Question

我有一个数据数组，其中包含有关人员和项目的一些信息：

person_id | project_id | action | time
--------------------------------------
        1 |          1 |      w |    1
        1 |          2 |      w |    2
        1 |          3 |      w |    2
        1 |          3 |      r |    3
        1 |          3 |      w |    4
        1 |          4 |      w |    4
        2 |          2 |      r |    2
        2 |          2 |      w |    3

我想用一些名为“first_time”和“first_time_project”的字段来扩充这些数据，这些字段共同确定第一次看到该人的任何操作，并且第一次开发人员看到对项目的任何操作。最后，数据应如下所示：

person_id | project_id | action | time | first_time | first_time_project
------------------------------------------------------------------------
        1 |          1 |      w |    1 |          1 |                  1
        1 |          2 |      w |    2 |          1 |                  2
        1 |          3 |      w |    2 |          1 |                  2
        1 |          3 |      r |    3 |          1 |                  2
        1 |          3 |      w |    4 |          1 |                  2
        1 |          4 |      w |    4 |          1 |                  4
        2 |          2 |      r |    2 |          2 |                  2
        2 |          2 |      w |    3 |          2 |                  2

我这样做的天真的方式是编写几个循环：

for (pid in unique(data$person_id)) {
    data[data$pid==pid, "first_time"] = min(data[data$pid==pid, "time"])
    for (projid in unique(data[data$pid==pid, "project_id"])) {
        data[data$pid==pid & data$project_id==projid, "first_time_project"] = min(data[data$pid==pid & data$project_id==projid, "time"]
    }
}

现在，通过双嵌套循环看到这种情况会变得非常慢，这并不是天才。但是，我无法想办法在R中处理这个问题。我有点仿效SQL的group by选项。我知道也许可以提供帮助，但我无法弄清楚如何做多个切片。

有关如何将我的代码从冰川缓慢变为更快的某些提示？我现在对蜗牛感到高兴。

Answer 1

Hadley的plyr和transform（）的组合非常强大。如果我正确理解你的问题，那么：

foo <- ddply(foo, .(person_id), transform, first_time=min(time))
foo <- ddply(foo, .(person_id, project_id), transform, 
  first_time_project=min(time))

Answer 2

尝试ave：

transform(data, 
   first_time = ave(time, person_id, FUN = min),
   first_time_project = ave(time, person_id, project_id, drop = TRUE, FUN = min)
)

Answer 3

如果您正在寻找速度，那么data.table就是您的选择。

library(data.table)
DT <- data.table(foo)
DT[, first_time := min(time), by = person_id]
DT[, first_time_project := min(time), by = list(person_id, project_id)]

Answer 4

没有循环的快速而肮脏的解决方案

library(plyr)


# function to get first time by any person/project
fp <- function(dat) 
{
dat$first_time=min(dat$time)
ftp <- function(d) { d$first_time_project=min(d$time); return (d) }
dat=ddply(dat, .(project_id), ftp)
return (dat)
}


#this single call should give you the result you want
result=ddply(data, .(person_id), fp)

Answer 5

我能想到的快速方法：

foo <- data.frame(
       person_id=rep(1:5,each=6),
       project_id=sample(1:5,30,T),
       time=sample(1:30))

first_time <- aggregate(foo$time, list(foo$person_id), min)

foo$first_time <- first_time[ match(foo$person_id,first_time[,1]),2]

bar <- subset(foo, time==first_time)

foo$first_time_project <- bar$project_id[match(foo$person_id, bar$person_id)]

在多个不同切片上应用聚合函数

5 个答案: