Question

So I have this dataset which contains a count of how many people have given specific ratings for a range of products, i.e. there is one column for each rating (1-5) and each row contains the count.

  ID ratings_count_5 ratings_count_4 ratings_count_3 ratings_count_2 ratings_count_1
  2             599             624              78             357               4
  3             350             407              95             382             255
  4             454             368              52             245             512
  5             729             938             520             145             478
  6             548             176             431             313             459
  7             628               1               1               1               2

Does anyone know how I could find the median rating?

Answer 1

这完全取决于ID列中的值是什么意思以及如何定义中位数。我做了以下假设：

ID是product-id
..._count_i表示值i的评级
＆＃34;中位数＆＃34;表示计数列的加权和除以该产品的所有评级的总和

然后你得到＆＃34;中位数＆＃34;通过：

 df$sum = apply(df, 1, sum)
 df$median = (df$ratings_count_5 * 5 + df$ratings_count_4 * 4 +
              df$ratings_count_3 * 3 + ratings_count_2 * 2 + 
              df$ratings_count_1 * 1) / df$sum

Answer 2

如何将数据框转换为矩阵（如果不是矩阵），然后使用times的{{1}}参数应用加权中位数。

让我们调用原始数据rep()：

df

我认为这会给你想要的输出。

mat <- as.matrix(df[, -1])
median_rating <- apply(mat, 1, function(x) median(rep(5:1, times=x)))
cbind(df, median_rating)

Answer 3

您可以执行以下操作来获取每行的列索引

ID <- c(2,3)
ratings_count_5 <- c(599,350)
ratings_count_4 <- c(624,407)
ratings_count_3 <- c(78,95)
ratings_count_2 <- c(357,382)
ratings_count_1 <- c(4,255)

df <- data.frame(ID,ratings_count_5,ratings_count_4,ratings_count_3,ratings_count_2,ratings_count_1)

df$median <- median(unname(unlist(df[,-1])))
r <- df[,2:6]-df[,7]
index <- data.frame(NULL)

for(i in 1:nrow(r)){
  ind <- which.min(unlist(abs(r[i,])))
  index <- rbind(index,ind)
}

df <- cbind(df,index)
setnames(df,"X4L","col_index")
df

Answer 4

<强>解决方案

严重依赖dplyr：

library(dplyr)
library(tidyr)
df %>%
    gather(rating, freq, -ID) %>%
    arrange(rating) %>%
    group_by(ID) %>%
    mutate(cum_dist = cumsum(freq) / sum(freq),
           past_half = cum_dist >= 0.5) %>%
    filter(past_half) %>%
    top_n(-1, cum_dist) %>%
    select(ID, rating) %>%
    arrange(ID)

<强>结果

     ID          rating
  <dbl>           <chr>
1     2 ratings_count_4
2     3 ratings_count_4
3     4 ratings_count_4
4     5 ratings_count_4
5     6 ratings_count_3
6     7 ratings_count_5

注意

我使用以下代码生成df。在将来，我建议包括这样的内容，以便用户轻松复制。

df <- data.frame( ID = c(2, 3, 4, 5, 6, 7), ratings_count_5 = c(599, 350, 454, 729, 548, 628), ratings_count_4 = c(624, 407, 368, 938, 176, 1), ratings_count_3 = c(78, 95, 52, 520, 431, 1), ratings_count_2 = c(357, 382, 245, 145, 313, 1), ratings_count_1 = c(4, 255, 512, 478, 459, 2))

How can I find the median rating for each product, given the count across?

4 个答案: