在Python中,if语句的语法如下: ifcondition:# code block executed if condition is trueelse:# code block executed if condition is false 1. 2. 3. 4. 其中,condition是一个表达式,可以是任何能够返回布尔值(True或False)的表达式。如果condition为真(True),则执行if语句下的代码块;如果condition为假(False)...
float'log.log_main('error', is_send_email, log_content)raiseException(log_content)ifjudge_retry_conditon_typein('==','!=','>=','<=','>','<','in','not in'):returngroup_response_statusreturngroup_response_contentdefretry_condition_and_exec_result_dispose(retry_condition,exec_result)...
"""WHILE statement @see: https://docs.python.org/3/tutorial/controlflow.html @see: https://docs.python.org/3/reference/compound_stmts.html#the-while-statement The while loop executes as long as the condition remains true. In Python, like in C, any non-zero integer value is true; zero...
# coding:utf-8 def judge_value(num_value): """自定义异常函数 自定义异常函数,用于抛出大于一定值的异常 :param num_value:用于判断的值 :return:异常信息 """ if num_value > 10: # raise用于抛出自定义异常,格式为:raise 异常类型(异常注明) # 一旦触发则不再执行raise之后的代码 raise ValueError("...
while True: cn = CN(A,c,x_b,c_b,m,n) theta = theTa(cn,A,m) num += 1 Excel(A,c_b,x_b,c,theta,cn,m,n) result = judge(A,cn,m,n,x_b,c_b) if result != 0: break theta = theTa(cn,A,m) A = baseChange(A,cn,theta,m,n) ...
If that condition proves to be TRUE, the number will be considered for multiplication by 2. In this case, our condition is that the number can be divided by 2 (without having any remainder of the division operation). Now, in a more general sense, you can also use list comprehension to...
balance_condition = [["布","布"],["石头","石头"],["剪刀","剪刀"]]# 每次平局的情况 computer_win_number =0# 电脑赢的次数 people_win_number =0# 玩家赢得次数 #*** Begin ***# forxinrange(0,7): if[guess_list[x],people_guess[x]] == computer_win_condition[0]: computer_win_...
2,ACM 的 OnlineJudge,有点难度 3,ZOJ,浙江大学的网站 4,Kaggle,一个竞赛类网站 当然还有其他网站,就先推荐以上网站吧,经过更深入的学习,也许大家就能找到更适合自己的学习网站。 3,项目实战 git上有很多项目,都可以自己去做做试试,项目的选择可以根据自己的实际情况选,你要学习爬虫,就多练练爬虫的项目,想学...
Write a program which accepts a string as input to print "Yes" if the string is "yes" or "YES" or "Yes", otherwise print "No". Hints: Use if statement to judge condition. Solution s= raw_input() if s=="yes" or s=="YES" or s=="Yes": print "Yes" else: print "No" Ques...
check len(items) for error condition add user error message, which is necessarily more helpful that a generic "used more than once" (optional) class UseOnceExtendAction(argparse._ExtendAction): def __init__( self, option_strings, dest, nargs=None, const=None, default=None, type=None, cho...