Some small details of a Python lambda expression

Warm so know new, inadvertently found before experiment Lambda writing test code, the first reflection is, this is I write????!!

Hehe, think of xx language just put Lambda formally add in, Python early support, I can yell "Python is the best language" to find scold it?

Ha ha.

However, since having lambda, a lot of code has been done in one line. However, there are a lot of small details not for the attention of the general people, see the following code:

# coding=utf-8# Gabriel G=lambda x,y:x+ytotal=0;for i in range (1,101):      #print  i    total=g (total,i)     print total# Even if you define the same lambda method, they are not the same g1=lambda x,y:x+yg2=lambda x,y:x+yprint g1print id (G1) print  G2print id (G2) #lambda方法嵌套定义, the lambda method does not apply a variable reference and calls print  directly (lambda x,y:x+ ((lambda a:a+1) (y))) ( #lambda方法定义不可以回车, which means that code like for this cannot appear in the lambda method #error# lambda x,y:#     num =0#     for i in range (x,y+1) #          num+=i# all the values in the sequence parameters that meet the true criteria in the iteration are the # tuple and list Print filter (None, (1,0,true,false)) Print filter ( None,[1,0,true,false]) #字典的奇葩结果: Returns a list of keys and evaluates to the type of the first conforming element as the Datum print filter (none,{"a": 1, "B": 0, "C": True, "D ": false}") Print filter (none,{1: ' s ', 0: ' s ', True: ' s ', false: ' s '}) Print filter (none,{true: ' s ', false: ' s ', 1: ' s ', 0: ' s '}) Print filter (NOne,{0: ' s ', 1: ' s ', true: ' s ', False: ' s '} ' Print filter (none,{false: ' s ', 0: ' s ', 1: ' s ', true: ' s '}) print filter (none,{false: ' s ', 0: ' s ', 1: ' s ', True: ' s ', 2: ' s ', 1.23: ' s ', ' KK ': ' s ', 0.00: ' s '}) Print filter (lambda x:x%2,{1 : ' s ', 2: ' s ', 3: ' s ', 4: ' s '}) #取反print  filter (lambda x:not x,[1,0,true,false] ) #取奇数和偶数print  filter (lambda x:x%2, [1,2,3,4,5,6,7]) Print filter (lambda x: (x+1)%2, [ 1,2,3,4,5,6,7]) #print  filter (Lambda [x,y]:(x+y)%2, [[1,2],[3,4],[5,6],[7,1]) errorprint  Map (None,range) Print map (Lambda x:x*100,range (Ten)) Print map (lambda x:x*100+ (x%2) * (-1) + ((x+1)%2) *1,range (Ten) Print map (Lambda x:0,range ()) Print map (Lambda x:[x-1,x,x+1],range ( ) Print map (lambda x: (x-1,x,x+1), Range (Ten)) #元组会变成元素相同的列表print  map (None, (1,2,3,4,5)) #即使返回的列表相同 , the address is different, that is, two lists are not the same l1 = [1, 2, 3, 4, 5, 6]l2 = map (NONE, L1 ) Print l1,id (L1) print&Nbsp;l2,id (L2) 

The

Output results are as follows: (I will insert the output counterpart in the time of the day, and now please make sure to see it)

5050<function <lambda> at 0xb7487bfc>3074980860<function <lambda>  at 0xb7487f7c>30749817564 (1, true) [1, true][' A ',  ' C ',  ' B ',  ' d '][1][True ][1][1][1, 2,  ' KK ', 1.23][1, 3][0, false][1, 3, 5, 7][2, 4,  6][0,&NBSP;1,&NBSP;2,&NBSP;3,&NBSP;4,&NBSP;5,&NBSP;6,&NBSP;7,&NBSP;8,&NBSP;9][0,&NBSP;100,&NBSP;200,&NBSP;300,  400, 500, 600, 700, 800, 900][1, 99, 201, 299, 401,  499, 601, 699, 801, 899][0, 0, 0, 0, 0, 0, 0, 0, 0 ,  0][[-1, 0, 1], [0, 1, 2], [1, 2, 3], [2, 3, 4],  [3, 4, 5], [4, 5, 6], [5, 6, 7], [6, 7, 8],  [7, 8, 9], [8, 9, 10]][( -1, 0, 1),  (0, 1, 2),  (1,  2,&NBSP;3), (2, 3, 4),  (3, 4, 5),  (4, 5, 6),  (5, 6, 7),  (6,  7, 8),  (7, 8, 9),  (8, 9, 10)][1, 2, 3, 4, 5][1,  2, 3, 4, 5, 6] 3074533388[1, 2, 3, 4, 5, 6]  3074983436

Some small details of a Python lambda expression