3

我知道这是一个非常愚蠢的问题，但我无法解决它，这就是为什么要问...我如何从公共 ID 的大型数据集中提取行并利用这些行的方法并制作具有这些 ID 的列作为行名。例如

IDs Var2    
Ae4 2       
Ae4 4
Ae4 6
Bc3 3   
Bc3 5   
Ad2 8
Ad2 7

OutPut
Var(x)
Ae4 4
Bc3 4
Ad2 7.5
4

2 回答 2

8

如果你有一个大的data.frame，你可以使用data.table

一些替代ddply是aggregate和data.table

set.seed(001)
dat <- data.frame(ID = rep(LETTERS[1:5], each = 20), value = runif(1e6))


library(data.table)
DT <- data.table(dat)
DT[, mean(value), by=list(ID)]  # data.table approach

aggregate(.~ID, data=dat, mean) # aggregate (R Base function) approach




library(rbenchmark) # comparing performance
benchmark(DT[, mean(value), by=list(ID)],                 # data.table approach
          aggregate(.~ID, data=dat, mean),                # aggregate  approach
          ddply(dat, .(ID), summarize, mn = mean(value)), # ddply approach (Paul Hiemstra's answer)
          columns=c("test", "replications", "elapsed", "relative"),
          order='relative',
          replications=1)

                                            test replications elapsed relative
1               DT[, mean(value), by = list(ID)]            1    0.14    1.000
3 ddply(dat, .(ID), summarize, mn = mean(value))            1    0.58    4.143
2            aggregate(. ~ ID, data = dat, mean)            1    3.59   25.643

如您所见，最快的是data.table方法。

编辑

有一个 R 基础方法甚至比 更快data.table，让我们看看：

 unlist(lapply(split(dat$value, dat$ID), mean)) # another R Base approach

    benchmark(DT[, mean(value), by=list(ID)],                 # data.table approach
              aggregate(.~ID, data=dat, mean),                # aggregate  approach
              ddply(dat, .(ID), summarize, mn = mean(value)), # ddply approach (Paul Hiemstra's answer)
              unlist(lapply(split(dat$value, dat$ID), mean)), # lapply, split approach
              columns=c("test", "replications", "elapsed", "relative"),
              order='relative',
              replications=1)
                                            test replications elapsed relative
4 unlist(lapply(split(dat$value, dat$ID), mean))            1    0.06    1.000
1               DT[, mean(value), by = list(ID)]            1    0.10    1.667
3 ddply(dat, .(ID), summarize, mn = mean(value))            1    0.56    9.333
2            aggregate(. ~ ID, data = dat, mean)            1    3.28   54.667

Venables 和 Ripley (2000, pag.37) 建议组合unlist,lapply和split比仅使用更快sapply，在这个特定示例中它甚至比data.table

参考：

Venables, WN 和 Ripley, BD (2000)。S 编程。施普林格。统计和计算 ISBN 0-387-98966-8（alk.paper）

扩大规模（Matthew Dowle 编辑）

更多组

dat <- data.frame(ID = as.character(as.hexmode(1:2000)), value = runif(1e6))
DT <- as.data.table(dat)
benchmark(
  DT[, mean(value), by=ID],
  aggregate(.~ID, data=dat, mean),
  ddply(dat, .(ID), summarize, mn = mean(value)),
  unlist(lapply(split(dat$value, dat$ID), mean)),
columns=c("test", "replications", "elapsed", "relative"),
order='relative',
replications=3)
                                            test replications elapsed relative
1                     DT[, mean(value), by = ID]            3    0.33    1.000
4 unlist(lapply(split(dat$value, dat$ID), mean))            3    0.41    1.242
2            aggregate(. ~ ID, data = dat, mean)            3    7.69   23.303
3 ddply(dat, .(ID), summarize, mn = mean(value))            3   17.08   51.758

更多行

dat <- data.frame(ID = as.character(as.hexmode(1:2000)), value = runif(1e7))
DT <- as.data.table(dat)
benchmark(
  DT[, mean(value), by=ID],
  aggregate(.~ID, data=dat, mean),
  ddply(dat, .(ID), summarize, mn = mean(value)),
  unlist(lapply(split(dat$value, dat$ID), mean)),
columns=c("test", "replications", "elapsed", "relative"),
order='relative',
replications=3)
                                            test replications elapsed relative
1                     DT[, mean(value), by = ID]            3    3.18    1.000
4 unlist(lapply(split(dat$value, dat$ID), mean))            3    4.26    1.340
2            aggregate(. ~ ID, data = dat, mean)            3   90.28   28.390
3 ddply(dat, .(ID), summarize, mn = mean(value))            3  268.86   84.547

先设置一个键

system.time(setkey(DT,ID))
   user  system elapsed 
   0.71    0.03    0.75 
object.size(dat)
152.7 Mb              # Quite small. Easy for a 32bit PC with 2GB RAM.
object.size(DT)
152.7 Mb
benchmark(
  DT[, mean(value), by=ID],
  aggregate(.~ID, data=dat, mean),
  ddply(dat, .(ID), summarize, mn = mean(value)),
  unlist(lapply(split(dat$value, dat$ID), mean)),
columns=c("test", "replications", "elapsed", "relative"),
order='relative',
replications=3)
                                            test replications elapsed relative
1                     DT[, mean(value), by = ID]            3    0.95    1.000
4 unlist(lapply(split(dat$value, dat$ID), mean))            3    4.08    4.295
2            aggregate(. ~ ID, data = dat, mean)            3   91.76   96.589
3 ddply(dat, .(ID), summarize, mn = mean(value))            3  265.15  279.105

更多行

dat <- data.frame(ID = rep(1:2000,each=50000), value = runif(1e8))
DT <- as.data.table(dat)
system.time(setkey(DT,ID))
   user  system elapsed 
   2.10    0.25    2.34 
object.size(dat)
   1.1 Gb             # Comfortable for a 64bit PC with 8GB RAM
object.size(DT)
   1.1 Gb
benchmark(
  DT[, mean(value), by=ID],                 
  unlist(lapply(split(dat$value, dat$ID), mean)),
columns=c("test", "replications", "elapsed", "relative"),
order='relative',
replications=3)
                                            test replications elapsed relative
1                     DT[, mean(value), by = ID]            3    7.30    1.000
2 unlist(lapply(split(dat$value, dat$ID), mean))            3  184.83   25.319
于 2012-10-18T14:17:38.720 回答
1

plyr使用以下功能可以轻松完成此类事情ddply：

dat = data.frame(ID = rep(LETTERS[1:5], each = 20), value = runif(100))
> head(dat)
  ID      value
1  A 0.45800889
2  A 0.11221072
3  A 0.58833532
4  A 0.70056704
5  A 0.08337996
6  A 0.05195357

ddply(dat, .(ID), summarize, mn = mean(value))
  ID        mn
1  A 0.4960083
2  B 0.5809681
3  C 0.4512388
4  D 0.5079790
5  E 0.5397708

如果您的数据集很大，和/或唯一的数量ID很大，您可以使用data.table. 有关. _ _plyr

于 2012-10-18T13:49:58.910 回答