Mai*_*ura 33 r plyr dataframe data.table
我的数据集中有一列,其中时间段(Time)是从ab开始的整数.有时,任何特定组都可能缺少时间段.我想用这些行填写NA.以下是1(几千个)组的示例数据.
structure(list(Id = c(1, 1, 1, 1), Time = c(1, 2, 4, 5), Value = c(0.568780482159894,
-0.7207749516298, 1.24258192959273, 0.682123081696789)), .Names = c("Id",
"Time", "Value"), row.names = c(NA, 4L), class = "data.frame")
Id Time Value
1 1 1 0.5687805
2 1 2 -0.7207750
3 1 4 1.2425819
4 1 5 0.6821231
如您所见,时间3缺失.通常一个或多个可能会丢失.我可以自己解决这个问题,但恐怕我不会以最有效的方式做到这一点.我的方法是创建一个函数:
生成时间段从序min(Time)到max(Time)
然后做一个setdiff抓取缺失的Time值.
将该向量转换为a data.frame
拉出唯一标识符变量(Id以及上面未列出的其他变量),并将其添加到此data.frame中.
合并两者.
从功能返回.
因此整个过程将按如下方式执行:
# Split the data into individual data.frames by Id.
temp_list <- dlply(original_data, .(Id))
# pad each data.frame
tlist2 <- llply(temp_list, my_pad_function)
# collapse the list back to a data.frame
filled_in_data <- ldply(tlist2)
更好的方法来实现这一目标
Mat*_*wle 36
跟随Ben Barnes的评论并从他的开始mydf3:
DT = as.data.table(mydf3)
setkey(DT,Id,Time)
DT[CJ(unique(Id),seq(min(Time),max(Time)))]
Id Time Value Id2
[1,] 1 1 -0.262482283 2
[2,] 1 2 -1.423935165 2
[3,] 1 3 0.500523295 1
[4,] 1 4 -1.912687398 1
[5,] 1 5 -1.459766444 2
[6,] 1 6 -0.691736451 1
[7,] 1 7 NA NA
[8,] 1 8 0.001041489 2
[9,] 1 9 0.495820559 2
[10,] 1 10 -0.673167744 1
First 10 rows of 12800 printed.
setkey(DT,Id,Id2,Time)
DT[CJ(unique(Id),unique(Id2),seq(min(Time),max(Time)))]
Id Id2 Time Value
[1,] 1 1 1 NA
[2,] 1 1 2 NA
[3,] 1 1 3 0.5005233
[4,] 1 1 4 -1.9126874
[5,] 1 1 5 NA
[6,] 1 1 6 -0.6917365
[7,] 1 1 7 NA
[8,] 1 1 8 NA
[9,] 1 1 9 NA
[10,] 1 1 10 -0.6731677
First 10 rows of 25600 printed.
CJ代表Cross Join,请参阅?CJ.带有NAs 的填充是因为nomatch默认情况下是NA.设置nomatch为0取消不匹配.如果不是用NAs 填充,而是需要主流行,只需添加即可roll=TRUE.这比用NAs 填充更有效,然后填充NAs.请参阅rollin 的说明?data.table.
setkey(DT,Id,Time)
DT[CJ(unique(Id),seq(min(Time),max(Time))),roll=TRUE]
Id Time Value Id2
[1,] 1 1 -0.262482283 2
[2,] 1 2 -1.423935165 2
[3,] 1 3 0.500523295 1
[4,] 1 4 -1.912687398 1
[5,] 1 5 -1.459766444 2
[6,] 1 6 -0.691736451 1
[7,] 1 7 -0.691736451 1
[8,] 1 8 0.001041489 2
[9,] 1 9 0.495820559 2
[10,] 1 10 -0.673167744 1
First 10 rows of 12800 printed.
setkey(DT,Id,Id2,Time)
DT[CJ(unique(Id),unique(Id2),seq(min(Time),max(Time))),roll=TRUE]
Id Id2 Time Value
[1,] 1 1 1 NA
[2,] 1 1 2 NA
[3,] 1 1 3 0.5005233
[4,] 1 1 4 -1.9126874
[5,] 1 1 5 -1.9126874
[6,] 1 1 6 -0.6917365
[7,] 1 1 7 -0.6917365
[8,] 1 1 8 -0.6917365
[9,] 1 1 9 -0.6917365
[10,] 1 1 10 -0.6731677
First 10 rows of 25600 printed.
你可以用tidyr它.
使用tidyr::complete填写行的Time,默认情况下该值与填充NA.
我扩展了样本数据以显示它适用于多个Ids,甚至在Id整个范围内Time都不存在.
library(dplyr)
library(tidyr)
df <- tibble(
Id = c(1, 1, 1, 1, 2, 2, 2),
Time = c(1, 2, 4, 5, 2, 3, 5),
Value = c(0.56, -0.72, 1.24, 0.68, 1.46, 0.74, 0.99)
)
df
#> # A tibble: 7 x 3
#> Id Time Value
#> <dbl> <dbl> <dbl>
#> 1 1 1 0.56
#> 2 1 2 -0.72
#> 3 1 4 1.24
#> 4 1 5 0.68
#> 5 2 2 1.46
#> 6 2 3 0.74
#> 7 2 5 0.99
df %>% complete(nesting(Id), Time = seq(min(Time), max(Time), 1L))
#> # A tibble: 10 x 3
#> Id Time Value
#> <dbl> <dbl> <dbl>
#> 1 1 1 0.56
#> 2 1 2 -0.72
#> 3 1 3 NA
#> 4 1 4 1.24
#> 5 1 5 0.68
#> 6 2 1 NA
#> 7 2 2 1.46
#> 8 2 3 0.74
#> 9 2 4 NA
#> 10 2 5 0.99
请参阅马修道尔的回答（现在，希望在上面）。
这是使用data.table包的东西，当有多个 ID 变量时它可能会有所帮助。它也可能比 更快merge，具体取决于您想要的结果。我会对基准测试和/或建议的改进感兴趣。
首先，用两个ID变量创建一些要求更高的数据
library(data.table)
set.seed(1)
mydf3<-data.frame(Id=sample(1:100,10000,replace=TRUE),
Value=rnorm(10000))
mydf3<-mydf3[order(mydf3$Id),]
mydf3$Time<-unlist(by(mydf3,mydf3$Id,
function(x)sample(1:(nrow(x)+3),nrow(x)),simplify=TRUE))
mydf3$Id2<-sample(1:2,nrow(mydf3),replace=TRUE)
创建一个函数（这已被编辑- 查看历史）
padFun<-function(data,idvars,timevar){
# Coerce ID variables to character
data[,idvars]<-lapply(data[,idvars,drop=FALSE],as.character)
# Create global ID variable of all individual ID vars pasted together
globalID<-Reduce(function(...)paste(...,sep="SOMETHINGWACKY"),
data[,idvars,drop=FALSE])
# Create data.frame of all possible combinations of globalIDs and times
allTimes<-expand.grid(globalID=unique(globalID),
allTime=min(data[,timevar]):max(data[,timevar]),
stringsAsFactors=FALSE)
# Get the original ID variables back
allTimes2<-data.frame(allTimes$allTime,do.call(rbind,
strsplit(allTimes$globalID,"SOMETHINGWACKY")),stringsAsFactors=FALSE)
# Convert combinations data.frame to data.table with idvars and timevar as key
allTimesDT<-data.table(allTimes2)
setnames(allTimesDT,1:ncol(allTimesDT),c(timevar,idvars))
setkeyv(allTimesDT,c(idvars,timevar))
# Convert data to data.table with same variables as key
dataDT<-data.table(data,key=c(idvars,timevar))
# Join the two data.tables to create padding
res<-dataDT[allTimesDT]
return(res)
}
使用功能
(padded2<-padFun(data=mydf3,idvars=c("Id"),timevar="Time"))
# Id Time Value Id2
# [1,] 1 1 -0.262482283 2
# [2,] 1 2 -1.423935165 2
# [3,] 1 3 0.500523295 1
# [4,] 1 4 -1.912687398 1
# [5,] 1 5 -1.459766444 2
# [6,] 1 6 -0.691736451 1
# [7,] 1 7 NA NA
# [8,] 1 8 0.001041489 2
# [9,] 1 9 0.495820559 2
# [10,] 1 10 -0.673167744 1
# First 10 rows of 12800 printed.
(padded<-padFun(data=mydf3,idvars=c("Id","Id2"),timevar="Time"))
# Id Id2 Time Value
# [1,] 1 1 1 NA
# [2,] 1 1 2 NA
# [3,] 1 1 3 0.5005233
# [4,] 1 1 4 -1.9126874
# [5,] 1 1 5 NA
# [6,] 1 1 6 -0.6917365
# [7,] 1 1 7 NA
# [8,] 1 1 8 NA
# [9,] 1 1 9 NA
# [10,] 1 1 10 -0.6731677
# First 10 rows of 25600 printed.
在与原始数据合并之前，edited 函数将 globalID 拆分为组合 data.frame 中的组成部分。这应该（我认为）更好。
我的一般方法是使用freqTable <- as.data.frame(table(idvar1, idvar2, idvarN))then 拉出其中的行Freq==0，根据需要填充，然后堆栈回原始数据。