heart

Heart

依然范特西╮ 提交于 2020-02-01 16:22:12
笔芯 -- (利用python 绘图工具turtle 进行动态心形的绘制) import time import turtle time.sleep(1) # 画心形圆弧 def hart_arc(): for i in range(200): turtle.right(1) turtle.forward(2) def move_pen_position(x, y): turtle.hideturtle() # 隐藏画笔 turtle.up() # 提笔 turtle.goto(x, y) # 移动画笔到指定起始坐标(窗口中心为0,0) turtle.down() # 下笔 turtle.showturtle() # 显示画笔 love = "\n给林妹妹比心" signature = "brother Mao" # 初始化 turtle.setup(width=800, height=500) # 画布大小 turtle.color('blue', 'pink') # 画笔颜色 turtle.pensize(3) # 画笔粗细 turtle.speed(1) # 画笔速度 # 初始化画笔起始坐标 move_pen_position(x=0, y=-180) # 移动画笔位置 turtle.left(140) # 向左旋转140度 turtle.begin_fill() #

Oracle实现分组统计记录

旧街凉风 提交于 2019-12-23 02:26:19
今天刚上班不久,QQ滴滴的响个不停,看了下信息是一个网友要我帮忙下一个SQL语句,大体意思是:统计heart_active字段为不同情况的记录数,然后按时间来分组。 我想了下,心里想这好办,于是马上建了一个表,语句如下: CREATE TABLE rfid_fixed_heart (input_date date, heart_active VARCHAR2(2)); 接下来往rfid_fixed_heart表中插入了数据,heart_active字段为0和1, input_date中插入YYYY-MM-DD格式的数据。 后来就写了下面两个SQL给她,语句一和语句二有点区别,语句一快一统计出heart_active字段为不同情况的记录数,而语句二则只统计heart_active字段为0和1情况的记录数,两个语句的输出格式也有不同。具体如下: 语句一: SELECT a.input_date, a.heart_active, SUM(decode(a.heart_active, 1, 1, 0, 1)) FROM rfid_fixed_heart a GROUP BY a.heart_active, a.input_date ORDER BY a.input_date DESC; 语句二: SELECT a.input_date, SUM(decode(a.heart_active,

cnblogs鼠标点击爱心效果

旧街凉风 提交于 2019-12-09 14:16:19
(function (window, document, undefined) { var hearts = []; window.requestAnimationFrame = (function () { return window.requestAnimationFrame || window.webkitRequestAnimationFrame || window.mozRequestAnimationFrame || window.oRequestAnimationFrame || window.msRequestAnimationFrame || function (callback) { setTimeout(callback, 1000 / 60); } })(); init(); function init() { css(".heart{width: 10px;height: 10px;position: fixed;background: #f00;transform: rotate(45deg);-webkit-transform: rotate(45deg);-moz-transform: rotate(45deg);}.heart:after,.heart:before{content: '';width: inherit;height:

python表白代码1.0桃心输出

纵饮孤独 提交于 2019-12-06 15:25:42
python爱心表达函数初级版本:def my_heart(a,b=2): print("\n".join(["".join([(a[(x-y) % len(a)] \ if ((x*0.04)**2+(0.1*y)**2-1)**3-(x*0.04)**b*(0.1*y)**3 \ <=0 else " ") for x in range(-30,30)]) \ for y in range(30,-20,-1)]))my_heart("yanjiangyi",2)运行结果如下所示: 来源: https://www.cnblogs.com/Yanjy-OnlyOne/p/11992862.html

点击页面出现爱心源代码

心已入冬 提交于 2019-12-05 19:32:25
好东西,大家一起分享! 1 (function(window,document,undefined){ 2 var hearts = []; 3 window.requestAnimationFrame = (function(){ 4 return window.requestAnimationFrame || 5 window.webkitRequestAnimationFrame || 6 window.mozRequestAnimationFrame || 7 window.oRequestAnimationFrame || 8 window.msRequestAnimationFrame || 9 function (callback){ 10 setTimeout(callback,1000/60); 11 } 12 })(); 13 init(); 14 function init(){ 15 css(".heart{width: 10px;height: 10px;position: fixed;background: #f00;transform: rotate(45deg);-webkit-transform: rotate(45deg);-moz-transform: rotate(45deg);}.heart:after,.heart:before

python-文件的操作

廉价感情. 提交于 2019-12-05 18:11:09
python-文件的操作 一、文件操作的流程 python文件的操作流程如下: 例子: 现有如下一个文件my-hear Every night in my dreams 在梦中的每个夜晚 I see you,I feel you 我看见你,我感觉到你 That is how I know you go on 我就是这样知道你的 Far across the distance 远处 And spaces between us 以及我们之间的空间 You have come to show you go on 你是来带你去的 Near far 近距离 Wherever you are 无论你在哪里 I believe 我相信 That the heart does go on 心脏确实在继续 Once more you open the door 你又一次开门 And you're here in my heart 你在我心中 And my heart will go on and on 我的心会不停地跳动 Love can touch us one time 爱可以触摸我们一次 And last for a lifetime 而且会持续一辈子 And never let go till we're gone 在我们走之前永远不要放手 Love was when I loved you

网页背景蜘蛛丝特效

无人久伴 提交于 2019-12-05 07:10:18
博客园点击出现爱心❤特效 <script type="text/javascript"> (function(window,document,undefined){ var hearts = []; window.requestAnimationFrame = (function(){ return window.requestAnimationFrame || window.webkitRequestAnimationFrame || window.mozRequestAnimationFrame || window.oRequestAnimationFrame || window.msRequestAnimationFrame || function (callback){ setTimeout(callback,1000/60); } })(); init(); function init(){ css(".heart{width: 10px;height: 10px;position: fixed;background: #f00;transform: rotate(45deg);-webkit-transform: rotate(45deg);-moz-transform: rotate(45deg);}.heart:after,.heart:before{content

Parallel Processing/Parallel Programming in C#

匿名 (未验证) 提交于 2019-12-03 08:52:47
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 need to process more than one function that goes and performs results returning the same type of records. I am using c# under visual studio 2010 considering the functions i have are: class Search{ public list<wrecords> GetrecordsofAAA(string term); public list<wrecords> GetrecordsofBBB(string term); public list<wrecords> GetrecordsofCCC(string term); } and im calling the functions this way list<wrecords> records1 = Search.GetrecordsofAAA(heart); list<wrecords> records2 = Search.GetrecordsofBBB(heart); list<wrecords> records3 = Search

iOS HealthKit how to save Heart Rate (bpm) values? Swift

匿名 (未验证) 提交于 2019-12-03 03:10:03
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: How to use : HKUnit Sample type Unit type Unit name Unit string Heart Rate count/time Beats per Minute "count/min” 回答1: Swift : Heart Rate (bpm) save into healthkit store private func saveHeartRateIntoHealthStore(height:Double) -> Void { // Save the user's heart rate into HealthKit. let heartRateUnit: HKUnit = HKUnit.countUnit().unitDividedByUnit(HKUnit.minuteUnit()) let heartRateQuantity: HKQuantity = HKQuantity(unit: heartRateUnit, doubleValue: height) var heartRate : HKQuantityType = HKQuantityType.quantityTypeForIdentifier

Make the heart bounce in and out

匿名 (未验证) 提交于 2019-12-03 02:43:01
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 was able to make a heart in matlab as: n=100; x=linspace(-3,3,n); y=linspace(-3,3,n); z=linspace(-3,3,n); [X,Y,Z]=ndgrid(x,y,z); F=((-(X.^2) .* (Z.^3) -(9/80).*(Y.^2).*(Z.^3)) + ((X.^2) + (9/4).* (Y.^2) + (Z.^2)-1).^3); isosurface(F,0) lighting phong axis equal Would it be possible to make it bounce in and out? What approach might be taken? 回答1: try this step = 0.05; x = -1.5 : step : 1.5; y = -1 : step : 1; z = -1.5 : step : 1.5; [X,Y,Z] = meshgrid(x, y, z); f = (X.^2 + 9/4 .* Y.^2 + Z.^2 - 1).^3 - X.^2 .* Z.^3 - 9/80 .* Y.^2 .* Z.^3;