1

我正在尝试使用非 equi 连接将 data.table x 连接到 z。表 x 包含两列 X1 和 X2，它们用作与 z 中的列 Z1 连接的范围。当前代码成功地进行了非 equi 连接，但是某些列被删除或重命名。我想返回提供的“理想”data.table，而不是我目前拥有的那个，我必须重命名列或进一步连接数据以获得提供的“理想”数据。

> library(data.table)
> 
> x <- data.table(Id  = c("A", "B", "C", "C"),
+                 X1  = c(1L, 3L, 5L, 7L),
+                 X2 = c(8L,12L,9L,18L),
+                 XY  = c("x2", "x4", "x6", "x8"))
> 
> z <- data.table(ID = "C", Z1 = 5:9, Z2 = paste0("z", 5:9))
> 
> x
   Id X1 X2 XY
1:  A  1  8 x2
2:  B  3 12 x4
3:  C  5  9 x6
4:  C  7 18 x8
> z
   ID Z1 Z2
1:  C  5 z5
2:  C  6 z6
3:  C  7 z7
4:  C  8 z8
5:  C  9 z9
> 
> # suboptimal data return data format
> x[z, on = .(Id == ID, X1 <= Z1, X2 >= Z1)]
   Id X1 X2 XY Z2
1:  C  5  5 x6 z5
2:  C  6  6 x6 z6
3:  C  7  7 x6 z7
4:  C  7  7 x8 z7
5:  C  8  8 x6 z8
6:  C  8  8 x8 z8
7:  C  9  9 x6 z9
8:  C  9  9 x8 z9
> 
> # column names are Id, X1 and X2 from x which replaces ID and Z1. The contents of X1 and X2 are also changed to the original values of Z1.
> # XY and Z2 remain unchanged.
> 
> # I want to create the following table where the original column names and values are retained, while still joining the table in a non-equi way.
> 
> ideal <- data.table(ID = c("C", "C", "C", "C", "C", "C", "C", "C"),
+                     Z1 = c(5, 6, 7, 7, 8, 8, 9, 9),
+                     Z2 = c("Z5", "z6", "z7", "z7", "z8", "z8", "z9", "z9"),
+                     X1 = c(5, 5, 5, 7, 5, 7, 5, 7),
+                     X2 = c(9, 9, 9, 18, 9, 18, 9, 18),
+                     XY = c("x6", "x6", "x6", "x8", "x6", "x8", "x6", "x8"))
> 
> print(ideal)
   ID Z1 Z2 X1 X2 XY
1:  C  5 Z5  5  9 x6
2:  C  6 z6  5  9 x6
3:  C  7 z7  5  9 x6
4:  C  7 z7  7 18 x8
5:  C  8 z8  5  9 x6
6:  C  8 z8  7 18 x8
7:  C  9 z9  5  9 x6
8:  C  9 z9  7 18 x8
4

3 回答 3

2

正如@Humpelstielzchen 所评论的，可以通过手动选择所需的列来完成。但是必须使用前缀x.（xinx.指的是参数x而[.data.table不是 data.table 的名称）来恢复原始 data.table 中的列x。否则会产生不正确的输出。

# desired
x[z, .(ID, Z1, Z2, X1 = x.X1, X2 = x.X2, XY), on = .(Id == ID, X1 <= Z1, X2 >= Z1)]
#    ID Z1 Z2 X1 X2 XY
# 1:  C  5 z5  5  9 x6
# 2:  C  6 z6  5  9 x6
# 3:  C  7 z7  5  9 x6
# 4:  C  7 z7  7 18 x8
# 5:  C  8 z8  5  9 x6
# 6:  C  8 z8  7 18 x8
# 7:  C  9 z9  5  9 x6
# 8:  C  9 z9  7 18 x8

# undesired
x[z, on = .(Id == ID, X1 <= Z1, X2 >= Z1), .(ID, Z1, Z2, X1, X2, XY)]
#    ID Z1 Z2 X1 X2 XY
# 1:  C  5 z5  5  5 x6
# 2:  C  6 z6  6  6 x6
# 3:  C  7 z7  7  7 x6
# 4:  C  7 z7  7  7 x8
# 5:  C  8 z8  8  8 x6
# 6:  C  8 z8  8  8 x8
# 7:  C  9 z9  9  9 x6
# 8:  C  9 z9  9  9 x8

packageVersion('data.table')
# '1.13.2'

如果有许多其他列，这使得手动选择不可行，这里有一个解决方法，方法是在加入过程中x重新放置位置：z

DT <- z[x, on = .(ID=Id, Z1 >= X1, Z1 <= X2), nomatch = NULL]
#' since for non-equi conditions, the values are from RHS while
#' the column names were from LHS, we known that `Z1` and `Z1.1`
#' correspond to `X1` and `X2`.
setnames(DT, c('Z1', 'Z1.1'), c('X1', 'X2'))
DT[z, Z1 := i.Z1, on = .(ID, Z2)]
# > DT
#    ID X1 Z2 X2 XY Z1
# 1:  C  5 z5  9 x6  5
# 2:  C  5 z6  9 x6  6
# 3:  C  5 z7  9 x6  7
# 4:  C  5 z8  9 x6  8
# 5:  C  5 z9  9 x6  9
# 6:  C  7 z7 18 x8  7
# 7:  C  7 z8 18 x8  8
# 8:  C  7 z9 18 x8  9
于 2020-11-25T12:08:19.430 回答
0

复印一份？

copy_cols <- function(dt, nms) {
  dt[, paste0(".", nms) := lapply(.SD, copy), .SDcols = nms]
}

copy_cols(x, c("X1", "X2"))
copy_cols(z, "Z1")
x[z, on = .(Id == ID, .X1 <= .Z1, .X2 >= .Z1)][, c(".X1", ".X2") := NULL][]

输出

   Id X1 X2 XY Z1 Z2
1:  C  5  9 x6  5 z5
2:  C  5  9 x6  6 z6
3:  C  5  9 x6  7 z7
4:  C  7 18 x8  7 z7
5:  C  5  9 x6  8 z8
6:  C  7 18 x8  8 z8
7:  C  5  9 x6  9 z9
8:  C  7 18 x8  9 z9
于 2020-11-25T08:32:25.940 回答
0

我最终回答了我自己的问题。

data_table_tidy_join <- function(x,y, join_by){

    x <- data.table(x)
    y <- data.table(y)

    # Determine single join names
    single_join_names <- purrr::keep((stringr::str_split(join_by, "==|>=|<=")), ~length(.) == 1) %>% unlist()

    # cols from x that won't require as matching in i
    remove_from_x_names <- c(trimws(na.omit(stringr::str_extract(join_by, ".*(?=[=]{2})"))), single_join_names)

    # names need to keep
    x_names_keep_raw <- names(x)[!names(x) %in% remove_from_x_names]
    y_names_keep_raw <- names(y)

    # cols that exist in both x and y, but not being equi joined on
    cols_rename_index <- x_names_keep_raw[x_names_keep_raw %in% y_names_keep_raw]

    #rename so indexing works
    x_names_keep <- x_names_keep_raw
    y_names_keep <- y_names_keep_raw

    # give prefix to necessary vars
    x_names_keep[x_names_keep %in% cols_rename_index] <- paste("x.",cols_rename_index, sep ="")
    y_names_keep[y_names_keep %in% cols_rename_index] <- paste("i.",cols_rename_index, sep ="")

    # implement data.table call, keeping required cols
    joined_data <-
        x[y, on = join_by,
          mget(c(paste0("i.", y_names_keep_raw),paste0("x.", x_names_keep_raw))) %>% set_names(c(y_names_keep,x_names_keep)),
          mult = "all", allow.cartesian = TRUE, nomatch = NA] %>%
        as_tibble()

    return(joined_data)

}

> x <- data.table(Id  = c("A", "B", "C", "C"),
+                  X1  = c(1L, 3L, 5L, 7L),
+                  X2 = c(8L,12L,9L,18L),
+                  XY  = c("x2", "x4", "x6", "x8"))
>  
> z <- data.table(ID = "C", Z1 = 5:9, Z2 = paste0("z", 5:9))
>   
> data_table_tidy_join(x, z, join_by = c("Id == ID","X1 <= Z1", "X2 >= Z1"))
# A tibble: 8 x 6
  ID       Z1 Z2       X1    X2 XY   
  <chr> <int> <chr> <int> <int> <chr>
1 C         5 z5        5     9 x6   
2 C         6 z6        5     9 x6   
3 C         7 z7        5     9 x6   
4 C         7 z7        7    18 x8   
5 C         8 z8        5     9 x6   
6 C         8 z8        7    18 x8   
7 C         9 z9        5     9 x6   
8 C         9 z9        7    18 x8
于 2020-11-27T07:07:47.697 回答