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【pytest】命令行参数-durations统计用例运行时间
前言：写完一个项目的自动化用例之后，发现有些用例运行较慢，影响整体的用例运行速度，于是领导说找出运行慢的那几个用例优化下。--durations 参数可以统计出每个用例运行的时间，对用例的时间做个排序。 pytest -h 查看命令行参数，关于 --durations=N 参数的使用方式 --durations=NPytest-命令行参数--durations的使用
--durations用于计算每个用例执行的执行时间，并进行排序 > pytest -h --durations=N show N slowest setup/test durations (N=0 for all). --durations-min=N Minimal duration in seconds for inclusion in slowest list. Default 0.005 --durations=0时，显示所有【leetcode】1010. Pairs of Songs With Total Durations Divisible by 60
You are given a list of songs where the ith song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j withLeetcode 983. Minimum Cost For Tickets [Python]
题目给了状态转移公示： dp[i]表示第i天以及往后旅行需要的最小花费。我们需要每次花钱都覆盖尽可能多的天数。 dp[i] = dp[i+1] + 1daypass fee 或者是= dp[i+7] + 7daypass fee 或者是 = dp[i+30] + 30daypass fee三者中最小的。则dp[days[0]] 表示从第一个日期到后面的全部vue-video-player记录上次播放时间继续播放
最近项目需求记录上次视频播放时长，下次接着播放 1.html 代码如下（示例）： <video-player class="video-player-box" ref="videoPlayer" :playsinline="true" :options="playerOptions" @play="onPlayerPlay($event,'play')&【Pytest篇】pytest常用参数
1.查看执行最慢的n条用例 --durations=n if __name__ == '__main__': #最慢的2条测试用例 pytest.main(["--durations=2",'test_login_dlz.py']) 运行结果如下： 2.关闭禁用插件 -p no:插件名称 if __name__ == '__main__': pytest.main(["-p"[LeetCode] 1010. Pairs of Songs With Total Durations Divisible by 60
In a list of songs, the i-th song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with (time[i]【leetcode_easy_array】1010. Pairs of Songs With Total Durations Divisible by 60
problem 1010. Pairs of Songs With Total Durations Divisible by 60 参考 1. leetcode_1010. Pairs of Songs With Total Durations Divisible by 60； 完1010. Pairs of Songs With Total Durations Divisible by 60
问题： 给出一列歌曲所花时间的数组。 求任意两首歌合起来时间是60分钟的倍数的组队pair数。 Example 1: Input: [30,20,150,100,40] Output: 3 Explanation: Three pairs have a total duration divisible by 60: (time[0] = 30, time[2] = 150): total duration 180 (time[1] = 2spark窗口函数简单实现
版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/zhangfengBX/article/details/80659612 Window函数，可以统计最近一段时间的数据，使用Window函数加载成DStream：DStream.window("窗口长度","滑动间隔") reduceByKeyAndWindow 窗口长度：必须是Bath1013. Pairs of Songs With Total Durations Divisible by 60
time >=1 and <= 500, so sum >= 2 and sum <= 1000 there are 16 numbers in this interval divisible by 60 from 60 to 960, so the problem becomes find the number of pairs with sum of 60...960 class Solution { public int numPairsDivis