Question

R的新手，所以我只是围绕数据争论方面。尝试寻找类似的问题，但找不到它。

我想添加一个额外的列，即文章组之间每天分割的视图百分比。示例数据集

  views       date      article
  1578   2015-01-01       A
  616    2015-01-01       B
  575    2015-01-01       C
  1744   2015-01-02       A
  541    2015-01-02       B
  660    2015-01-02       C
  2906   2015-01-03       A
  629    2015-01-03       B
  643    2015-01-03       C

我期待的结果......

 views     percentage   date           article
  1578     56.99        2015-01-01       A
  616      22.25        2015-01-01       B
  575      20.77        2015-01-01       C
  1744     59.22        2015-01-02       A
  541      18.37        2015-01-02       B
  660      22.41        2015-01-02       C
  2906     69.55        2015-01-03       A
  629      15.06        2015-01-03       B
  643      15.39        2015-01-03       C

我知道通过使用子集拆分日期框架可以实现这一点，但我希望使用库有更简洁的方法吗？

谢谢！

Answer 1

library(dplyr)
df %>% group_by(date) %>% mutate( percentage = views/sum(views))
Source: local data frame [9 x 4]
Groups: date

  views       date article percentage
1  1578 2015-01-01       A  0.5698808
2   616 2015-01-01       B  0.2224630
3   575 2015-01-01       C  0.2076562
4  1744 2015-01-02       A  0.5921902
5   541 2015-01-02       B  0.1837012
6   660 2015-01-02       C  0.2241087
7  2906 2015-01-03       A  0.6955481
8   629 2015-01-03       B  0.1505505
9   643 2015-01-03       C  0.1539014

或者，如果每天可能有多篇相同的文章：

df %>% group_by(date) %>% mutate(sum = sum(views)) %>% 
group_by(date, article) %>% mutate(percentage = views/sum) %>% 
select(-sum)

Answer 2

如果df是您的data.frame，您可以执行以下操作：

library(data.table)
setDT(df)[,percentage:=signif(100*views/sum(views),4),by=date][]
#   views       date article percentage
#1:  1578 2015-01-01       A      56.99
#2:   616 2015-01-01       B      22.25
#3:   575 2015-01-01       C      20.77
#4:  1744 2015-01-02       A      59.22
#5:   541 2015-01-02       B      18.37
#6:   660 2015-01-02       C      22.41
#7:  2906 2015-01-03       A      69.55
#8:   629 2015-01-03       B      15.06
#9:   643 2015-01-03       C      15.39

或基地R：

df$percentage = signif(100*with(df, ave(views, date, FUN=function(x) x/sum(x))),4)

数据：

df = structure(list(views = c(1578L, 616L, 575L, 1744L, 541L, 660L, 2906L, 629L, 643L), date = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L), .Label = c("2015-01-01", "2015-01-02", "2015-01-03" ), class = "factor"), article = structure(c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c("A", "B", "C"), class = "factor"), percentage = c(56.99, 22.25, 20.77, 59.22, 18.37, 22.41, 69.55, 15.06, 15.39)), .Names = c("views", "date", "article", "percentage"), class = "data.frame", row.names = c(NA, -9L))

计算R中每个组的每个时间序列观察的百分比

2 个答案: