    mid_dict = {key: value for x in dict_list for key, value in x.items()}  ###格式写法。列表与字典的结构
    #print (mid_dict)
    ordered_dict = OrderedDict(sorted(mid_dict.items(), key=lambda t: int(t[1]), reverse=True))
    #print (type(ordered_dict),ordered_dict)
    #<class 'collections.OrderedDict'> OrderedDict([('xgei-0/0/1/2', '20'), ('xgei-0/0/1/15', '12'), ('xgei-0/0/1/1', '9')])
    for x in ordered_dict:
        New_D[x]=mid_dict[x]
    New_List.append(New_D)
    print (New_List)
    return New_List

特殊:
newlist = sorted(data, key=lambda k: k['PricingOptions']["Price"], reverse=True)
print(newlist)
#[{'PricingOptions': {'Price': 227243.14, 'Agents': [4056270]}}, {'PricingOptions': {'Price': 51540.72, 'Agents': [4056270]}}]