Question

我的数据如下：

id <- c(1,1,1,1,1,1,2,2,2,2,3,3,3,3,3,3,3,3,3,4,4,4)
start <- c(NA, NA, NA, 1, NA, NA, NA, NA, 1, NA, NA, NA, 1, NA, NA, NA, NA, NA, 1, NA, NA, NA)
e <- as.data.frame(cbind(id, start))

我想在每次从start == 1开始或者有一个新的id时，用累积和的类比来填充NAs。我做了一个for循环，但我的实际数据太长了，for-loop在最近的几天内结束了。有没有办法加快我的解决方案？我的目标变量可以复制如下：

e$target <- NA
for (i in 2:length(e$id)){
  if (e$id[i]!=e$id[i-1]){
    e$target[i] <- NA
  } else {
    e$target[i] <- e$target[i-1]+1
    if (!is.na(e$start[i]==1)){
      e$target[i] <- 0
    }
  }
}

Answer 1

我们可以使用data.table

执行此操作

library(data.table)
setDT(e)[,  target1 := seq_len(.N)-1,.(grp = cumsum(!is.na(start)), id)]
e[e[, c(.I[all(is.na(start))], .I[seq_len(which.max(!is.na(start))-1)]),
                  id]$V1, target1 := NA]
e
#    id start target target1
# 1:  1    NA     NA      NA
# 2:  1    NA     NA      NA
# 3:  1    NA     NA      NA
# 4:  1     1      0       0
# 5:  1    NA      1       1
# 6:  1    NA      2       2
# 7:  2    NA     NA      NA
# 8:  2    NA     NA      NA
# 9:  2     1      0       0
#10:  2    NA      1       1
#11:  3    NA     NA      NA
#12:  3    NA     NA      NA
#13:  3     1      0       0
#14:  3    NA      1       1
#15:  3    NA      2       2
#16:  3    NA      3       3
#17:  3    NA      4       4
#18:  3    NA      5       5
#19:  3     1      0       0
#20:  4    NA     NA      NA
#21:  4    NA     NA      NA
#22:  4    NA     NA      NA

Answer 2

您可以尝试tidyverse。使用fill向下拖动最新的非NA条目，然后用它们的长度序列替换这些值（-1是使序列从0开始）

library(tidyverse)

e %>% 
 group_by(id) %>% 
 mutate(target = start) %>% 
 fill(target) %>% 
 mutate(target = replace(target, !is.na(target), seq(length(target[!is.na(target)]))-1), 
        target = replace(target, start == 1, 0))

Answer 3

另一个data.table选项是：

library(data.table)
setDT(e)[, subgroup := cumsum(start==1 &  !is.na(start)), by = id]
e[ , target2 := cumsum(is.na(start)), by = .(id, subgroup)][subgroup == 0, target2 := NA_integer_]

 #   id start target subgroup target2
 #1:  1    NA     NA        0      NA
 #2:  1    NA     NA        0      NA
 #3:  1    NA     NA        0      NA
 #4:  1     1      0        1       0
 #5:  1    NA      1        1       1
 #6:  1    NA      2        1       2
 #7:  2    NA     NA        0      NA
 #8:  2    NA     NA        0      NA
 #9:  2     1      0        1       0
#10:  2    NA      1        1       1
#11:  3    NA     NA        0      NA
#12:  3    NA     NA        0      NA
#13:  3     1      0        1       0
#14:  3    NA      1        1       1
#15:  3    NA      2        1       2
#16:  3    NA      3        1       3
#17:  3    NA      4        1       4
#18:  3    NA      5        1       5
#19:  3     1      0        2       0
#20:  4    NA     NA        0      NA
#21:  4    NA     NA        0      NA
#22:  4    NA     NA        0      NA

填充0-1列，累计总和在1s之间

3 个答案: