14. Python custom sorting and two sort methods

Data_list = []data_list.append ({' Softname ': ' 1 ', ' Version ': ' 1.2.2.2 '}) data_list.append ({' Softname ': ' 7 ', ' Version ': ' 1.2.2.2 '}) data_list.append ({' Softname ': ' 5 ', ' Version ': ' 1.2.2.2 '}) data_list.append ({' Softname ': ' 2 ', ' Version ': ' 1.2.2.2 '}) data_list.append ({' Softname ': ' 3 ', ' Version ': ' 1.2.2.2 '}) data_list.append ({' Softname ': ' 9 ', ' Version ': ' 1.2.2.2 '}) #升序data_list. Sort (Key=lambda obj:obj.get (' Softname '), reverse=false) print data_list#[{' softname ': ' 1 '}, { ' Softname ': ' 2 '}, {' Softname ': ' 3 '}, {' Softname ': ' 5 '}, {' Softname ': ' 7 '}, {' Softname ': ' 9 '}] #降序data_list. Sort (key= Lambda obj:obj.get (' softname '), reverse=true) print data_list#[{' softname ': ' 9 '}, {' Softname ': ' 7 '}, {' Softname ': ' 5 '}, {' Softname ': ' 3 '}, {' Softname ': ' 2 '}, {' Softname ': ' 1 '}] to sort list, Python provides two method method 1. Sort by the list's built-in function List.sort List.sort (Func=none, Key=none, Reverse=false) Python instance:123456>>> list = [2,5,8,9,3] >>> list [2,5,8,9,3] >>> list.sort () >>> lis t [2, 3, 5, 8, 9] Method 2. Sort by the sequence type function sorted (list) (from 2.4 start) Python instance:12345>>> list = [2,5,8,9,3] >>> list [2,5,8,9,3] >>> sorted (list) [2, 3, 5, 8, 9] The difference between the two methods: sorted (list) Returns an object that can be used as an expression. The original list does not change, creating a new sorted list object. List.sort () does not return an object, changing the original list. Other sort instances: Instance 1: Forward Sort 1234>>>l = [2,3,1,4]>>>l.sort () >>>l>>>[1,2,3,4] Instance 2: Reverse Sort 1234>>>l = [2,3,1,4]>>>l.sort (reverse=true) >>>l>>>[4,3,2,1] Instance 3: Sort the second keyword 1234>>>l = [(' B ', 6), (' A ', 1), (' C ', 3), (' d ', 4)]>>>l.sort (Lambda x,y:cmp (x[1],y[1])) >>>l >>>[(' A ', 1), (' C ', 3), (' d ', 4), (' B ', 6)] Instance 4: Sort the second keyword 1234>>>l = [(' B ', 6), (' A ', 1), (' C ', 3), (' d ', 4)]& Gt;>>l.sort (Key=lambda x:x[1]) >>>l>>>[(' A ', 1), (' C ', 3), (' d ', 4), (' B ', 6)] Instance 5: Sort the second keyword 1234 5>>>l = [(' B ', 2), (' A ', 1), (' C ', 3), (' d ', 4)]>>>import Operator>>>l.sort (key= Operator.itemgetter (1)) >>>l>>>[(' A ', 1), (' B ', 2), (' C ', 3), (' d ', 4)] Instance 6: (DSU method: Decorate-sort-undercorate) 123456>>>l = [(' B ', 2), (' A ', 1), (' C ', 3), (' d ', 4)]>>>a = [(x[1],i,x) for i,x in enumerate (  L)] #i can confirm the stable sort>>>a.sort () >>>l = [s[2] for S in a]>>>l>>>[(' A ', 1), (' B ', 2), (' C ', 3), (' d ', 4)] above gives the method of ordering the list in 6, where instance 3.4.5.6 can be used to sort an item in list item as the comparison keyword. Comparison of efficiencies: CMP < DSU < key by experiment , Method 3 is slower than method 6, method 6 is slower than Method 4, Method 4 and Method 5 are basically quite a few keywords comparison sort: instance 7:1234>>>l = [(' d ', 2), (' A ', 4), (' B ', 3), (' C ', 2)]>>> L.sort (Key=lambda x:x[1]) >>> l>>>[(' d ', 2), (' C ', 2), (' B ', 3), (' A ', 4)] we see that at this time the sort of L is ordered only by the second keyword, What if we want to order the second keyword and then sort it with the first keyword? There are two methods of example 8:1234>>> L = [(' d ', 2), (' A ', 4), (' B ', 3), (' C ', 2)]>>> L.sort ( Key=lambda x: (X[1],x[0])) >>> l>>>[(' C ', 2), (' d ', 2), (' B ', 3), (' A ', 4)] instance 9:1234>>> L = [(' d ', 2), (' A ', 4), (' B ', 3), (' C ', 2)]>>> L.sort (Key=operator.itemgetter (1,0)) >>> l>>>[(' C ', 2), ( ' d ', 2), (' B ', 3), (' A ', 4)] Why does instance 8 work? The reason is that the comparison of the tuple is from left to right, compares the first, if equal, compares the second 

14. Python custom sorting and two sort methods