1、須要用到的函數python
須要用到的函數不多,基本上就是sorted()和lambda。git
sorted()函數原型:算法
sorted(iterable[,key][,reverse])
可見sorted()能夠接受3個參數,須要排序的變量必不可少,而後是key指定排序的元素,reverse指定是否逆序。express
key:一個只有一個參數的函數,這個函數會被用在序列裏的每個元素上,所產生的結果將是排序算法依賴的對比關鍵字。好比說,在對一些字符串排序時,能夠用key=len來實現根據字符串的長度進行排序。這個參數的默認值是恆等函數,也就是默認用元素本身的值來排序 (key=lambda x: x)。函數
lambda實際上就是匿名函數ui
通常形式:spa
lambda arguments: expression
寫成函數形式就是code
def <lambda>(arguments): return expression
當lambda和sorted()的key參數結合就有意想不到的效果了。排序
2、各類排序索引
1.簡單list排序
lis = ['a', 'b', 'c'] print(sorted(lis)) # ['a', 'b', 'c'] print(sorted(lis, reverse=True)) # ['c', 'b', 'a']
2.dict的key排序
dic = {'c': 1, 'b': 2, 'a': 3} #字典排序默認按照字典的key排序,返回的也是字典的key print(sorted(dic)) # ['a', 'b', 'c'] print(sorted(dic, reverse=True)) # ['c', 'b', 'a']
3.dict的value排序
dic = {'c': 1, 'b': 2, 'a': 3} print(sorted(dic, key=lambda k: dic[k])) #dic([k])就是value # ['c', 'b', 'a'] print(sorted(dic, key=lambda k: dic[k], reverse=True)) # ['a', 'b', 'c']
4.list內嵌套list排序
lis = [[4, 2, 9], [1, 5, 6], [7, 8, 3]] print(sorted(lis, key=lambda k: k[0])) #列表內的沒項代入key後的表達式,根據返回值排序 # [[1, 5, 6], [4, 2, 9], [7, 8, 3]] print(sorted(lis, key=lambda k: k[1])) # [[4, 2, 9], [1, 5, 6], [7, 8, 3]] print(sorted(lis, key=lambda k: k[2])) # [[7, 8, 3], [1, 5, 6], [4, 2, 9]] print(sorted(lis, key=lambda k: k[0], reverse=True)) # [[7, 8, 3], [4, 2, 9], [1, 5, 6]]
5.dict內嵌套dict排序
dic = { 'a': {'x': 3, 'y': 2, 'z': 1}, 'b': {'x': 2, 'y': 1, 'z': 3}, 'c': {'x': 1, 'y': 3, 'z': 2}, } print(sorted(dic, key=lambda k: dic[k]['x'])) # ['c', 'b', 'a'] print(sorted(dic, key=lambda k: dic[k]['y'])) # ['b', 'a', 'c'] print(sorted(dic, key=lambda k: dic[k]['z'])) # ['a', 'c', 'b'] print(sorted(dic, key=lambda k: dic[k]['x'], reverse=True)) # ['a', 'b', 'c']
6.list內嵌套dict排序
lis = [ {'x': 3, 'y': 2, 'z': 1}, {'x': 2, 'y': 1, 'z': 3}, {'x': 1, 'y': 3, 'z': 2}, ] print(sorted(lis, key=lambda k: k['x'])) # [{'z': 2, 'x': 1, 'y': 3}, {'z': 3, 'x': 2, 'y': 1}, {'z': 1, 'x': 3, 'y': 2}] print(sorted(lis, key=lambda k: k['y'])) # [{'z': 3, 'x': 2, 'y': 1}, {'z': 1, 'x': 3, 'y': 2}, {'z': 2, 'x': 1, 'y': 3}] print(sorted(lis, key=lambda k: k['z'])) # [{'z': 1, 'x': 3, 'y': 2}, {'z': 2, 'x': 1, 'y': 3}, {'z': 3, 'x': 2, 'y': 1}] print(sorted(lis, key=lambda k: k['x'], reverse=True)) # [{'z': 1, 'x': 3, 'y': 2}, {'z': 3, 'x': 2, 'y': 1}, {'z': 2, 'x': 1, 'y': 3}]
7.dict內嵌套list排序
dic = { 'a': [1, 2, 3], 'b': [2, 1, 3], 'c': [3, 1, 2], } print(sorted(dic, key=lambda k: dic[k][0])) # ['a', 'b', 'c'] print(sorted(dic, key=lambda k: dic[k][1])) # ['b', 'c', 'a'] print(sorted(dic, key=lambda k: dic[k][2])) # ['c', 'b', 'a'] print(sorted(dic, key=lambda k: dic[k][0], reverse=True)) # ['c', 'b', 'a']
實例
一、給定一個只包含大小寫字母,數字的字符串,對其進行排序,保證:
全部的小寫字母在大寫字母前面;
全部的字母在數字前面;
全部的奇數在偶數前面;
>>> s = "Sorting1234" >>> "".join(sorted(s, key=lambda x: (x.isdigit(), x.isdigit() and int(x) % 2 == 0, x.isupper(), x.islower(), x))) 'ginortS1324'
二、排序一副FranchDeck。流暢的python書中示例
import collections Card = collections.namedtuple('Card', ['rank', 'suit']) #命名元組 class FrenchDeck: ranks = [str(n) for n in range(2, 11)] + list('JQKA') #2到10和JQKA suits = 'spades diamonds clubs hearts'.split() #split默認按空格生成列表 def __init__(self): self._cards = [Card(rank, suit) for suit in self.suits for rank in self.ranks] #生成全部52張牌 def __len__(self): return len(self._cards) #能夠用Len()方法 def __getitem__(self, position): return self._cards[position] #能夠用索引或切片 suit_values = dict(spades=3, hearts=2, diamonds=1, clubs=0) #設置花色優先級 def spades_high(card): rank_value = FrenchDeck.ranks.index(card.rank) #找到當前Card所在的索引:2的索引爲0 return rank_value * len(suit_values) + suit_values[card.suit] #返回當前牌對應的優先級數字,若是沒有乘於len(suit_values)那麼('2', 'spades')和('4', 'diamonds')的優先級數字就同樣了,這裏也能夠直接乘於4或5,6.... #排序 >>> import Frenchcard >>> deck = Frenchcard.FrenchDeck() >>> for i in sorted(deck, key=b.spades_high): #sorted返回一個列表。key後的函數會代入deck列表裏的每一項返回一個優先級數字,並以這個優先級數字排序 ... print(i) Card(rank='2', suit='clubs') Card(rank='2', suit='diamonds') Card(rank='2', suit='hearts') Card(rank='2', suit='spades') Card(rank='3', suit='clubs') Card(rank='3', suit='diamonds') .............