Python：如何按多个值对字典列表进行排序？
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Python: How to sort a list of dictionaries by several values?
提问by Joko
I want to sort a list at first by a value and then by a second value. Is there an easy way to do this? Here is a small example:
我想首先按一个值对列表进行排序，然后再按第二个值排序。是否有捷径可寻？这是一个小例子：
A = [{'name':'john','age':45},
{'name':'andi','age':23},
{'name':'john','age':22},
{'name':'paul','age':35},
{'name':'john','age':21}]
This command is for sorting this list by 'name':
此命令用于按以下方式排序此列表'name'：
sorted(A, key = lambda user: user['name'])
But how I can sort this list by a second value? Like 'age'in this example.
但是我如何按第二个值对这个列表进行排序？就像'age'在这个例子中一样。
I want a sorting like this (first sort by 'name'and then sort by 'age'):
我想要这样的排序（首先排序'name'，然后排序'age'）：
andi - 23
john - 21
john - 22
john - 45
paul - 35
Thanks!
谢谢！
采纳答案by jamylak
>>> A = [{'name':'john','age':45},
{'name':'andi','age':23},
{'name':'john','age':22},
{'name':'paul','age':35},
{'name':'john','age':21}]
>>> sorted(A, key = lambda user: (user['name'], user['age']))
[{'age': 23, 'name': 'andi'}, {'age': 21, 'name': 'john'}, {'age': 22, 'name': 'john'}, {'age': 45, 'name': 'john'}, {'age': 35, 'name': 'paul'}]
This sorts by a tuple of the two attributes, the following is equivalent and much faster/cleaner:
这按两个属性的元组排序，以下等效且更快/更清晰：
>>> from operator import itemgetter
>>> sorted(A, key=itemgetter('name', 'age'))
[{'age': 23, 'name': 'andi'}, {'age': 21, 'name': 'john'}, {'age': 22, 'name': 'john'}, {'age': 45, 'name': 'john'}, {'age': 35, 'name': 'paul'}]
From the comments: @Bakuriu
来自评论：@Bakuriu
I bet there is not a big difference between the two, but
itemgetteravoids a bit of overhead because it extracts the keys and make thetupleduring a single opcode(CALL_FUNCTION), while calling thelambdawill have to call the function, load the various constants(which are other bytecodes) finally call the subscript (BINARY_SUBSCR), build thetupleand return it... that's a lot more work for the interpreter.
我敢打赌两者之间没有太大区别，但
itemgetter避免了一点开销，因为它tuple在单个操作码（CALL_FUNCTION）期间提取密钥并制作，而调用lambda将不得不调用该函数，加载各种常量（它们是其他字节码) 最后调用下标 (BINARY_SUBSCR)，构建tuple并返回它......这对解释器来说是更多的工作。
To summarize: itemgetterkeeps the execution fully on the Clevel, so it's as fast as possible.
总结一下：itemgetter让执行完全保持在C水平上，所以它尽可能快。
回答by Jon Clements
from operator import itemgetter
sorted(your_list, key=itemgetter('name', 'age'))
回答by vvladymyrov
Here is the alternative general solution - it sorts elements of dict by keys and values. The advantage of it - no need to specify keys, and it would still work if some keys are missing in some of dictionaries.
这是替代的通用解决方案 - 它按键和值对 dict 的元素进行排序。它的优点 - 无需指定键，如果某些字典中缺少某些键，它仍然可以工作。
def sort_key_func(item):
""" helper function used to sort list of dicts
:param item: dict
:return: sorted list of tuples (k, v)
"""
pairs = []
for k, v in item.items():
pairs.append((k, v))
return sorted(pairs)
