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Intersection of two Counters

由 二次信任 提交于 2020-01-13 08:52:10
问题 I'm trying to find the shared elements (and the shared number of occurrences) between two lists. For example, the intersection of these two lists: a = [1, 1, 2, 3, 4, 5, 6, 7, 8, 1] b = [1, 1, 3, 5, 7, 9] should return Counter({1: 2, 3: 1, 5: 1, 7: 1}) or something similar, e.g. {1: 2, 3: 1, 5: 1, 7: 1} or [1, 1, 3, 5, 7] (order of the list doesn't matter). I already have an approach that works: cnts_a = Counter(a) cnts_b = Counter(b) cnts_a_b = Counter() # counter for the shared values for

read time stamp counter

由 萝らか妹 提交于 2020-01-13 04:23:33
　　在Pentium以上的CPU中，提供了一条机器指令RDTSC（Read Time Stamp Counter）来读取这个时间戳的数字，并将其保存在EDX:EAX寄存器对中。由于EDX:EAX寄存器对恰好是Win32平台下C++语言保存函数返回值的寄存器，所以我们可以把这条指令看成是一个普通的函数调用。像这样： inline unsigned __int64 GetCycleCount() { __asm RDTSC } 但是不行，因为RDTSC不被C++的内嵌汇编器直接支持，所以我们要用_emit伪指令直接嵌入该指令的机器码形式0X0F、0X31，如下： inline unsigned __int64 GetCycleCount() { __asm _emit 0x0F __asm _emit 0x31 } 来源： https://www.cnblogs.com/gaoqichao/archive/2012/12/10/2812050.html

Get the index (counter) of an 'ng-repeat' item with AngularJS?

由 假如想象 提交于 2020-01-11 18:18:17
问题 I am using AngularJS and its ng-repeat directive to display a sequence of questions. I need to number each question starting with 1 . How do I display and increment such a counter with ng-repeat ? Here is what I have so far: <ul> <li ng-repeat="question in questions | filter: {questionTypesId: questionType, selected: true}"> <div> <span class="name"> {{ question.questionText }} </span> </div> <ul> <li ng-repeat="answer in question.answers"> <span class="name"> {{answer.selector}}. {{ answer

Datetime does not show up

由 旧巷老猫 提交于 2020-01-11 14:06:12
问题 I am trying to get timestamp to show- I have tried the onCreate query in different ways and also tried to to have addTime as a value in addPrime . Nothing seems to work. My intention is for the app to show previous primes and the time that they were found. The intention for the app is for the user to be able to close/kill the app and resume counting from last found prime number when restarting the app, if you have any hints how also this would be possible I would be grateful. This is the

Datetime does not show up

由 戏子无情 提交于 2020-01-11 14:05:40
问题 I am trying to get timestamp to show- I have tried the onCreate query in different ways and also tried to to have addTime as a value in addPrime . Nothing seems to work. My intention is for the app to show previous primes and the time that they were found. The intention for the app is for the user to be able to close/kill the app and resume counting from last found prime number when restarting the app, if you have any hints how also this would be possible I would be grateful. This is the

Datetime does not show up

由 我与影子孤独终老i 提交于 2020-01-11 14:05:10
问题 I am trying to get timestamp to show- I have tried the onCreate query in different ways and also tried to to have addTime as a value in addPrime . Nothing seems to work. My intention is for the app to show previous primes and the time that they were found. The intention for the app is for the user to be able to close/kill the app and resume counting from last found prime number when restarting the app, if you have any hints how also this would be possible I would be grateful. This is the

how to create a counter in a dr Java program

由 梦想与她 提交于 2020-01-11 11:36:12
问题 /*This is a quiz program that will ask the user 10 questions. the user will answer * these questions and will be scored out of 10.*/ class Quiz { public static void main(String args[]) { // Instructions System.out.println("instructions"); System.out.println(" "); System.out .println("1. You wll be asked ten questions through out the quiz."); System.out .println("2. The first question will appear, you will have to answer that question for the next question to appear."); System.out .println("3.

a to the power of b without (a**b), Python

由 淺唱寂寞╮ 提交于 2020-01-10 06:10:53
问题 Struggling with an exercise that asks me to write a**b without this operator. Tried to write something myself but am not getting correct results. Instead of one value am getting two, both incorrect. Seems like the counter doesnt really increase. May I ask for help? Thanks! def powerof(base,exp): result=1 counter=0 # until counter reaches exponent, go on if counter<=exp: # result multiplies itself by base, starting at 1 result=result*base # increase counter counter=counter+1 return result

a to the power of b without (a**b), Python

由 北城余情 提交于 2020-01-10 06:10:15
问题 Struggling with an exercise that asks me to write a**b without this operator. Tried to write something myself but am not getting correct results. Instead of one value am getting two, both incorrect. Seems like the counter doesnt really increase. May I ask for help? Thanks! def powerof(base,exp): result=1 counter=0 # until counter reaches exponent, go on if counter<=exp: # result multiplies itself by base, starting at 1 result=result*base # increase counter counter=counter+1 return result

less10 loop循环

由 萝らか妹 提交于 2020-01-08 22:03:22
less .loop(@counter) when (@counter > 0) { .loop((@counter - 1)); // 递归调用自身 4 3 2 1 0 width: (10px * @counter); // 每次调用时产生的样式代码 50px 40px 30px 20px 10px } div { .loop(5); // 调用循环 } .loop1(@counter) when (@counter > 0) { h@{counter}{ padding: (10px * @counter); }// 每次调用时产生的样式代码 .loop1((@counter - 1)); // 递归调用自身 } div { .loop1(6); // 调用循环 } .loop2(@counter) when (@counter < 7) { h@{counter}{ padding: (10px * @counter); }// 每次调用时产生的样式代码 .loop2((@counter + 1)); // 递归调用自身 } div { .loop2(1); // 调用循环 } css div { width: 10px; width: 20px; width: 30px; width: 40px; width: 50px; } div h6 { padding: 60px
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