Python:通过计算阶乘来学习lambda和reduce这两个函数的使用
从上学时开始,通常是用C来求阶乘,今天无事,用Python写了一下,主要在于学习lambda和reduce这两个函数的使用。
其实lambda就是个匿名函数,它本身是一个表达式,而def为一个语句,这就是说lambda可以用于函数中做为参数等,但def这个语句不能。 就是用函数function对序列,如list中的元素进行处理,每次处理两个数据项(一个是前次处理的结果,一个是序列中的下一个元素),如此反复的递归处理,最后对整个序列求出一个单一的返回值。 运行结果:
实现:
- #!/usr/bin/env python
- #-*- coding: utf-8 -*-
- import time
- def test_factorial_reduce():
- '''''
- Function:使用reduce函数
- Input:NONE
- Output: NONE
- author: socrates
- blog:http://blog.csdn.net/dyx1024
- date:2012-02-19
- '''
- time_begin = time.clock()
- print reduce(lambda x,y:x*y, range(1, long(raw_input("please input a number ( > 0):"))))
- print "Use time: %s" % (time.clock() - time_begin)
- return;
- def test_factorial_math():
- '''''
- Function:使用math库函数
- Input:NONE
- Output: NONE
- author: socrates
- blog:http://blog.csdn.net/dyx1024
- date:2012-02-19
- '''
- import math
- time_begin = time.clock()
- print math.factorial(long(raw_input("please input a number ( > 0):")))
- print "Use time: %s" % (time.clock() - time_begin)
- if __name__ == '__main__':
- print '*' * 50 + "Use reduce" + '*' * 50
- test_factorial_reduce()
- print '*' * 50 + "Use math api" + '*' * 50
- test_factorial_math()
测试:
- **************************************************Use reduce**************************************************
- please input a number ( > 0):50
- 608281864034267560872252163321295376887552831379210240000000000
- Use time: 3.3163365735
- **************************************************Use math api**************************************************
- please input a number ( > 0):50
- 30414093201713378043612608166064768844377641568960512000000000000
- Use time: 3.70409004496
我们感兴趣的不在程序本身,我想更多地会关注lambda和reduce这两个函数,我们来看一下:
lambda
手册中这样描述:
- lambda
- An anonymous inline function consisting of a single expression which is evaluated when the function is called. The syntax to create a lambda function is lambda [arguments]: expression
我们的 lambda x,y:x*y 语句等价于下面这个函数:
- def my_func(x, y):
- return x * y
reduce
手册中这样描述:
- reduce(function, iterable[, initializer])
- Apply function of two arguments cumulatively to the items of iterable, from left to right, so as to reduce the iterable to a single value. For example, reduce(lambda x, y: x+y, [1, 2, 3, 4, 5]) calculates ((((1+2)+3)+4)+5). The left argument, x, is the accumulated value and the right argument, y, is the update value from the iterable. If the optional initializer is present, it is placed before the items of the iterable in the calculation, and serves as a default when the iterable is empty. If initializer is not given and iterable contains only one item, the first item is returned.
改写一下上面的程序,很快可以理解这句话的意思:
- def my_func(x, y):
- print 'x = %d, y = %d' %(x, y)
- return x * y
- print reduce(my_func, range(1, long(raw_input("please input a number ( > 0):"))))
- please input a number ( > 0):50
- x = 1, y = 2
- x = 2, y = 3
- x = 6, y = 4
- x = 24, y = 5
- x = 120, y = 6
- x = 720, y = 7
- x = 5040, y = 8
- x = 40320, y = 9
- x = 362880, y = 10
- x = 3628800, y = 11
- x = 39916800, y = 12
- x = 479001600, y = 13
- x = 6227020800, y = 14
- x = 87178291200, y = 15
- x = 1307674368000, y = 16
- x = 20922789888000, y = 17
- x = 355687428096000, y = 18
- x = 6402373705728000, y = 19
- x = 121645100408832000, y = 20
- x = 2432902008176640000, y = 21
- x = 51090942171709440000, y = 22
- x = 1124000727777607680000, y = 23
- x = 25852016738884976640000, y = 24
- x = 620448401733239439360000, y = 25
- x = 15511210043330985984000000, y = 26
- x = 403291461126605635584000000, y = 27
- x = 10888869450418352160768000000, y = 28
- x = 304888344611713860501504000000, y = 29
- x = 8841761993739701954543616000000, y = 30
- x = 265252859812191058636308480000000, y = 31
- x = 8222838654177922817725562880000000, y = 32
- x = 263130836933693530167218012160000000, y = 33
- x = 8683317618811886495518194401280000000, y = 34
- x = 295232799039604140847618609643520000000, y = 35
- x = 10333147966386144929666651337523200000000, y = 36
- x = 371993326789901217467999448150835200000000, y = 37
- x = 13763753091226345046315979581580902400000000, y = 38
- x = 523022617466601111760007224100074291200000000, y = 39
- x = 20397882081197443358640281739902897356800000000, y = 40
- x = 815915283247897734345611269596115894272000000000, y = 41
- x = 33452526613163807108170062053440751665152000000000, y = 42
- x = 1405006117752879898543142606244511569936384000000000, y = 43
- x = 60415263063373835637355132068513997507264512000000000, y = 44
- x = 2658271574788448768043625811014615890319638528000000000, y = 45
- x = 119622220865480194561963161495657715064383733760000000000, y = 46
- x = 5502622159812088949850305428800254892961651752960000000000, y = 47
- x = 258623241511168180642964355153611979969197632389120000000000, y = 48
- x = 12413915592536072670862289047373375038521486354677760000000000, y = 49
- 608281864034267560872252163321295376887552831379210240000000000
相关推荐
IT之家 2020-03-11
graseed 2020-10-28
zbkyumlei 2020-10-12
SXIAOYI 2020-09-16
jinhao 2020-09-07
impress 2020-08-26
liuqipao 2020-07-07
淡风wisdon大大 2020-06-06
yoohsummer 2020-06-01
chenjia00 2020-05-29
baike 2020-05-19
扭来不叫牛奶 2020-05-08
hxmilyy 2020-05-11
黎豆子 2020-05-07
xiongweiwei00 2020-04-29
Cypress 2020-04-25
冰蝶 2020-04-20