arr = [i for i in range(10), 1,[]] #注意, i for in xx 这个必须放在第一个位置,否则要先定义i, 如: arr = [i for i in range(5), j for j in range(5), []] 这是错误的 i = 0 j = 0 arr = [i for i in range(5), j for j in range(5), []] 这是正确的 c、del...
#Getthemodulenamespace(dict)early;thisispartofthetypecheck modns=mod.__dict__ #Parseitintopackagenameandmodulename,e.g.'foo.bar'and'whatever' i=modname.rfind(".") ifi>=0: pkgname,modname=modname[:i],modname[i+1:] else: pkgname=None ...
第一步写出who where what 的NFA 画出状态图,添加空步骤ε 使用ε-closure 算法写出所有可能的状态 步骤一,初始化阶段q0 DNF NFA 字符集:whatore 步骤二,链路w经过 步骤ε有h 的NFA 集合为S2 步骤三,链路h经过步骤ε 分别有S4 S5 S6 集合NFA 步骤四,链路o a e 分别经过步骤步骤ε 已经有确定的状态机S7...
Re: CONTEST - What is the (best) solution? On Wed, 02 Feb 2005 02:35:03 -0800, python wrote:[color=blue] > Each pair in a dictionary is separated by CRLF and in each dictionary > numbers of pairs can be different. > I need to read only the the first and the last dictionaries....