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R 如何通过公共ID从大型数据集中提取行，并利用这些行的方式生成具有这些ID的列

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R 如何通过公共ID从大型数据集中提取行，并利用这些行的方式生成具有这些ID的列,r,data.table,plyr,R,Data.table,Plyr,我知道这是一个非常愚蠢的问题，但我无法解决它，这就是为什么问。。。 如何通过公共ID从大型数据集中提取行，并利用这些行的方法生成一个列，其中包含这些ID作为行名。 e、 g 使用plyr功能ddply可以轻松完成此类操作： 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

我知道这是一个非常愚蠢的问题，但我无法解决它，这就是为什么问。。。 如何通过公共ID从大型数据集中提取行，并利用这些行的方法生成一个列，其中包含这些ID作为行名。 e、 g

使用
plyr
功能
ddply
可以轻松完成此类操作：

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，并且/或者唯一
ID
的数量很大，则可以使用
data.table
。有关
plyr

的更多详细信息，请参阅。如果您有一个大的data.frame，则可以使用
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

ddply
的一些替代方案是
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
Venables和Ripley（2000年，第37页）提出，将
未列出
、
拉普拉
和
拆分
相结合比仅使用
sapply
更快，在这个特定的例子中，它甚至比
数据更快


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


参考：

维纳布尔斯，W.N.和里普利，B.D.（2000年）。S编程。斯普林格。统计和计算
ISBN 0-387-98966-8（碱性纸）

放大（从Matthew Dowle编辑）
更多群组


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



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



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 


更多行


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



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



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 



dat非常感谢您对Paul Hiemstra的帮助，感谢您的论文推荐，我以后也会注意接受答案。对不起，我没有接受之前的答案，尽管它们都很好用。没问题！只是提到它正在发生：）+1也。但单次运行的0.06 vs 0.10既不显著也不稳健。
unlist
、
lappy
和
split
方法不可缩放。它返回一个不同的结果。