我有一个示例dataFrame
dF <- structure(list(status = structure(c(1L, 1L, 1L, 4L, 1L, 3L, 1L,
1L, 2L, 4L, 4L, 2L), .Label = c("complete", "go", "no go", "revise"
), class = "factor"), group = structure(c(1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c("101", "102", "103"), class =
"factor"),
date = structure(c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L,
3L, 4L), .Label = c("1", "2", "3", "4"), class = "factor")),
.Names = c("status",
"group", "date"), row.names = c(NA, -12L), class = "data.frame")
我希望在每个组中比较dF$status[2]
到dF$status[1]
和dF$status[3]
到dF$status[2]
等等。我可以通过一个简单的函数和ddply()
:
state_change_function <- function(x){
tmp <- integer(length = nrow(x))
for(i in 2:nrow(x)){
if(x$statu[i] == x$status[i-1]){
tmp[i] <- "no change"
} else {
tmp[i] <- "state change"
}
}
return(tmp)
}
state_change <- ddply(dF, .(group), state_change_function)
这提供了一个非常简单的输出,然后我可以melt()
使用reshape
包并附加到我的dF
作为新列。
> state_change
group V1 V2 V3 V4
1 101 0 no change no change state change
2 102 0 state change state change no change
3 103 0 state change no change state change
我的问题是当组中的行数不同时。例如,如果dF
突然丢失了一行`dF $ group == 102&#34;,
dF1 <- structure(list(status = structure(c(1L, 1L, 1L, 4L, 3L, 1L, 1L,
2L, 4L, 4L, 2L), .Label = c("complete", "go", "no go", "revise"
), class = "factor"), group = structure(c(1L, 1L, 1L, 1L, 2L,
2L, 2L, 3L, 3L, 3L, 3L), .Label = c("101", "102", "103"), class = "factor"),
date = structure(c(1L, 2L, 3L, 4L, 2L, 3L, 4L, 1L, 2L, 3L,
4L), .Label = c("1", "2", "3", "4"), class = "factor")), .Names =
c("status",
"group", "date"), row.names = c(NA, -11L), class = "data.frame")
然后运行相同的函数会导致错误:
state_change <- ddply(dF1, .(group), state_change_function)
Error in list_to_dataframe(res, attr(.data, "split_labels"), .id, id_as_factor) :
Results do not have equal lengths
我在SO上发现了一个使用不同功能的部分解决方案:
state_change_function <- function(data){
output <- integer(length(rrsIdeas)-1)
for(i in seq_along(output)){
output[[i]] <- (data$status[i] == data$status[i+1])
}
return(output)
}
state_change <- ddply(dF1, .(group), state_change_function)
并提供不同的输出:
> state_change
group V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17
1 101 1 1 0 NA NA NA NA NA NA NA NA NA NA NA NA NA NA
2 102 0 1 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
3 103 0 1 0 NA NA NA NA NA NA NA NA NA NA NA NA NA NA
我对此输出的问题是,melt()
更加困难,并且在没有太多工作的情况下附加到原始dF1
因为组102
包含数据101
或102
的几列。这特别困难,因为我有超过1500个组,我应用这个函数nrow()
可能会随着时间的推移而改变。
我想要的是一个功能,可以将每一行与一组中的上一行进行比较,并且 - 理想情况下 - 输出一个像
这样的数据框架group V1
101 0
101 no change
101 no change
101 state change
102 0
102 state change
102 state change
102 no change
etc...
但是,如果某些组的行数少于其他组,则可以限制该组的dataFrame中的行数。
我在这里和其他地方寻求帮助,但没有找到我正在寻找的东西。我确信这是可能的,我可能会忽略一些非常简单的事情。
感谢您的帮助。
答案 0 :(得分:3)
包data.table
的解决方案:
library(data.table)
setDT(dF1)[,V1:=c("0",ifelse(head(status,-1)!=status[-1],'change','no change')),group]
# status group date V1
# 1: complete 101 1 0
# 2: complete 101 2 no change
# 3: complete 101 3 no change
# 4: revise 101 4 change
# 5: no go 102 2 0
# 6: complete 102 3 change
# 7: complete 102 4 no change
# 8: go 103 1 0
# 9: revise 103 2 change
#10: revise 103 3 no change
#11: go 103 4 change
答案 1 :(得分:3)
以下是使用dplyr
的解决方案:
library(dplyr)
dF$status <- as.character(dF$status)
dF %>%
group_by(group) %>%
mutate(change = ifelse(status == lag(status), "no change", "change"))