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R语言 | 条形图绘制
# cabbage_exp数据内容 > cabbage_exp Cultivar Date Weight sd n se 1 c39 d16 3.18 输出图片 3 绘制堆积条形图 演示数据 同上，以gcookbook包中的cabbage_exp数据集为例，该数据集包含两个分类变量Cultivar和Date和一个连续变量Weight。 # cabbage_exp数据内容 > cabbage_exp Cultivar Date Weight sd n se 1 c39 d16 3.18 ggplot(cabbage_exp,aes(x = Date,y = Weight,fill = Cultivar)) + geom_col() 输出图片 反转图例顺序 我们可以通过guides( # cabbage_exp数据内容 > cabbage_exp Cultivar Date Weight sd n se 1 c39 d16 3.18
3.3K20编辑于 2022-03-29 - 来自专栏渔夫
Java中Synchronized的用法
Thread2:cabbage1 Thread2:cabbage2 Thread2:cabbage3 Thread2:cabbage4 Thread2:cabbage5 Thread2:cabbage6 Thread2:cabbage7 Thread2:cabbage8 Thread2:cabbage9 Thread1:cabbage9 Thread2:apple0 Thread1:cabbage9 Thread1:cabbage1 Thread1:cabbage2 Thread1:cabbage3 Thread1:cabbage4 Thread1:cabbage5 Thread1:cabbage6 Thread1:cabbage7 Thread1:cabbage8 Thread1:cabbage9 Thread2:cabbage9 Thread2:cabbage9 Thread2:cabbage9 Thread2:cabbage9 Thread2:cabbage9 Thread2:cabbage9 Thread2:cabbage9 Thread2:cabbage9 Thread2:cabbage9
3.5K30发布于 2019-07-31 - 来自专栏Coding迪斯尼
python高级算法与数据结构:使用treap实现双索引1
: (Cabbage, 77) left: None right: None 对比右旋转前后输出的二叉树看，旋转后的二叉树打印信息的确跟上面我们旋转后对应的图像是一致的。 接下来我们实现左旋转，先把上图中cabbage节点对应的值改成75，这样它与父节点就违背了小堆性质： 我们要做的是：1，把cabbage节点向“左”旋转到beer的位置；2，beer的父节点设置为cabbage ;3:beer的右孩子设置为cabbage的左孩子;4，cabbage的左孩子变成beer；左旋转后二叉树应该成形如下： 从上图看，左旋转后，字符串依然保持二叉树排序性，同时数值的排放也遵守小堆原则 _root = x y.right = x.left x.left = y 为了测试上面代码实现，我们首先把cabbage的值修改，然后调用上面代码: cabbage : (Cabbage, 75) right: None (Cabbage, 75) parent: (Flour, 10) left: (Beer, 76) right: (Eggs, 129) (Beer
52520发布于 2021-09-16 水果蔬菜检测数据集VOC+YOLO格式9637张94类别
txt文件个数)：9637 标注类别数：94 标注类别名称(注意yolo格式类别顺序不和这个对应，而以labels文件夹classes.txt为准):["Bitter melon","Brinjal","Cabbage bell_pepper","bittergourd","blueberries","bottle","bottlegourd","bread","brinjal","broccoli","butter","cabbage eggplant","eggs","fig","flour","garlic","ginger","goat_cheese","grape","green_beans","ground_beef","haft-cabbage bottle 框数 = 7 bottlegourd 框数 = 110 bread 框数 = 51 brinjal 框数 = 288 broccoli 框数 = 297 butter 框数 = 128 cabbage 2185 ginger 框数 = 17 goat_cheese 框数 = 95 grape 框数 = 23 green_beans 框数 = 131 ground_beef 框数 = 80 haft-cabbage
53520编辑于 2025-07-17- 来自专栏Coding迪斯尼
python高级算法与数据结构:使用treap实现双索引2
: Node = Node("Cabbage", 159) pork : Node = Node("Pork", 56) flour.left = butter flour.right = water butter.left = bacon butter.right = eggs water.left = milk eggs.left = cabbage , 159) right: None (Cabbage, 159) parent: (Eggs, 129) left: None right: None (Water, 32) parent: (Flour : Node = Node("Cabbage", 159) pork: Node = Node("Pork", 56) beet : Node = Node("Beet", 81) , 159) right: None (Cabbage, 159) parent: (Eggs, 129) left: None right: None (Water, 32) parent: (Flour
65440发布于 2021-10-11 - 来自专栏刷题笔记
【HBUOJ】吃货阿生
Sample Input 1 5 3 water 100 flour 20 cabbage 71 pork 12 bean 5 2 water 20 flour 5 3 water 70 cabbage 54 pork 10 5 water 1 flour 1 cabbage 1 pork 2 bean 1 Sample Output 1 YES water 9 flour 14 cabbage 16
48240发布于 2019-11-08 智慧厨房菜品识别检测数据集VOC+YOLO格式5790张16类别
Braised_Pork_Meatballs_in_Brown_Sauce","Buddha_Jumps_Over_the_Wall","Dongpo_Pork","Fish_with_Pickled_Cabbage_and_Chili Braised_Pork_Meatballs_in_Brown_Sauce 框数 = 394 Buddha_Jumps_Over_the_Wall 框数 = 344 Dongpo_Pork 框数 = 386 Fish_with_Pickled_Cabbage_and_Chili Lake Vinegar Fish (西湖醋鱼) Buddha Jumps Over the Wall (佛跳墙) Steamed Sea Bass (清蒸鲈鱼) Fish with Pickled Cabbage
37900编辑于 2025-07-16- 来自专栏PPV课数据科学社区
数据挖掘知识脉络与资源整理(九)–柱形图
= weight)) + geom_bar(stat = "identity", fill = "lightblue", colour = "black") 2、 涉及分组变量的条形图 ggplot(cabbage_exp 另外我们发现,上面图形都是两两一组,那如果有一组就一个值怎么办,那么图形会确失一条bar,然后加宽,具体看图 ce <- cabbage_exp[1:5, ] ggplot(ce, aes(x = Date 用position = position_dodge(0.7)来控制 ggplot(cabbage_exp, aes(x = Date, y = Weight, fill = Cultivar)) + geom_bar(stat = "identity", width = 0.5, position = "dodge") ggplot(cabbage_exp, aes(x = Date, y = Weight 修改一下 ggplot(cabbage_exp, aes(x = Date, y = Weight, fill = Cultivar)) + geom_bar(stat = "identity") +
5.1K100发布于 2018-04-25 - 来自专栏程序员成长指北
使用 JavaScript 编写更好的条件语句
printVegetablesWithQuantity('cabbage'); // I like cabbage printVegetablesWithQuantity('cabbage', 20); // 'I like cabbage // 'I have bought a large quantity' 现在，我们有： 1 if/else 语句过滤非法条件 3 级嵌套if语句 (条件 1, function printVegetablesWithQuantity(vegetable, quantity) { const vegetables = ['potato', 'cabbage ; } //results printVegetablesWithQuantity('cabbage'); // We have 1 cabbage! ', quantity: 2 }); // cabbage 在上面的例子中，如果vegetable 存在，我们想要打印 vegetable name, 否则打印"unknown"。
2.3K30发布于 2021-04-21 - 来自专栏Reck Zhang
ACMSGURU 184 - Patties
Patties Problem Description TsPetya is well-known with his famous cabbage patties. Petya has P grams of flour, M milliliters of milk and C grams of cabbage. Petya knows that he needs K grams of flour, R milliliters of milk and V grams of cabbage to cook one
31230发布于 2021-08-11 [数据集][目标检测]目标检测数据集大白菜数据集VOC格式1557张
不包含分割路径的txt文件和yolo格式的txt文件，仅仅包含jpg图片和对应的xml) 图片数量(jpg文件个数)：1557 标注数量(xml文件个数)：1557 标注类别数：1 标注类别名称:["cabbage "] 每个类别标注的框数： cabbage count = 2309 使用标注工具：labelImg 标注规则：对类别进行画矩形框 重要说明：这个是大白菜数据集，注意是大白菜不是小白菜，由于图片太少截取了
18610编辑于 2025-07-19- 来自专栏米扑专栏
Python 学习入门（17）—— args, kwargs
, "arg2":"two"} #test4(1, **kwargs) test4(arg1=1, **kwargs) test5("apple", "banana", "cabbage ") test6(apple="fruit", cabbage="vagetable") mylist = ['aa', 'bb', 'cc'] test7(*mylist) keyword arg: myargs, 3 arg1: 1 arg2: two arg3: 3 arg1: 1 arg2: two arg3: 3 0 -> apple 1 -> banana 2 -> cabbage cabbage = vagetable apple = fruit a= aa & b= bb & c= cc 参考推荐： How to use *args and **kwargs in Python
53730发布于 2019-02-19 - 来自专栏生信技能树
了解绘制条形图和折线图的细节
A： cabbage_exp#带白菜的生长情况 Cultivar Date Weight sd n se 1 c39 d16 3.18 0.9566144 A：使用geom_bar()函数，并映射一个变量给fill参数（注意和簇状条形图的区别，这里不能设置position='dodge'） cabbage_exp Cultivar Date Weight 默认条件下条形的堆积顺序与图例顺序是一致的 ggplot(cabbage_exp,aes(x=Date,y=Weight,fill=Cultivar))+geom_col() #2.可以通过guides ()进行调整并指定图例对应需要的调整的图形属性 ggplot(cabbage_exp,aes(x=Date,y=Weight,fill=Cultivar))+geom_col()+ guides(fill A：使用geom_col(position='fill')实现 #1.绘制百分比堆积图 ggplot(cabbage_exp,aes(x=Date,y=Weight,fill=Cultivar))+
8.9K10编辑于 2022-06-08 - 来自专栏pythonista的日常
Python进阶之强大的装饰器 Decorators （三）
in kwargs.items(): print('{0} = {1}'.format(name, value)) f1('twtrubiks', apple='fruit', cabbage in kwargs.items(): print('{0} = {1}'.format(name, value)) f1('twtrubiks', apple='fruit', cabbage in kwargs.items(): print('{0} = {1}'.format(name, value)) f1('twtrubiks', apple='fruit', cabbage
40020发布于 2020-06-16 农作物害虫分类数据集12846张27类别
)：12846 分类类别数：27 类别名称:["ants","aphids","armyworms","bees","beetle","brown_marmorated_stink_bugs","cabbage_loopers aphids":"蚜虫" "armyworms":"粘虫" "bees":"蜜蜂" "beetle":"甲虫" "brown_marmorated_stink_bugs":"棕色土拨鼠蝽" "cabbage_loopers aphids 图片数：473 armyworms 图片数：480 bees 图片数：500 beetle 图片数：415 brown_marmorated_stink_bugs 图片数：502 cabbage_loopers
33000编辑于 2025-07-20多种蝴蝶识别分割数据集labelme格式784张9类别
"Crimson-patched Longwing","Common Buckeye","American Cooper","Mourning Cloak","Giant Swallowtail","Cabbage Buckeye count = 89 American Cooper count = 85 Mourning Cloak count = 92 Giant Swallowtail count = 88 Cabbage
19600编辑于 2025-07-16- 来自专栏个人编程笔记
C#-TreeView
Pear); TreeNode vegetables = new TreeNode(); vegetables.Name = "蔬菜"; TreeNode cabbage = new TreeNode(); cabbage.Name = "卷心菜"; TreeNode waterspinach = new TreeNode(); waterspinach.Name = "空心菜"; vegetables.Children.Add(cabbage); vegetables.Children.Add(
1.2K10编辑于 2023-03-03 食物种类检测数据集VOC+YOLO格式5576张20类别
标注类别数：20 标注类别名称(注意yolo格式类别顺序不和这个对应，而以labels文件夹classes.txt为准):["bellpepper","bittergourd","broccoli","cabbage radish","shrimp","tofu","tomato"] 每个类别标注的框数： bellpepper 框数 = 706 bittergourd 框数 = 706 broccoli 框数 = 714 cabbage
30800编辑于 2025-07-22- 来自专栏pythonista的日常
Python进阶之强大的装饰器 Decorators （二）
in kwargs.items(): print('{0} = {1}'.format(name, value)) f1('twtrubiks', apple='fruit', cabbage in kwargs.items(): print('{0} = {1}'.format(name, value)) f1('twtrubiks', apple='fruit', cabbage
39310发布于 2020-06-16 - 来自专栏小陈飞砖
设计模式之组合模式
{ menuItems = new ArrayList<MenuComponent>(); addItem("KFC Cake Breakfast", "boiled eggs&toast&cabbage DinerMenu() { menuItems = new MenuComponent[Max_Items]; addItem("vegetable Blt", "bacon&lettuce&tomato&cabbage
29420编辑于 2022-06-25
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