引用单元格（R）中的先前值

Question

I have the following data.frame:

head(data.c)

     mark high_mark mark_cum
      5        0        0
      7        1        1
      7        1        2
      NA       0        2
      7        1        3
      7        1        4

As there are NAs I need to construct an additional column of normal sequence from 1:length(mark). However, if it is NA vector cell has to take a previous value. So it must look like this:

     mark high_mark mark_cum mark_seq
      5        0        0       1 
      7        1        1       2
      7        1        2       3
      NA       0        2       3 
      7        1        3       4
      7        1        4       5
      NA       0        4       5

Answer 1

1) cumsum This solution uses the fact that each mark_seq element equals the cumulative number of non-NA elements in mark at that point.

transform(data.c, mark_seq = cumsum(!is.na(mark)))

giving:

  mark high_mark mark_cum mark_seq
1    5         0        0        1
2    7         1        1        2
3    7         1        2        3
4   NA         0        2        3
5    7         1        3        4
6    7         1        4        5
7   NA         0        4        5


data.c <- read.table(text = Lines, header = TRUE)

2) na.locf Here is a second solution using seq_along and na.locf (from zoo). It creates a sequence the same length as the number of non-NA elements in mark and uses replace to put them in the spots where the non-NA elements exist. Then na.locf is used to fill in the NAs with the prior values.

library(zoo)

transform(data.c, mark_seq=na.locf(replace(mark, !is.na(mark), seq_along(na.omit(mark)))))

3) mark_cum It was not stated in the question how the input column mark_cum is constructed but in the sample output in the question the mark_seq column equals the mark_cum column plus 1 so if that is always the case then an easy solution is:

transform(data.c, mark_seq = mark_cum + 1)

Note: We used this as the input:

Lines <- "mark high_mark mark_cum 
      5        0        0        
      7        1        1       
      7        1        2       
      NA       0        2        
      7        1        3       
      7        1        4       
      NA       0        4"