Question

我想在每组（grp）之后插入新行，并用下一行值填充某些新列，并用上一行值填充一些新列

我正在尝试使用：

x<-rbind(setDT(DF), DF[,.SD[.N], grp][, color := shift(color,1L, type = "lag")][, Lat:= shift(Lat,1L, type = "lead")])[order(id)]

在DF上：

a <- c(1,2,3,4,5,6,7,8,9,10)
b <- c(10,20,30,40,50,60,70,80,90,100)
c <- c("a","a","b","b","b","a","a","b","c","c")
d <- c(11,23,67,89,90,100,101,123,200,290)
df <- data.frame(color=a, Lat=b, grp=c, id=d)

我可能使用了不正确的shift（），似乎无法正常工作。

谢谢

预期结果为：

   color Lat grp  id
1      1  10   a  11
2      2  20   a  23
new row with color from previous row, Lat from next row, grp from previous and if from next
3      3  30   b  67
4      4  40   b  89
5      5  50   b  90
new row as before
6      6  60   a 100
7      7  70   a 101
new row as before
8      8  80   b 123
new row as before
9      9  90   c 200
10    10 100   c 290

Answer 1

我们创建两个lead列，即'Lat'和'id'的下一行值，然后再对{grp'的rleid进行分组。这里，rleid正在检查'grp'的相邻元素是否相同。如果不同，它将为该元素分配一个新的ID

library(data.table)
setDT(df)[, c("LatN", "idN") := shift(.SD, type = 'lead'), .SDcols = c('Lat', 'id')]

通过'grp'的last获得对所选列的rleid观察

tmp <- df[, .(color = color[.N], Lat = LatN[.N], id = idN[.N], grp = grp[.N]), 
       .(grp1 = rleid(grp))][, grp1 := NULL]

rbind包含原始数据集列，order带有“颜色”，并删除任何NA行

na.omit(rbind(df[,.(color, Lat, id, grp)], tmp)[order(color)])
#     color Lat  id grp
# 1:     1  10  11   a
# 2:     2  20  23   a
# 3:     2  30  67   a
# 4:     3  30  67   b
# 5:     4  40  89   b
# 6:     5  50  90   b
# 7:     5  60 100   b
# 8:     6  60 100   a
# 9:     7  70 101   a
#10:     7  80 123   a
#11:     8  80 123   b
#12:     8  90 200   b
#13:     9  90 200   c
#14:    10 100 290   c

在组之后插入一行并填充下一行的值

1 个答案: