Question

我有以下数据集：

Name         Make_Miss       Half        
Player A         1             1                
Player B         1             1                
Player A         0             2                
Player A         0             1                
Player A         1             1                
Player B         0             2

名称是玩家的名字，Make_Miss是玩家是否进行了射击，而Half是射击的一半。我目前正在使用以下代码来计算前半部分的计数。

代码：

dt[ , Player_First_Made := .N, by = list(dt$Name == "Player A" & dt$Half == 1 & dt$Make_Miss == 1)]

输出：

Name         Make_Miss       Half        Player_First_Made
Player A         1             1                2
Player B         1             1                4
Player A         0             2                4
Player A         0             1                4
Player A         1             1                2
Player B         0             2                4

这里发生的事情是，无论玩家A在Make_Miss列中输入为0，Player_First_Made列中的相应行都会被分配与列表中的条件不匹配的镜头数（即名称）！=玩家A或半！= 1或Make_Miss！= 1）;但是，我的愿望如下：

Name         Make_Miss       Half        Player_First_Made
Player A         1             1                2
Player B         1             1                4
Player A         0             2                2
Player A         0             1                2
Player A         1             1                2
Player B         0             2                4

我希望与Name = Player A匹配的行总是具有他们在上半场制作的镜头数量。是否有某种data.table语法我可以指定这个赋值？

Answer 1

正如@ chinsoon12指出的那样，你提供的数据并不合理。但是，这是一个使用dplyr的方法，我认为会给你你想要的东西......

library(dplyr)

# Make some data
DATA <- data.frame(Name = c("Player A", "Player B", "Player C",
 "Player A", "Player A", "Player B"), Make_Miss = c(1,1,0,0,1,0),
 Half = c(1,1,2,1,2,2))

# Use dplyr to calculate the sums of 'Half' for each player
OUT <- DATA %>% group_by(Name) %>% mutate(Player_First_Made = sum(Half))

# Check the output
> OUT
# A tibble: 6 x 4
# Groups:   Name [3]
  Name     Make_Miss  Half Player_First_Made
  <fct>        <dbl> <dbl>             <dbl>
1 Player A      1     1                 4
2 Player B      1     1                 3 
3 Player C      0     2                 2
4 Player A      0     1                 4 
5 Player A      1     2                 4 
6 Player B      0     2                 3

如果这不是您想要的，那么请编辑您的问题以使其更清晰。

Answer 2

执行此操作的data.table方法是：

dat[Half == 1, .(Player_First_Made = sum(Make_Miss)), .(Name)
    ][dat, on = c('Name')]

第一行计算每个玩家（sum(Make_Miss)）在上半场（.(Name)）拍摄的次数（Half == 1）。

第二行将结果聚合表从上面的步骤连接回原始数据集。

这是我使用的示例数据：

dat <-
  data.table(
    Name = c('A', 'B'),
    Make_Miss = round(runif(30, 0, 1)),
    Half = round(runif(30, 1, 2))
  )

根据r

2 个答案: