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PYTHON编程问题,只求大神来

# -*- coding: GBK -*- def tostr(linelist): line = "[" + " ".join(linelist)+"]" return line def get_Listfromfile(filename): linelists = [] with open(filename) as f: for line in f: linelist = line.split() if linelist: linelists.a...

Python 3.6.1 (default, Mar 22 2017, 06:17:05) [GCC 6.3.0 20170321] on linuxType "help", "copyright", "credits" or "license" for more information.>>> from random import randint>>> nums = [randint(1, 200) for __ in range(5)]>>> p...

# -*- coding: gbk -*-def print_prime_factors(num): if num < 2: print '请输入大于 1 的整数用于质因数分解' return print '输出:', prime_num = 2 while prime_num

分为三步 产生随机数 筛选3位数,小于1000 排序输出 import randomx=[random.randint(1,10000) for i in xrange(1000)]y=[i for i in x if i

# 如果是对象的话,就像下面这样( ( '2010-05-09', # t[0][0] '\xe5\x.....' # t[0][1] ), [ 671379.0 # t[1][0] ])

def is_prime(n): i = 2 while(i < n): if n % i == 0: break i += 1 if n == i: return True else: return False p = 0for i in range(3,101): if is_prime(i): print i, p += 1 if p % 10 == 0: print '\n'

有一本《集体智慧编程》的书,里面有详细的讲解,而且有python的示例代码。 建议你看看。

用!表示阶乘,不重置的话,计算出来的是1!+1!2!+1!2!3!+1!2!3!4!+1!2!3!4!5!, 所以是34863。因为不重置,每次都在前一项的基础上乘了一个i!,但是我们只需要在前一项的基础上乘以i就行了,所以多乘了。重置的话,虽然是正确的,但是多此一举嘛...

红外接收头出来的是脉冲信号,需要单片机判断脉冲时间。输出的就是数字信号,或者叫电平信号,通过总线送到液晶屏的RAM里。

1 astr = "Iamatestcast" 2 print astr 3 lens = len(astr) 4 cnt = {} 5 for i in range (0, lens): 6 cnt.setdefault(ord(astr[i]), 0) 7 cnt[ord(astr[i])] += 1 8 9 maxCnt = 0 10 maxLetter = ord(astr[0]) 11 for k in cnt.keys(): 12 if...

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