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':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9}[s]

def str2int(s):
    def char2num(s):
        return {'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':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,'213'))))
print (str2int('1231'))
print (str2int_2('321'))

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