我有这两个表，我需要创建未出现的列。
这是我之前问过的问题的延伸。 有一定的修改，所以我发布另一个问题。
df <- data.table::as.data.table(structure(list(types=c("x","y","x","x","x","x","x","z","z","y","z","z"),class = c("a", "c", "v", "f", "r", "b", "t", "o","z","c","l","e"), value = c(0.76, 0.91, 1.94, 0.37, 1.35, 0.75, 1.95, 1.69,0,0.9,0.1,0.17), vehicle = c("we", "df", "rt", "yh", "uj", "er", "ed", "we","x","rt","xt","ca"), carbon = c(0.984, 0.27, 0.419, 0.469, 0.132, 0.865, 0.562, 0.133,0.05,0.001,0.7,0.37), cap = c("3", "2", "1", "6", "y", "t", "4", "6","e","4",6,"e"), up = c(4, 2, 3, "d", "t", "y", "u", "i","d","y",4,"t"), down = c("t", "e", "r", 3, 4, 5 ,2,1,2,5,1,2), amt = c(34, 23, 12, 67, 87, 43, 23, 12,34,15,14,45)), row.names = c(NA, -12L), class = c("data.table", "data.frame" )));
map <- data.table::as.data.table(structure(list(types=c("x","x","x","y","z","z"),up = c("d", "y", 4,"","",""), vehicle = c("yh", "er", "we","","",""), exercise = c("ty", 45, "k","cf","th","sh"),class = c("","","","c","",""),cap = c("","","","",6,"e"),down = c("","","","",1,2)), class = c("data.table", "data.frame"), row.names = c(NA, -6L)));
简而言之，需要在
exercisedf需要满足的条件是。
需要使用
df根据向量
upvehicleclasscapdowndfexercisedf例如
复杂的是映射表的每一列中并非所有项目都出现
map示例 1。 匹配列
upvehicle示例 2.
在
mapclassccclassdfexercisecf示例 3.
同样
使用
capdowmmapexercise下面有更多解释。
不会有任何冲突，并且
df还有一个附加列
types对于
xupvehicleyclasszcapdown可以使用或不使用
types所以如果
upvehiclecapdownclassmap仅考虑那些在
map理想情况下，
dfexercise<-c("k","cf","","ty","",45,"","th","sh","cf","th","sh")对于 x，列向上并且车辆将始终存在。对于你来说 列类将始终存在。对于 z 列上限和下限 会一直存在
这允许我们按类型拆分地图表，然后删除空列。因此，我们有三个一致且不同的映射表，我们与原始映射表进行自然连接。然后我们将所有三个连接的锻炼结果与 fcoalesce 结合起来。
maps <- split(map, by = "types")
df[, exercise := fcoalesce(lapply(maps, \(x) {
x[, .SD, .SDcols = \(x) !all(x == "")][df, on = .NATURAL]$exercise
}))]
结果
df
types class value vehicle carbon cap up down amt exercise
1: x a 0.76 we 0.984 3 4 t 34 k
2: y c 0.91 df 0.270 2 2 e 23 cf
3: x v 1.94 rt 0.419 1 3 r 12 <NA>
4: x f 0.37 yh 0.469 6 d 3 67 ty
5: x r 1.35 uj 0.132 y t 4 87 <NA>
6: x b 0.75 er 0.865 t y 5 43 45
7: x t 1.95 ed 0.562 4 u 2 23 <NA>
8: z o 1.69 we 0.133 6 i 1 12 th
9: z z 0.00 x 0.050 e d 2 34 sh
10: y c 0.90 rt 0.001 4 y 5 15 cf
11: z l 0.10 xt 0.700 6 4 1 14 th
12: z e 0.17 ca 0.370 e t 2 45 sh
数据
df <- data.table::as.data.table(structure(list(types=c("x","y","x","x","x","x","x","z","z","y","z","z"),class = c("a", "c", "v", "f", "r", "b", "t", "o","z","c","l","e"), value = c(0.76, 0.91, 1.94, 0.37, 1.35, 0.75, 1.95, 1.69,0,0.9,0.1,0.17), vehicle = c("we", "df", "rt", "yh", "uj", "er", "ed", "we","x","rt","xt","ca"), carbon = c(0.984, 0.27, 0.419, 0.469, 0.132, 0.865, 0.562, 0.133,0.05,0.001,0.7,0.37), cap = c("3", "2", "1", "6", "y", "t", "4", "6","e","4",6,"e"), up = c(4, 2, 3, "d", "t", "y", "u", "i","d","y",4,"t"), down = c("t", "e", "r", 3, 4, 5 ,2,1,2,5,1,2), amt = c(34, 23, 12, 67, 87, 43, 23, 12,34,15,14,45)), row.names = c(NA, -12L), class = c("data.table", "data.frame" )));
map <- data.table::as.data.table(structure(list(types=c("x","x","x","y","z","z"),up = c("d", "y", 4,"","",""), vehicle = c("yh", "er", "we","","",""), exercise = c("ty", 45, "k","cf","th","sh"),class = c("","","","c","",""),cap = c("","","","",6,"e"),down = c("","","","",1,2)), class = c("data.table", "data.frame"), row.names = c(NA, -6L)));