python学习笔记 map&&reduce

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1.map

1)map其实相当对吧运算符进行一个抽象，返回的是一个对象，但是这里不知道为什么不可以对一个map返回变量打印两次，难道是因为回收了？

def f(x):
    return x*x

tmp = map(f,range(6))
tmps = map(str,range(6))
print (list(tmp))
#print (list(tmps))
print (type(range(6)))#range返回的就是range类型<class 'range'>
print (set(tmps))
print(list(tmps))#此时的输出不在是range6的List而是一个空的[]

2）reduce 需要两个以上的参数才能使用，一般是作用域一个list上的，比如下面的求和

 tadd = reduce(add,range(6))
 print (tadd)
 tadd = reduce(add, [x for x in range(6) if x%2 != 0])
 print (tadd)

reduce把一个函数作用在一个序列[x1, x2, x3, ...]上，这个函数必须接收两个参数，reduce把结果继续和序列的下一个元素做累积计算，其效果就是：

reduce(f, [x1, x2, x3, x4]) = f(f(f(x1, x2), x3), x4)

 # -*- coding: utf-8 -*-
 """
 Spyder Editor

 This is a temporary script file.
 """
 from functools import reduce
 tadd = reduce(add,range(6))
 print (tadd)
 tadd = reduce(add, [x for x in range(6) if x%2 != 0])
 print (tadd)
 tadd = reduce(fn,[x for x in range(9) if x%2 != 0])
 print (tadd)

3）对于一些版本的简化：

 from functools import reduce
 def f(x):
     return x*x

 def add(x1,x2):
     return x1+x2
 def fn(x1,x2):
     return 10*x1+x2

 def char2num(s):
     return {'':0,'':1,'':2,'':3,'':4,'':5,'':6,'':7,'':8,'':9}[s]

 def str2int(s):
     def char2num(s):
         return {'':0,'':1,'':2,'':3,'':4,'':5,'':6,'':7,'':8,'':9}[s]

     def fn(x1,x2):
         return 10*x1+x2
     return reduce(fn,map(char2num,s))
 def str2int_2(s):
     return reduce(lambda x,y:x*10+y,map(char2num,s))

 tmp = map(f,range(6))
 tmps = map(str,range(6))
 print (list(tmp))
 #print (list(tmps))
 print (type(range(6)))#range返回的就是range类型<class 'range'>
 print (set(tmps))
 print(list(tmps))#此时的输出不在是range6的List而是一个空的[]

 tadd = reduce(add,range(6))
 print (tadd)
 tadd = reduce(add, [x for x in range(6) if x%2 != 0])
 print (tadd)
 tadd = reduce(fn,[x for x in range(9) if x%2 != 0])
 print (tadd)
 print (reduce(fn,(map(char2num,''))))
 print (str2int(''))
 print (str2int_2(''))

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