[英]Comparing rows within groups in a dataframe
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")
我想将dF$status[2]
与dF$status[1]
以及dF$status[3]
与dF$status[2]
等进行比较,以此类推。 我可以使用简单的函数和ddply()
相对容易地做到这一点:
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)
这给出了一个非常简单的输出,然后我可以将其与reshape
包一起melt()
并作为新列连接到我的dF
。
> 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
我的问题是当组中行数不同时。 例如,如果dF
突然丢失了“ dF $ group == 102”所在的一行,
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
我发现了使用不同功能的SO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
我的输出问题是,在不进行大量工作的情况下,将melt()
附加到我的原始dF1
上要困难得多,因为组102
数据在101
或102
的少数列中。 这特别困难,因为我有超过1500个组要应用此函数,其nrow()
可能随时间变化。
我想拥有的功能是将每行与组中的前一行进行比较,并理想地输出类似
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。
我已经在这里和其他地方搜索了帮助,但没有找到我想要的东西。 我确信这是可能的,并且我可能忽略了一些非常简单的事情。
谢谢你的帮助。
带有package 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
这是使用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"))
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