sorted

单例方式 通过类的绑定方法 class Person(): _instance=None def __init__(self,port,host): self.port=port self.host=host @classmethod def get_sigoleton(cls): import settings if not cls._instance: cls._instance = cls(settings.PORT, settings.HOST) return cls._instance s1=Person.get_sigoleton() s2=Person.get_sigoleton() s3=Person.get_sigoleton() print(s1) print(s2) s4=Person('33306','192.168.1.1') print(s4) 通过装饰器 def get_sigoleton(cls): import settings instance=cls(settings.PORT,settings.HOST) def wrapper(*args,**kwargs): if len(args)!=0 or len(kwargs)!=0 res=cls(*args,**kwargs) return res else: return instance

python sort(),sorted()的参数key array = [ 1 , 2 , 8 , 4 , 5 , 8 , 3 , 5 , - 1 ] #key参数，要指向一个函数，返回该计算方法得到的、排序用的权值. print ( sorted ( array , key = lambda x : x ) ) #[-1, 1, 2, 3, 4, 5, 5, 8, 8] print ( sorted ( array , key = lambda x : - x ) ) #[8, 8, 5, 5, 4, 3, 2, 1, -1] print ( sorted ( array , key = lambda x : ( 4 - x ) ** 2 ) ) #[4, 5, 3, 5, 2, 1, 8, 8, -1] array . sort ( key = lambda x : - x ) print ( array ) #[8, 8, 5, 5, 4, 3, 2, 1, -1] 来源： https://blog.csdn.net/weixin_42764391/article/details/100540071

使用lambda函数取出字典中的value：d = {'a': 24, 'g': 52, 'i': 12, 'k': 33}print(list(map(lambda x:x[1],d.items())))[24, 52, 12, 33]使用sort函数排序 d_1=sorted(d.items(),key=lambda x:x[1],reverse=False) print('d_1:',d_1)d_1: [('i', 12), ('a', 24), ('k', 33), ('g', 52)] d_2=sorted(d.items(),key=lambda x:x[1],reverse=True)print('d_2:',d_2)d_2: [('g', 52), ('k', 33), ('a', 24), ('i', 12)] """按照每个元组元素的长度排序"""l = [(1,5,3),(1,3,6,3),(1,1,2,4,5,6),(1,9)]def func(item): return len(item)l1=sorted(l,key=func)print(l1)[(1, 9), (1, 5, 3), (1, 3, 6, 3), (1, 1, 2, 4, 5, 6)]l2=sorted(l,key=lambda x:len(x))print(l2) [(1, 9), (1

I need to have an automatically sorted-by-values map in Java - so that It keeps being sorted at any time while I'm adding new key-value pairs or update the value of an existing key-value pair, or even delete some entry. Please also have in mind that this map is going to be really big (100's of thousands, or even 10's of millions of entries in size). So basically I'm looking for the following functionality: Supposed that we had a class 'SortedByValuesMap' that implements the aforementioned functionality and we have the following code: SortedByValuesMap<String,Long> sorted_map = new

1.背景 　 在python中，通常需要使用排序函数。而对字典针对一个键值进行排序会经常使用到。记录sorted 函数的键值排序用法。 2.代码 　　data 一个list，list 中的元素由字典组成 dic { "label"：True "seqence"：“I love Python” "rate"：0.5 "length"： 13 } 　　 如果需要对每个字典中的rate值进行从低到高排序，代码如下： datas = sorted(datas, key=lambda d: d["rate"]) 来源： https://www.cnblogs.com/siyuan1998/p/11343147.html