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HDU4870:Rating(DP)「建议收藏」
Every user who has registered in “TopTopTopCoder” system will have a rating, and the initial value of rating equals to zero. rating will be min(X+50,1000). Her/His rating will be max(X-100,0) otherwise. She uses the account with less rating in each contest.
34820编辑于 2022-07-06 - 来自专栏Porschev[钟慰]的专栏
Asp.Net开发等级星使用(Jquery Rating)
"> <html xmlns="http://www.w3.org/1999/xhtml"> <head runat="server"> <title>Porschev----Jquery Rating StringBuilder sb = new StringBuilder(); sb.Append("<ul style=\"width: 125px;\" class=\"rating
95470发布于 2018-01-16 - 来自专栏CSDNToQQCode
HarmonyOS-UIAbitity-Rating——【坚果派-红目香薰】
示例代码 HarmonyOS-UIAbitity-Rating 提供在给定范围内选择评分的组件。 接口 Rating(options?: { rating: number, indicator? : boolean }) 参数： 参数名 参数类型 必填 默认值 参数描述 rating number 是 0 设置并接收评分值。 示例代码 @Entry @Component struct Index { @State rating: number = 1 @State indicator: boolean = false ).fontSize(20) Rating({ rating: this.rating, indicator: this.indicator }) .stars(5) .stepSize(0.5) .onChange((value: number) => { this.rating = value })
25210编辑于 2024-02-03 - 来自专栏包子铺里聊IT
Design a Food Rating System
system that can do the following: Modify the rating of a food item listed in the system. String highestRated(String cuisine) Returns the name of the food item that has the highest rating for this.cuisine = cuisine; this.rating = rating; } @Override public = null; if (o.rating == this.rating) { return this.food.compareTo(o.food) ; } return o.rating - this.rating; } } // food -> Food private Map
46710编辑于 2022-11-11 - 来自专栏机器学习炼丹之旅
Johnny and Another Rating Drop（思维）
两个数的二进制不同的个数称为两个数的不公平度，求1-n个数所有相邻的两个数的不公平度的总和。
35020编辑于 2022-06-29 - 来自专栏CDN及云技术分享
用9行python代码演示推荐系统里的协同过滤算法
In [5～7]: rating_c = rating[rating_toby[rating.title].isnull().values & (rating.critic ! = 'Toby')] rating_c['similarity'] = rating_c['critic'].map(sim_toby.get) rating_c['sim_rating'] = rating_c.similarity * rating_c.rating 接下来先得到其他人对这3部的电影打分rating_c。 * rating_c.rating得出。 ') rating_toby = rp['Toby'] sim_toby = rp.corrwith(rating_toby) rating_c = rating[rating_toby[rating.title
1.6K10编辑于 2024-03-08 - 来自专栏程序那些事
数据分析实际案例之：pandas在餐厅评分数据中的使用
餐厅评分数据简介 数据的来源是UCI ML Repository，包含了一千多条数据，有5个属性，分别是： userID： 用户ID placeID：餐厅ID rating：总体评分 food_rating /data/restaurant_rating_final.csv' df = pd.read_csv(path) df userID placeID rating food_rating service_rating food_rating rating placeID 132560 1.00 0.50 132561 1.00 0.75 132564 1.25 1.25 132572 1.00 1.00 132583 [:10] food_rating rating diff placeID 132667 2.000000 1.250000 -0.750000 132594 1.200000 0.600000 placeID')['rating'].std() # Filter down to active_titles rating_std_by_place = rating_std_by_place.loc
2.3K20编辑于 2022-02-25 - 来自专栏快乐阿超
React优化技巧——避免state层级过高
{rating} setRating={setRating} /> </> ); } function ProductRating({ rating, setRating }) { return ( <> <label htmlFor="<em>rating</em>">Rating</label> <input type="number" id= "<em>rating</em>" onChange={(e) => setRating(parseInt(e.target.value))} /> {rating &&
Rating ); return ( <> <label htmlFor="<em>rating</em>">Rating</label> <input type="number" &&
Rating is {rating}
} </> ); } 如果state仅传递给一个子组件，那么级别太高了。23510编辑于 2024-09-15 - 来自专栏SnailTyan
Leetcode 1395. Count Number of Teams
Version 1 class Solution: def numTeams(self, rating: List[int]) -> int: count = 0 n = len(rating) for i in range(n): for j in range(i+1, n): for k in range(j+1, n): if rating[i] < rating[j] and rating[j] < rating[k]: count += 1 elif rating[i] > rating[j] and rating[j] > rating[k]: += 1 for j in range(i+1, n): if rating[i] > rating[j]:
50540发布于 2021-07-29 - 来自专栏Michael阿明学习之路
LeetCode 1395. 统计作战单位数（蛮力法）
每 3 个士兵可以组成一个作战单位，分组规则如下： 从队伍中选出下标分别为 i、j、k 的 3 名士兵，他们的评分分别为 rating[i]、rating[j]、rating[k] 作战单位需满足： rating [i] < rating[j] < rating[k] 或者 rating[i] > rating[j] > rating[k] ，其中 0 <= i < j < k < n 请你返回按上述条件可以组建的作战单位数量 示例 3： 输入：rating = [1,2,3,4] 输出：4 提示： n == rating.length 1 <= n <= 200 1 <= rating[i] <= 10^5 来源：力扣（ < n-1; ++j) for(k = j+1; k < n; ++k) if((rating[i] < rating[j] && rating[j] < rating[k ]) ||(rating[i] > rating[j] && rating[j] > rating[k])) sum++; return sum;
54020发布于 2020-07-13 - 来自专栏R语言数据分析指南
ggplot2绘制正负分布条形图
) 数据清洗 df <- chocolate %>% group_by(company_location) %>% summarise(n = n(),min_rating = min(rating ),max_rating = max(rating), avg_rating = mean(rating, na.rm = T)) %>% mutate(company_location = fct_reorder(company_location, avg_rating)) %>% filter(n > 3) %>% mutate(rating_diff = avg_rating - mean) %>% filter(abs(rating_diff) >0.05) 数据可视化 df %>% ggplot() + geom_col(aes(x = rating_diff (aes(x = rating_diff,y = company_location, color=ifelse(rating_diff > 0,"#E11B4D","#8456BA
1.2K10编辑于 2022-09-21 - 来自专栏机器学习算法原理与实践
用Spark学习矩阵分解推荐算法
, rating=4.995227969811873), Rating(user=38, product=304, rating=2.5159673379104484), Rating(user=38, product=1014, rating=2.165428673820349), Rating(user=38, product=322, rating=1.7002266119079879), Rating ), Rating(user=38, product=23, rating=1.0590775012913558), Rating(user=38, product=327, rating=1.0335651317559753 =20, rating=2.9892138653406635), Rating(user=25, product=20, rating=1.7558472892444517), Rating(user= 7, product=20, rating=1.523935609195585), Rating(user=286, product=20, rating=1.3746309116764184), Rating
1.8K30发布于 2018-08-14 Elo 评分系统 和 TrueSkill库
期望胜率（Expected Score） 例子 Python代码 import math def probability(rating1, rating2): """ 计算 rating1 对 rating2 的胜率 :param rating1: 评分1 :param rating2: 评分2 :return: 胜率 """ return 1.0 / (1 + math.pow(10, (rating2 - rating1) / 400.0)) def elo_rating(Ra, Rb, outcome, K=30): """ A = Rating() B = Rating() # 默认mu = 25，sigma=8.3 print("A: ",A) # A: trueskill.Rating(mu=25.000, sigma , quality, rate r1 = Rating() # 1P's skill r2 = Rating() # 2P's skill r3 = Rating() # 3P's skill t1
1.1K10编辑于 2025-07-14- 来自专栏人工智能LeadAI
基于协同过滤的推荐引擎（实战部分）
那么首先要做的处理就是添加一列预测列，这一列里我们将rating列复制出一列，叫predict_rating，部分rating置零，当作要预测的评分，我们的程序就计算为零的rating，然后对比predict_rating 和rating的差距。 这是又一个拦路虎，自认为比较理想的是每个都有1/3的predict_rating是0，用来做预测，想到下面个plan： 1、excel复制rating，粘贴，重命名为predict_rating，看数据发现相同 import time real_rating = full_data['rating'] # 原rating predict_rating = np.array(full_data['rating'] ，如果有就同时把这两个rating分别加入预测列的rating和对照列的rating中，没有就两个都不加入，这样就实现了和“取出电影-用户矩阵中都不为零的两列”同样的效果。
1.2K70发布于 2018-03-08 - 来自专栏Flutter
【Flutter】评级对话框组件
使用 添加依赖 rating_dialog: ^2.0.0 引入 import 'package:rating_dialog/rating_dialog.dart'; 运行命令：「flutter packages : ${response.rating}, ' 'comment: ${response.comment}'); if (response.rating < 3.0) { print('response.rating: ${response.rating}'); } else { Container(); } : ${response.rating}, ' 'comment: ${response.comment}'); if (response.rating < 3.0 ) { print('response.rating: ${response.rating}'); } else { Container();
5.4K50发布于 2021-06-09 - 来自专栏小馒头学Python
【Python数据分析五十个小案例】电影评分分析：使用Pandas分析电影评分数据，探索评分的分布、热门电影、用户偏好
'rating'], bins=bins, labels=labels, right=False)# 统计各评分区间的电影数量rating_distribution = df['rating_category 我们可以根据rating对电影进行排序，找出评分最高的前10部电影。 # 计算每种类型的平均评分genre_avg_rating = df.groupby('genre')['rating'].mean().sort_values(ascending=False)print (genre_avg_rating)# 绘制电影类型的平均评分plt.figure(figsize=(10, 6))genre_avg_rating.plot(kind='bar', color='purple = df.groupby('year')['rating'].mean()# 绘制年度平均评分趋势plt.figure(figsize=(10, 6))yearly_avg_rating.plot(kind
1K00编辑于 2024-11-28 - 来自专栏大数据，java，Python和前端
基于ChatGPT辅助优化协同过滤算法SQL实现的实践与思考
* u2.rating) as dot_product, SQRT(SUM(u1.rating * u1.rating)) * SQRT(SUM(u2.rating * u2.rating)) u1.user_id as user1, u2.user_id as user2, u1.product_id, u1.rating as rating1, u2.rating as rating2 FROM user_ratings_filtered u1 INNER JOIN user_ratings_filtered u2 = u2.user_id)SELECT user1, user2, COUNT(*) as common_products, SUM(rating1 * rating2) / * rating)) FROM user_ratings WHERE user_id = user1) * (SELECT SQRT(SUM(rating * rating)) FROM
26810编辑于 2025-09-11 - 来自专栏Lansonli技术博客
2021年大数据Spark（二十八）：SparkSQL案例三电影评分数据分析
读取电影评分数据，从本地文件系统读取 val rawRatingsDS: Dataset[String] = spark.read.textFile("data/input/rating ), 2) AS avg_rating, COUNT(movieId) AS cnt_rating |FROM | view_temp_ratings 200 |ORDER BY | avg_rating DESC, cnt_rating DESC |LIMIT "), 2).as("avg_rating"), count($"movieId").as("cnt_rating") ) .orderBy($"avg_rating".desc, $"cnt_rating".desc) // 获取前10 .limit(10)
1.7K20发布于 2021-10-09 - 来自专栏网络日志
最佳多列索引公式
BY release_date DESC; 当索引为 (country, rating, release_date) 时： index on (country, rating, release_date 增加一个字段 is_high_rating，当评分大于 8.0 时，is_high_rating 为 1，否则为 0。 例如 ORDER BY release_date DESC, rating ASC 对应的索引应该是 (release_date DESC, rating ASC) 或 (release_date ASC , rating DESC)。 例如我们要将查询按照 release_date 升序、rating 降序排序的前五个结果： SELECT * FROM films ORDER BY release_date DESC, rating
1K10编辑于 2024-06-22 - 来自专栏数据小魔方
左手用R右手Python——CSS网页解析实战
构建网页解析函数： getcontent<-function(url){ myresult=data.frame() title=subtitle=author=category=price=rating ###提取作者、副标题、评价、评分、价格： author_text=subtext=eveluate_text=rating_text=price_text=rep('',length) = 0){ rating_text[i]=result %>% html_nodes(sprintf("ol > li:nth-of-type(%d) span.rating-average =c(rating,rating_text) price=c(price,price_text) ###打印任务状态： print(sprintf("page <-as.numeric(myresult$rating) myresult$eveluate_nums<-as.numeric(myresult$eveluate_nums) DT::datatable
1.2K50发布于 2018-04-11
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