ni

Python: “global name 'time' is not defined”

匿名 (未验证) 提交于 2019-12-03 02:24:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I'm writing a silly program in python for a friend that prints "We are the knights who say 'Ni'!". then sleeps for 3 seconds, and then prints "Ni!" twenty times at random intervals using the random module's uniform() method. Here's my code: from time import sleep import random def knights_of_ni(): generator = random.Random() print "We are the knights who say 'ni'." sleep(3) for i in range(0,20): print "Ni!" sleep(generator.uniform(0,2)) I've tried to import this module by typing in the interpreter from silly import knights_of_ni() and import

How to debug: w3wp.exe process was terminated due to a stack overflow (works on one machine but not another)

匿名 (未验证) 提交于 2019-12-03 02:13:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: The problem I have an ASP.NET 4.0 application that crashes with a stack overflow on one computer, but not another. It runs fine on my development environment. When I move the site to the production server, it throws a stack overflow exception (seen in event log) and the w3wp.exe worker process dies and is replaced with another. What I've tried so far For reference, I used the debug diagnostic tool to try to determine what piece of code is causing the overflow, but I'm not sure how to interpret the output of it. The output is

CPU usage of w3wp rises to 100% on a ASP.NET MVC4 web application

匿名 (未验证) 提交于 2019-12-03 01:20:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: We have a ASP.NET MVC4 application which uses Entity Framework 4.0. When we deploy on the production server, the CPU rises after some time (~2-5h) until nearly 100%. The memory also rises then until maximum. After a while the application pool will be resetted automatically because of the high memory. Server: Windows 2008 R2 Standard SP1 IIS: 7 (V 7.5.7600.16385) Only one application pool and one webapplication is running. Stacktrace from Debug Diagnostic Tool of the thread which consumes most of the CPU: SNIReadSyncOverAsync(SNI_ConnWrapper*

undefined method `gsub' for nil:NilClass (NoMethodError

匿名 (未验证) 提交于 2019-12-03 01:03:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I have the below code snippet: line_sub = Regexp.new(/\s+|"|\[|\]/) tmp = Array.new # reading a file while line = file.gets ... tmp[0],tmp[1] = line.to_s.scan(/^.*$/).to_s.split('=') #remove unwanted characters tmp.collect! do |val| val.gsub(line_sub, "") end ... end but when I run the code I get the error: undefined method `gsub' for nil:NilClass (NoMethodError) something seems to be wrong here: tmp.collect! do |val| val.gsub(line_sub, "") end Any idea? 回答1: try this way for your solution tmp.collect! do |val| if val.present? # or unless

Tomcat thowing NoSuchMethodError for Nio2Endpoint.setSocketOptions after start serving

匿名 (未验证) 提交于 2019-12-03 00:58:01
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I have just updated a server where an exception started to happen after the application starts and tomcat starts serving requests. I changed the connection to verify if there was an isue with it as well but the same happened: 19-Apr-2018 20:31:38.833 SEVERE [http-nio2-8080-Acceptor-0] org.apache.tomcat.util.net.Nio2Endpoint.setSocketOptions java.lang.NoSuchMethodError: java.nio.ByteBuffer.clear()Ljava/nio/ByteBuffer; at org.apache.tomcat.util.net.SocketBufferHandler.reset(SocketBufferHandler.java:146) at org.apache.tomcat.util.net

MATLAB转换为Python

匿名 (未验证) 提交于 2019-12-02 22:51:30
SMOP SMOP 将matlab翻译成python。 SMOP 生成人类可读的蟒蛇,这似乎也比八度快。 速度有多快? 表1显示了“移动家具”的计时结果。似乎对于该程序,转换为python导致加速大约两倍,并且 将 SMOP 运行时库 编译 runtime.py 实现了额外的两倍加速 。 这个伪基准测量标量性能,而我的解释是标量计算对八度组不太感兴趣。 源代码 Working example $ cd smop/smop $ python main.py solver.m $ python solver.py solver.m solver.m a.py 01 function mv = solver(ai,af,w) 01 def solver_(ai,af,w, nargout = 1 ): 02 nBlocks = max(ai(:)); 02 nBlocks=max_(ai[:]) 03 [m,n] = size(ai); 03 m,n=size_(ai, nargout = 2 ) 02 Matlab uses round brackets both for array indexing and for function calls. To figure out which is which, SMOP computes local use-def information

python之集合(set)学习

匿名 (未验证) 提交于 2019-12-02 22:11:45
集合(set) 集合是一个无序的不重复元素序列,使用大括号({})、set()函数创建集合, 注意 :创建一个空集合必须用set()而不是{},因为{}是用来创建一个空字典。   集合是无序的、不重复的、没有索引的 1 a = {'hello','ni','hao','hi','ni','hao'} 2 3 print(a) # 输出结果没有重复项,且无序 4 # print(a[1]) # TypeError: 'set' object does not support indexing 输出结果: {'ni', 'hao', 'hello', 'hi'} 添加集合元素   添加单个元素: 1 a = {'hello','ni','hao','hi','ni','hao'} 2 print(a) 3 4 a.add('wo') 5 print(a) 6 7 a.add('hi') # 如果元素已存在,则不进行任何操作。 8 print(a) 输出结果: {'hi', 'ni', 'hao', 'hello'} {'hello', 'hi', 'wo', 'ni', 'hao'} {'hello', 'hi', 'wo', 'ni', 'hao'}   添加多个元素、列表元素、字典元素 1 a = {'hello','ni','hao','hi','ni','hao'} 2

「BZOJ4242」水壶

半腔热情 提交于 2019-12-02 21:56:46
传送门 Luogu团队题链接 解题思路 考虑以图中的建筑为点建一张图,那么我们就只要求出这个图的 \(\text{Kruskal}\) 重构树然后按套路搞就行了。 重点是优化连边,因为直接连边边数就会太多 贼JB多 。 考虑 \(\text{BFS}\) ,我们把所有建筑都作为源点跑 \(\text{BFS}\) ,求出每个点的距离他最近的建筑以及这段距离。 然后只要两个相邻点的最近建筑不同,就把这两个最近建筑连边,边数就优化到了 \(O(4\times H\times W)\) 级别(思想参照 这题 ) 细节注意事项 这题我调了一上午,原因有: 太菜:并查集忘记合并。 太菜:忘记倍增。 太菜:忘记判 \(-1\) 。 参考代码 #include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <cctype> #include <cmath> #include <ctime> #include <queue> #define rg register using namespace std; template < typename T > inline void read(T& s) { s = 0; int f = 0; char c

ios swift中??空合运算符(Nil Coalescing Operator)

妖精的绣舞 提交于 2019-11-30 13:04:21
空合运算符 (a ?? b)将对可选类型 a 进行空判断,如果 a 不为nil就返回,否则就返回一个默认值 b。表达式 a 必须是 Optional 类型。默认值 b 的类型必须要和 a 存储值的类型保持一致。 空合运算符a??b实际上是对如下三目运算的的简短表达方法 a != nil ? a!:b 两者之间是等价的 例如: let defaultColorName = "red" var userDefinedColorName: String?//默认值为nil var colorNameToUser =userDefinedColorName ?? defaultColorName //userDefinedColorName 的值为空 所以 colorNameToUser的值为"red" 如果你分配一个非空值(non-nil)给 userDefinedColorName,再次执行空合运算,运算结果为封包在 userDefaultColorName 中的值,而非默认值。 userDefinedColorName = "green" colorNameToUser =userDefinedColorName ?? defaultColorName //userDefinedColorName 非空,因此colorNameToUser 的值为"green" 来源: https:/

【结论】区间和的和

 ̄綄美尐妖づ 提交于 2019-11-30 05:47:20
区间和的和 题目大意: 给出一个数组,求出所有区间和的总和 输入样例: 3 1 2 3 输出样例: 20 数据范围: 对于30%的数据: 1 ⩽ n ⩽ 100 1\leqslant n\leqslant 100 1 ⩽ n ⩽ 1 0 0 对于50%的数据: 1 ⩽ n ⩽ 1000 1\leqslant n\leqslant 1000 1 ⩽ n ⩽ 1 0 0 0 对于100%的数据: 1 ⩽ n ⩽ 100000 1\leqslant n\leqslant 100000 1 ⩽ n ⩽ 1 0 0 0 0 0 解题思路: 直接枚举所有区间会TLE 那我们从每个数下手 每个数可能在多少个区间中出现呢 首先可以发现区间的最前端要小于等于 i ( i 种 可 能 ) i(i种可能) i ( i 种 可 能 ) ,最后端要大于等于 i ( n − i + 1 种 可 能 ) i(n-i+1种可能) i ( n − i + 1 种 可 能 ) 于是我们直接那两个数相乘然后乘上 a [ i ] a[i] a [ i ] 在求和即可 代码: # include <cstdio> using namespace std ; long long n , a , ans ; int main ( ) { scanf ( "%ld" , & n ) ; for ( long long i = 1