anonymous function-lambda

anonymous function-lambda

The anonymous function lambda [arg1 [, arg2,..... argn]]:expression Lambda is just an expression, and the function body is much simpler than def. The body of a lambda is an expression, not a block of code.    Only a finite amount of logic can be encapsulated in a lambda expression.    The lambda function has its own namespace and cannot access parameters outside its own argument list or in the global namespace.    Although the lambda function appears to be only a single line, it is not equivalent to a C or C + + inline function, which is designed to call small functions without consuming stack memory to increase operational efficiency.     Exp:dic = {' K1 ': Ten, ' K2 ': +, ' K3 ': +, ' K4 ': +} print (max (Dic,key=lambda K:dic[k])) The built-in function has the following key parameters:    Max min Filter map sorted above can be combined with lambda to complete some functions exp2: Existing two tuple tup1,tup2, use anonymous function production list lst Tup1 = ((' A '), (' B ')) Tup2 = ((' C '), (' d ')) LST = [{' A ': ' C '},{' B ': ' d '}] 1: LST written by yourself = lambda x, y: [{(x[i][0]): y[i][0]} for I in rang E (len (x))] Print (LST (tup1,tup2)) [{' A ': ' C '}, {' B ': ' d '}] notation 2:tup1 = ((' A '), (' B ')) Tup2 = ((' C '), (' d ') # lst = [{' A ': ' C '},{' B ': ' d '}] # ret = Zip (tup1,tup2) # def func (t): # return {t[0]:t[1]} # func = Lam BDA T:{t[0]:t[1]} # res = map (lambda t:{t[0]:t[1]},zip (tup1,tup2)) Print (list (map (Lambda T:{t[0]:t[1]},zip (tup1,tup2 ))) Exp3. The following codeWhat is the output?    Please give an answer and explain. def multipliers (): return [lambda x:i*x for I in range (4)] Print ([M (2) for M in Multipliers ()]) please modify the multiplier    s is defined to produce the desired result. Resolution: Def multipliers (): return [lambda x:i *x for I in range (4)] #返回列表推导式 [Lambda x:i *x,lambda x: I *x,lambda x:i *x,lambda x:i *x] Print ([M (2) for M in Multipliers ()]) # [M (2) to M in [Lambda x:i *x , Lambda x:i *x,lambda x:i *x,lambda x:i *x]] # MULTIPLIERSW () for [Lambda x:i *x,lambda x:i *x,lambda x:i *  X,lambda x:i *x] # [M (2) for M in [Lambda x:i *x,lambda x:i *x,lambda x:i *x,lambda x:i *x]] # m is Lambda X:i *m, 4 times # m (2) [Lambda x:i *2,lambda x:i *2,lambda x:i *2,lambda x:i]] #而此时multip The For Loop in Liers () is finished, and the last value of I is 3 # so [Lambda x:3 *2,lambda x:3 *2,lambda x:3 *2,lambda x:3 *]] #[6, 6, 6 , 6] exp4. What is the output of the following code?        Please give an answer and explain. def multipliers (): Return (lambda x:i *x for I in range (4)) #返回一个生成器 g = (lambda x:i *x for I in range (4) print ([M (2) to M in Multipliers ()]) # [M (2)-M in Mu  Ltipliers ()] # [M (2) for M in (Lambda x:i *x-I in range (4)] # [M (2) for M in (Lambda x:0 *x,lambda x : 1 *x,lambda x:2 *x,lambda x:3 *x] # [2 * 0,,2 * 1, 2 * 2,2 * 3] #[0,2,4,6]

　　

anonymous function-lambda