如何计算条件不同列的行

Question

我需要计算每个影片（列）的评分为4或大于4的行数。然后将它除以评分总数。如何才能做到这一点 ？ 请看下面的图片以获取一个简单的想法。

最终的结果应该是这样的

0.7000000, 'The Shawshank Redemption' 
0.5333333, 'Star Wars IV - A New Hope' 
0.5000000, 'Gladiator' 
0.4444444, 'Blade Runner' 
0.4375000, 'The Silence of the Lambs' 

Answer 1

ratings<-data.frame(User=c("John","Maria","Anton","Roger","Martina","Ana","Sergi","Marc","Jim","Chris") 
        ,Star.Wars.IV...A.New.Hope=c(1,5,NA,NA,4,2,NA,4,5,4) 
        ,Star.Wars.VI...Return.of.the.Jedi=c(5,3,NA,3,3,4,NA,NA,1,2) 
        ,Forrest.Gump=c(2,NA,NA,NA,4,4,3,NA,NA,2) 
        ) 
ratings 

 User Star.Wars.IV...A.New.Hope Star.Wars.VI...Return.of.the.Jedi Forrest.Gump 
1  John       1         5   2 
2 Maria       5         3   NA 
3 Anton      NA        NA   NA 
4 Roger      NA         3   NA 
5 Martina       4         3   4 
6  Ana       2         4   4 
7 Sergi      NA        NA   3 
8  Marc       4        NA   NA 
9  Jim       5         1   NA 
10 Chris       4         2   2 

如果你想包括NA S中的总收视率统计：

colSums(ratings[,-1]>=4,na.rm=T)/nrow(ratings) 

 Star.Wars.IV...A.New.Hope Star.Wars.VI...Return.of.the.Jedi      Forrest.Gump 
           0.5        0.2        0.2 

如果你想排除从总收视率NA count：伯爵：

colMeans(ratings[,-1]>=4,na.rm=T) 

 Star.Wars.IV...A.New.Hope Star.Wars.VI...Return.of.the.Jedi      Forrest.Gump 
        0.7142857143      0.2857142857      0.4000000000 

Answer 2

您可以使用colMeans计算百分比和stack结果以长格式：

示例数据帧：

df = data.frame(user = c("A", "B", "C", "D"), 
       movieA = c(4,2,NA,5), 
       movieB = c(1,1,NA,4)) 

stack(colMeans(df[-1] >= 4, na.rm = T)) 

#  values ind 
#1 0.6666667 movieA 
#2 0.3333333 movieB 

要看到这是如何工作的：

df[-1] >= 4      # returns a boolean matrix where ratings >= 4 gives TRUE 

#  movieA movieB 
#[1,] TRUE FALSE 
#[2,] FALSE FALSE 
#[3,]  NA  NA 
#[4,] TRUE TRUE 

和布尔向量的平均值TRUEs的比例（与NA删除），所以计算出的平均值与colMeans所有列会给你个你需要。

Answer 3

该数据不是一般整洁的格式。 df是具有某些临时值的数据帧。

library(dplyr) 
df <- data_frame(user = letters[1:10], 
      m1 = c(1,5,NA,NA,4,2,NA,4,5,4), 
      m2 = c(5,3,NA,3,3,4,NA,NA,1,2), 
      m3 = c(2,NA,NA,NA,4,4,3,NA,NA,NA)) 
df 
# A tibble: 10 × 4 
# user m1 m2 m3 
# <chr> <dbl> <dbl> <dbl> 
#1  a  1  5  2 
#2  b  5  3 NA 
#3  c NA NA NA 
#4  d NA  3 NA 
#5  e  4  3  4 
#6  f  2  4  4 
#7  g NA NA  3 
#8  h  4 NA NA 
#9  i  5  1 NA 
#10  j  4  2 NA 

让我们把它转换为key:value对即movie:rating，在这种情况下。

library(tidyr)  
df <- gather(df, movie, rating, -user) 
df 
# A tibble: 30 × 3 
# user movie rating 
# <chr> <chr> <dbl> 
#1  a m1  1 
#2  b m1  5 
#3  c m1  NA 
#4  d m1  NA 
#5  e m1  4 
#6  f m1  2 
#7  g m1  NA 
#8  h m1  4 
#9  i m1  5 
#10  j m1  4 
# ... with 20 more rows 

现在很容易总结。

df %>% group_by(movie) %>% summarise(countp = mean(rating>=4, na.rm=T)) 
# A tibble: 3 × 2 
# movie countp 
# <chr> <dbl> 
#1 m1 0.7142857 
#2 m2 0.2857143 
#3 m3 0.5000000