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.
"> <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
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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
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。
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]:
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;
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
1795 non-null object Company Location 1795 non-null object Rating Specific Bean Origin or Bar Name：产品名称 REF：不祥 Review Date： Cocoa Percent：可可含量 Company Location：公司地址 Rating Origin"]).apply(np.mean) best_been_data.sort_values(by="Rating",inplace=True) print(best_been_data[- best_coco = dataset_nona[["Cocoa Percent","Rating"]] best_coco.columns = best_coco.columns.map(lambda 散点图 可以看出巧克力质量和含可可量没有明显的关系 探索分析 print(dataset_nona.groupby(["Review Date"]).apply(lambda x:x["Rating"]
) 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
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();
PG</rating <stars 10</stars <description Talk about a US-Japan war</description </movie <movie ="Trigun" <type Anime, Action</type <format DVD</format <episodes 4</episodes <rating PG</rating </description </movie <movie title="Ishtar" <type Comedy</type <format VHS</format <rating PG ": print ("Rating:", self.rating) elif self.CurrentData == "stars": print ("Stars:", self.stars = movie.getElementsByTagName('rating')[0] print ("Rating: %s" % rating.childNodes[0].data) description
, 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
练习6：计算movie的rating分布情况 SELECT DISTINCT MIN("RATING_COUNT") OVER( ) AS "MIN", MAX("RATING_COUNT") OVER( ) AS "MAX", AVG("RATING_COUNT") OVER( ) AS "AVG", SUM("RATING_COUNT") OVER( ) AS "SUM", MEDIAN ("RATING_COUNT") OVER( ) AS "MEDIAN", STDDEV("RATING_COUNT") OVER( ) AS "STDDEV", COUNT(*) OVER( 明细情况： SELECT "RATING_COUNT", COUNT(1) as "MOVIE_COUNT" FROM ( SELECT "MOVIEID", COUNT(1) as "RATING_COUNT 练习8：统计用户投票得分情况 SELECT "RATING", COUNT(1) as "RATING_COUNT" FROM "MOVIELENS"."
[1240] 练习6：计算movie的rating分布情况 SELECT DISTINCT MIN("RATING_COUNT") OVER( ) AS "MIN", MAX("RATING_COUNT ") OVER( ) AS "MAX", AVG("RATING_COUNT") OVER( ) AS "AVG", SUM("RATING_COUNT") OVER( ) AS "SUM", MEDIAN("RATING_COUNT") OVER( ) AS "MEDIAN", STDDEV("RATING_COUNT") OVER( ) AS "STDDEV", COUNT(* "RATING_COUNT", COUNT(1) as "MOVIE_COUNT" FROM ( SELECT "MOVIEID", COUNT(1) as "RATING_COUNT" FROM public.aa.movielens.hdb::data.RATINGS" GROUP BY "MOVIEID" ) GROUP BY "RATING_COUNT" ORDER BY "RATING_COUNT
. // A is the highest rating; it is greater than any other valid rating. // C is the lowest rating; it (string rating) { if (rating == null) { throw new ArgumentNullException ("rating"); } string v = rating.ToUpper(CultureInfo.InvariantCulture); if (v.Length ", "rating"); } Rating = v; } public int CompareTo(object obj) { return -string.Compare(this.Rating, other.Rating, StringComparison.OrdinalIgnoreCase); } public
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。
rating_data = sc.textFile("u.data") print rating_data.first() 196 242 3 881250949 num_ratings = rating_data.count [2])) #最高得分 max_rating = ratings.reduce(lambda x,y:max(x,y)) #最低得分 min_rating = ratings.reduce(lambda x,y:min(x,y)) #评价得分 mean_rating = ratings.reduce(lambda x,y:x+y) / num_ratings #中位数 median_rating = print "Max rating: %d" % max_rating print "Average rating: %2.2f" % mean_rating print "Median rating : %d" % median_rating print "Average # of rating per user:%2.2f" % ratings_per_user print "Average #
= df[df['Rating']==5]['Review Text'].apply(clean_data) create_cloud(rating5) ? 3、等级是4的词云 rating4= df[df['Rating']==4]['Review Text'].apply(clean_data) create_cloud(rating4) ? 4、等级是3的词云 rating3= df[df['Rating']==3]['Review Text'].apply(clean_data) create_cloud(rating3) ? 5、等级是2的词云 rating2= df[df['Rating']==2]['Review Text'].apply(clean_data) create_cloud(rating2) ? 6、等级是1的词云 rating1= df[df['Rating']==1]['Review Text'].apply(clean_data) create_cloud(rating1) ?
} } return-1; } function initEvent() { var rating = document.getElementById("rating"); //取id为rating的标签 var tds = rating.getElementsByTagName ("td"); //在id为rating标签下去名字为td的值 for (var i =0; i < tds.length; i++) { = document.getElementById("rating"); //不用担心和initEvent中的rating、tds变量冲突，因为他们是两个函数，只是写在了一起而已。 var tds = rating.getElementsByTagName("td"); var index = indexOf(tds, this);
; private int duration; } 构建初始化data.sql: INSERT INTO movie(id, title, director, rating, duration King of the Monsters', ' Michael Dougherty', 'PG-13', 132); INSERT INTO movie(id, title, director, rating VALUES(3, 'Captain Marvel', 'Anna Boden', 'PG-13', 123); INSERT INTO movie(id, title, director, rating duration) VALUES(4, 'Dumbo', 'Tim Burton', 'PG', 112); INSERT INTO movie(id, title, director, rating LIKE 'PG%'; 我们可以这样使用： List<Movie> findByRatingStartsWith(String rating); 测试代码如下： List<Movie> results
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