我有一个数据集,序列中有一些缺失值:
Seq<-c(1,2,3,4,6,7,10,11,12,18,19,20)
Data<-c(3,4,5,4,3,2,1,2,3,5,4,3)
DF<-data.frame(Seq, Data)
我想在此数据集中添加行,近似于我缺少值的位置,并使用NA填充数据。因此,每当我有一个大于2的间隙时,我会添加一个NA行(如果间隙很大,则添加多行)。结果看起来像这样:
NewSeq<-c(1,2,3,4,6,7,8.5,10,11,12,14,16,18,19,20)
NewData<-c(3,4,5,4,3,2,NA,1,2,3,NA,NA,18,19,20)
NewDF<-data.frame(NewSeq,NewData)
所以我忽略了差距只是&lt; 2,但是每当有间隙时我都会添加NA行&gt; 2.如果还有一个&gt;添加NA行后2个间隙,我添加另一个,直到填补了间隙。
答案 0 :(得分:1)
不是很优雅,但我会这样做:
Seq<-c(1,2,3,4,6,7,10,11,12,18,19,20)
Data<-c(3,4,5,4,3,2,1,2,3,5,4,3)
DF<-data.frame(Seq, Data)
first <- DF$Seq
second <- DF$Data
for(i in length(first):2) {
gap <- first[i] - first[i - 1]
if(gap > 2) {
steps <- ifelse(gap %% 2 == 1, gap %/% 2, (gap %/% 2) -1)
new_values_gap <- gap / (steps + 1)
new_values <- vector('numeric')
for(j in 1:steps) {
new_values <- c(new_values, first[i - 1] + j * new_values_gap)
}
first <- c(first[1:i - 1], new_values, first[i:length(first)])
second <- c(second[1:i - 1], rep(NA, length(new_values)), second[i:length(second)])
}
}
NewDF <- data.frame(NewSeq = first, NewData = second)
> NewDF
## NewSeq NewData
## 1 1.0 3
## 2 2.0 4
## 3 3.0 5
## 4 4.0 4
## 5 6.0 3
## 6 7.0 2
## 7 8.5 NA
## 8 10.0 1
## 9 11.0 2
## 10 12.0 3
## 11 14.0 NA
## 12 16.0 NA
## 13 18.0 5
## 14 19.0 4
## 15 20.0 3
答案 1 :(得分:1)
似乎适用于您的示例,但不确定它将如何对我尚未见过的数据执行。您需要根据您想要考虑不同差异区间的方式来调整ifelse语句中的区间。
Seq<-c(1,2,3,4,6,7,10,11,12,18,19,20)
Data<-c(3,4,5,4,3,2,1,2,3,5,4,3)
DF<-data.frame(Seq, Data)
diffs <- diff(Seq)
inds <- which(diffs > 2)
new.vals <- sapply(inds, function(x)
if(diffs[x] %% 2 != 0){
seq(Seq[x]+1.5, Seq[x+1]-1.5,1.5)
}else{
seq(Seq[x]+2, Seq[x+1]-2,2)
})
add.length <- unlist(lapply(new.vals, function(x) length(x)))
Seq.new <- c(Seq, unlist(new.vals))
id <- c(seq_along(Seq),
rep(inds+0.5,add.length))
Seq.new <- Seq.new[order(id)]
Data.new <- c(Data, rep(NA, sum(add.length)))
id <- c(seq_along(Seq),
rep(inds+0.5,add.length))
Data.new <- Data.new[order(id)]
NewDF <- data.frame(Seq.new, Data.new)