python学习笔记 map&&reduce

Posted 2020-09-04

---恢复内容开始---

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

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

View Code

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)

 1 # -*- coding: utf-8 -*-
 2 """
 3 Spyder Editor
 4 
 5 This is a temporary script file.
 6 """
 7 from functools import reduce
 8 tadd = reduce(add,range(6))
 9 print (tadd)
10 tadd = reduce(add, [x for x in range(6) if x%2 != 0])
11 print (tadd)
12 tadd = reduce(fn,[x for x in range(9) if x%2 != 0])
13 print (tadd)

View Code

3）对于一些版本的简化：

 1 from functools import reduce
 2 def f(x):
 3     return x*x
 4 
 5 def add(x1,x2):
 6     return x1+x2
 7 def fn(x1,x2):
 8     return 10*x1+x2
 9 
10 def char2num(s):
11     return {‘0‘:0,‘1‘:1,‘2‘:2,‘3‘:3,‘4‘:4,‘5‘:5,‘6‘:6,‘7‘:7,‘8‘:8,‘9‘:9}[s]
12 
13 def str2int(s):
14     def char2num(s):
15         return {‘0‘:0,‘1‘:1,‘2‘:2,‘3‘:3,‘4‘:4,‘5‘:5,‘6‘:6,‘7‘:7,‘8‘:8,‘9‘:9}[s]
16         
17     def fn(x1,x2):
18         return 10*x1+x2
19     return reduce(fn,map(char2num,s))
20 def str2int_2(s):
21     return reduce(lambda x,y:x*10+y,map(char2num,s))
22 
23 tmp = map(f,range(6))
24 tmps = map(str,range(6))
25 print (list(tmp))
26 #print (list(tmps))
27 print (type(range(6)))#range返回的就是range类型<class ‘range‘>
28 print (set(tmps))
29 print(list(tmps))#此时的输出不在是range6的List而是一个空的[]
30 
31 tadd = reduce(add,range(6))
32 print (tadd)
33 tadd = reduce(add, [x for x in range(6) if x%2 != 0])
34 print (tadd)
35 tadd = reduce(fn,[x for x in range(9) if x%2 != 0])
36 print (tadd)
37 print (reduce(fn,(map(char2num,‘213‘))))
38 print (str2int(‘1231‘))
39 print (str2int_2(‘321‘))

View Code

 

---恢复内容结束---