Python 實現字典排序、按照list中字典的某個key排序

1.給字典按照value按照從大到小排序

排序

dict = {'a':21, 'b':5, 'c':3, 'd':54, 'e':74, 'f':0}
new_dict = sorted(dict.items(), key=lambda d:d[1], reverse = True)
print(new_dict)

輸出：

[('e', 74), ('d', 54), ('a', 21), ('b', 5), ('c', 3), ('f', 0)]

2. python按照list中的字典的某key排序:

例子:

s=[
    {"no":28,"score":90},
    {"no":25,"score":90},
    {"no":1,"score":100},
    {"no":2,"score":20},
]
print("original s: ",s)

單級排序，僅按照score排序

new_s = sorted(s,key = lambda e:e.__getitem__('score'))
print("new s: ", new_s)

多級排序,先按照score，再按照no排序

new_s_2 = sorted(new_s,key = lambda e:(e.__getitem__('score'),e.__getitem__('no')))
print("new_s_2: ", new_s_2)

輸出:

# 原有數據
[('e', 74), ('d', 54), ('a', 21), ('b', 5), ('c', 3), ('f', 0)]

# 單級排序
[{'no': 2, 'score': 20},
 {'no': 28, 'score': 90},
 {'no': 25, 'score': 90},
 {'no': 1, 'score': 100}]

# 多級排序
[{'no': 2, 'score': 20},
 {'no': 25, 'score': 90},
 {'no': 28, 'score': 90},
 {'no': 1, 'score': 100}]