通过根据组条件创建重复项来扩展 data.frame (2)

Expand data.frame by creating duplicates based on group condition (2)

从@AndrewGustar answer/code开始:

1) 如果我的输入 data.frame 和 ID 值不按顺序并且也可以自我复制怎么办?

示例data.frame:

df = read.table(text = 'ID  Day Count   Count_group
18  1933    6   11
33  1933    6   11
37  1933    6   11
18  1933    6   11
16  1933    6   11
11  1933    6   11
111 1932    5   8
34  1932    5   8
60  1932    5   8
88  1932    5   8
18  1932    5   8
33  1931    3   4
13  1931    3   4
56  1931    3   4
23  1930    1   1
6   1800    6   10
37  1800    6   10
98  1800    6   10
52  1800    6   10
18  1800    6   10
76  1800    6   10
55  1799    4   6
6   1799    4   6
52  1799    4   6
133 1799    4   6
112 1798    2   2
677 1798    2   2
778 888     4   6
111 888     4   6
88  888     4   6
10  888     4   6
37  887     2   3
26  887     2   3
8   886     1   2
56  885     1   1', header = TRUE)

Count 列显示每个 DayID 个值的总数,Count_group 列显示每个 ID 个值的总和每个 DayDay - 1.

例如1933 = Count_group 11 因为 Count 6 (1933) + Count 5 (1932),依此类推。

我需要做的是为每个 Count_group 创建重复的观察并将它们添加到其中,以便每个 Count_group 显示其 DayDay - 1 .

例如Count_group = 11 由 Day 1933 和 1932 的 Count 值组成。所以这两天都需要包含在 Count_group = 11 中。下一个是 Count_group = 8,由1932年和1931年等组成...

期望的输出:

    ID  Day   Count Count_group
    18  1933    6   11
    33  1933    6   11
    37  1933    6   11
    18  1933    6   11
    16  1933    6   11
    11  1933    6   11
    111 1932    5   11
    34  1932    5   11
    60  1932    5   11
    88  1932    5   11
    18  1932    5   11
    111 1932    5   8
    34  1932    5   8
    60  1932    5   8
    88  1932    5   8
    18  1932    5   8
    33  1931    3   8
    13  1931    3   8
    56  1931    3   8
    33  1931    3   4
    13  1931    3   4
    56  1931    3   4
    23  1930    1   4
    23  1930    1   1
    6   1800    6   10
    37  1800    6   10
    98  1800    6   10
    52  1800    6   10
    18  1800    6   10
    76  1800    6   10
    55  1799    4   10
    6   1799    4   10
    52  1799    4   10
    133 1799    4   10
    55  1799    4   6
    6   1799    4   6
    52  1799    4   6
    133 1799    4   6
    112 1798    2   6
    677 1798    2   6
    112 1798    2   2
    677 1798    2   2
    778 888     4   6
    111 888     4   6
    88  888     4   6
    10  888     4   6
    37  887     2   6
    26  887     2   6
    37  887     2   3
    26  887     2   3
    8   886     1   3
    8   886     1   2
    56  885     1   2
    56  885     1   1

这是一个保持上述 ID 值的解决方案。

#first add grouping variables
df$smalldaygroup <- c(0,cumsum(sapply(2:nrow(df),function(i) df$Day[i]!=df$Day[i-1]))) #individual days
df$bigdaygroup <- c(0,cumsum(sapply(2:nrow(df),function(i) df$Day[i]<df$Day[i-1]-1))) #blocks of consecutive days

#duplicate individual days except the first in each big group
df2 <- lapply(split(df,df$bigdaygroup),function(x) 
  split(x,x$smalldaygroup)[c(1,rep(2:length(split(x,x$smalldaygroup)),each=2))])

#change the Count_group to previous value in alternate entries
df2 <- lapply(df2,function(L) lapply(1:length(L),function(i) {
  x <- L[[i]]
  if(!(i%%2)) x$Count_group <- L[[i-1]]$Count_group[1]
  return(x)
}))

df2 <- do.call(rbind,unlist(df2,recursive=FALSE)) #bind back together

head(df2,20) #ignore rownames!
       ID  Day Count Count_group
01.1   18 1933     6          11
01.2   33 1933     6          11
01.3   37 1933     6          11
01.4   18 1933     6          11
01.5   16 1933     6          11
01.6   11 1933     6          11
02.7  111 1932     5          11
02.8   34 1932     5          11
02.9   60 1932     5          11
02.10  88 1932     5          11
02.11  18 1932     5          11
03.7  111 1932     5           8
03.8   34 1932     5           8
03.9   60 1932     5           8
03.10  88 1932     5           8
03.11  18 1932     5           8
04.12  33 1931     3           8
04.13  13 1931     3           8
04.14  56 1931     3           8
05.12  33 1931     3           4