如何按组缩小此数据框？

Question

我有以下

t <- structure(list(name = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), .Label = c("Alice", "Bob", 
"Jane Doe", "John Doe"), class = "factor"), school = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L), .Label = c("Alice School", 
"Bob School", "Someother School", "Someschool College"), class = "factor"), 
    group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("A", "B"), class = "factor"), 
    question = structure(c(2L, 4L, 6L, 8L, 1L, 3L, 5L, 7L, 2L, 
    4L, 6L, 8L, 1L, 3L, 5L, 7L, 2L, 4L, 6L, 8L, 1L, 3L, 5L, 7L, 
    2L, 4L, 6L, 8L, 1L, 3L, 5L, 7L), .Label = c("q1", "q2", "q3", 
    "q4", "q5", "q6", "q7", "q8"), class = "factor"), mark = c(0L, 
    0L, 0L, 0L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 
    1L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 1L, 0L, 1L, 0L, 1L, 
    1L), subject = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
    1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 
    2L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c("C", "M"), class = "factor")), .Names = c("name", 
"school", "group", "question", "mark", "subject"), row.names = c(7L, 
15L, 23L, 31L, 3L, 11L, 19L, 27L, 8L, 16L, 24L, 32L, 4L, 12L, 
20L, 28L, 6L, 14L, 22L, 30L, 2L, 10L, 18L, 26L, 5L, 13L, 21L, 
29L, 1L, 9L, 17L, 25L), class = "data.frame") 

，我需要产生其中每个学生具有用于每个受试者一个组合标记的数据帧。这个组合只是每个问题上的标记总和。所以，例如，Jane Doe将在主题C上有3个，在主题M上有2个。我已经用Reduce和其他方法足够长的时间来敲打我的头。我可以用一种非常程序化的方式来解决这个问题，但是如果我可以用单线（或近似）来做到这一点，我会更开心。我敢肯定它可以做...

Answer 1

你说过你的问题;要group_by学生，主题和计算总和

library(tidyverse) 
asdf %>% 
    group_by(name, subject) %>% 
    summarise(score = sum(mark)) 

Answer 2

这里一个data.table解决方案：

library(data.table) 
setDT(t)[, sum(mark), by = list(name, subject)] 

Answer 3

和公正的完整性，基础R：

aggregate(mark ~ name + subject, data=t, sum) 

这是说“汇总响应变量mark由分组变量name和subject，使用sum a是聚合功能“。