一个业余小项目,用于监控指定股票的成交指标,触发事件的主动通知 源码 !https://github.com/jeffchen81/stock-starer 设计说明 场景 定时任务1:通过网上Open API,抓取指定股票的即时交易信息,格式化后存入原始数据表 定时任务2:根据指标公式,计算出新的指标数据,存入指标数据表 定时任务3:判断指标数据是否触发提醒事件,如触发就发邮件给用户 (可选)可视化图表:指定股票的指标趋势图 技术选型 开发语言:pyt 大专栏  Stock Starer股票盯…

1.前言 话说最近一段时间受疫情的影响,股市各种妖魔横行.本人作为一个入股市不满三年的小韭菜,就有幸见证了好几次历史,也是满心惊喜,就权当是接受资本市场的再教育了吧. 小韭菜的炒股方法其实很简单,这两年多来一直死守着一只股票,不满仓,也不空仓,逢低就买点,逢高就卖点.靠着这种守株待兔的笨方法,没有赚到大钱,但好在也不至于亏钱,目前的累计收益率是25.83%. 年初复工前窝在家,天天抱着个手机,实时看着股票行情,心情也跟着起飞下降,好不惬意:复工上班后,不能一直抱着手机看,突然还有点不习惯了. 怎…

周末继续写博客,算起来,关于rabbitMQ这个中间件的研究已经持续至两个星期了,上一篇文章使用sring amqp实现了同步和异步的消息接收功能.这一节继续实用spring amqp实现一个股票交易系统的主要逻辑.这个例子更为复杂也更具代表意义,因为它是现实世界中的例子. stock trading这个例子包含一个服务端(server),它用于向指定的Topic Exchange发送股票数据.还要很多客户端(clients),它们从特定的Queue(绑定了routing patten的Queu…

一 ［抄题］: If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit. Input: [7, 1, 5, 3, 6, 4] Output: 5 max. difference = 6-1 = 5 (not 7-1 = 6, as selli…

Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit. Exam…

You are given an array prices where prices[i] is the price of a given stock on the ith day. You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock. Return the maximum p…

Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit. 这道题相…

Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit. Have…

题意:给定一个序列,第i个元素代表第i天这支股票的价格,问在最佳时机买入和卖出能赚多少钱?只买一次,且仅1股,假设本钱无限. 思路:要找一个最低价的时候买入,在最高价的时候卖出利润会最大.但是时间是不能冲突的,比如说在明天买入,今天卖出.因此,对于今天的价格,应该要找到今天之前的该股的最低价,买入,今天卖出. 其实就是要为序列中的一个元素A[k],找到另一个元素A[e],位置满足e<k,结果使得A[k]-A[e]最大. 用动态规划解决,从左扫起,遇到一个元素就更新当前最小值,再用当前元素去减这个…

Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit. Exam…