我怎样才能更快地对小组观察结果进行排名? [英] How can I rank observations in-group faster?

问题描述

我有一个非常简单的问题，但我可能没有考虑矢量有效地解决问题.我尝试了两种不同的方法，并且它们已经在两台不同的计算机上循环很长时间了.我希望我能说比赛使比赛变得更加激动人心，但是...令人沮丧.

I have a really simple problem, but I'm probably not thinking vector-y enough to solve it efficiently. I tried two different approaches and they've been looping on two different computers for a long time now. I wish I could say the competition made it more exciting, but ... bleh.

我有很长的数据(每人多行，每人观察一行)，我基本上想要一个变量，它告诉我已经观察过该人的频率.

I have long data (many rows per person, one row per person-observation) and I basically want a variable, that tells me how often the person has been observed already.

我有前两列，并希望第三列:

person  wave   obs
pers1   1999   1
pers1   2000   2
pers1   2003   3
pers2   1998   1
pers2   2001   2

现在我正在使用两个循环方法.两者都非常慢(15万行).我确定我会丢失一些东西，但是我的搜索查询还没有真正帮助到我(很难说明问题).

Now I'm using two loop-approaches. Both are excruciatingly slow (150k rows). I'm sure I'm missing something, but my search queries didn't really help me yet (hard to phrase the problem).

感谢任何指针！

# ordered dataset by persnr and year of observation
person.obs <- person.obs[order(person.obs$PERSNR,person.obs$wave) , ]

person.obs$n.obs = 0

# first approach: loop through people and assign range
unp = unique(person.obs$PERSNR)
unplength = length(unp)
for(i in 1:unplength) {
   print(unp[i])
   person.obs[which(person.obs$PERSNR==unp[i]),]$n.obs = 
1:length(person.obs[which(person.obs$PERSNR==unp[i]),]$n.obs)
    i=i+1
   gc()
}

# second approach: loop through rows and reset counter at new person
pnr = 0
for(i in 1:length(person.obs[,2])) {
  if(pnr!=person.obs[i,]$PERSNR) { pnr = person.obs[i,]$PERSNR
  e = 0
  }
  e=e+1
  person.obs[i,]$n.obs = e
  i=i+1
  gc()
}

推荐答案

带有 dplyr 软件包.

数据表:

library(data.table)
setDT(foo)[, rn := 1:.N, by = person]  # setDT(foo) is needed to convert to a data.table

或具有新的rowid函数( v1.9.7 + ，目前仅在开发版本)

Or with the new rowid function (v1.9.7+, currently thus only available in the development version)

setDT(foo)[, rn := rowid(person)]   

都给:

> foo
   person year rn
1:  pers1 1999  1
2:  pers1 2000  2
3:  pers1 2003  3
4:  pers2 1998  1
5:  pers2 2011  2

如果您想获得真实排名，则应使用frank函数:

If you want a true rank, you should use the frank function:

setDT(foo)[, rn := frank(year, ties.method = 'dense'), by = person]

dplyr:

library(dplyr)
# method 1
foo <- foo %>% group_by(person) %>% mutate(rn = row_number())
# method 2
foo <- foo %>% group_by(person) %>% mutate(rn = 1:n())

两者都给出相似的结果:

both giving a similar result:

> foo
Source: local data frame [5 x 3]
Groups: person [2]

  person  year    rn
  (fctr) (dbl) (int)
1  pers1  1999     1
2  pers1  2000     2
3  pers1  2003     3
4  pers2  1998     1
5  pers2  2011     2

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