Question

所以，我有两个表示旧地址和当前地址的数据集。

move

我需要的是将新数据（main）与旧数据（main）合并，并在合并后更新id。

我想知道它是否可以在一次操作中使用它？

更新基于idspace，这是个人标识符。

x，y，> main idspace id x y move 198 1238 33 4 stay 4 1236 4 1 move # this one is updated 1515 1237 30 28 move是位置ID。

所以，我需要的输出是

merge(main, move, by = c('id'), all = T, suffixes = c('old', 'new'))

我不知道如何做到这一点。

像

这样的东西

> dput(main)
structure(list(idspace = structure(c(2L, 3L, 1L), .Label = c("1515", 
"198", "641"), class = "factor"), id = structure(c(3L, 1L, 2L
), .Label = c("1236", "1237", "1238"), class = "factor"), x = structure(c(2L, 
3L, 1L), .Label = c("30", "33", "36"), class = "factor"), y = structure(c(3L, 
1L, 2L), .Label = c("12", "28", "4"), class = "factor"), move =     structure(c(2L, 
1L, 1L), .Label = c("move", "stay"), class = "factor")), .Names = c("idspace", 
"id", "x", "y", "move"), row.names = c(NA, -3L), class = "data.frame")

> dput(move)
structure(list(idspace = structure(1L, .Label = "4", class = "factor"), 
id = structure(1L, .Label = "1236", class = "factor"), x = structure(1L,  .Label = "4", class = "factor"), 
    y = structure(1L, .Label = "1", class = "factor"), move = structure(1L, .Label = "move", class = "factor")), .Names = c("idspace", 
"id", "x", "y", "move"), row.names = c(NA, -1L), class = "data.frame")`

然而，这是错误的，因为我需要手工做很多操作。

任何解决方案？

数据

{{1}}

Answer 1

使用data.table的加入+更新功能：

require(data.table) # v1.9.6+
setDT(main) # convert data.frames to data.tables by reference
setDT(move)

main[move, on=c("id", "move"),  # extract the row number in 'main' where 'move' matches
       c("idspace", "x", "y") := .(i.idspace, i.x, i.y)] # update cols of 'main' with 
                                                         # values from 'i' = 'move' for 
                                                         # those matching rows


main
#    idspace   id  x  y move
# 1:     198 1238 33  4 stay
# 2:       4 1236  4  1 move
# 3:    1515 1237 30 28 move

这会就地更新main 。

Answer 2

这是一个dplyr解决方案：

# If you want both old and new
dplyr::full_join(main, move)

# If you want both old and new with a suffix column
main$suffix <- "old"
move$suffix <- "new"
dplyr::full_join(main, move)

# If you want new only
new       <- dplyr::left_join(main,move,by="id") # could also use  %>%
main[!is.na(new$move.y),1]   <- new[!is.na(new$move.y),6]  
main[!is.na(new$move.y),3:4] <- new[!is.na(new$move.y),7:8]

Answer 3

我想我找到了一种用

解决这个问题的简单方法

main = as.matrix(main)
move = as.matrix(move)

main[main[,'id'] %in% move[,'id'], ] <- move

哪个匹配id，保持id有序，只更改匹配的rows。它似乎适用于整个数据集。

R - 合并和更新主数据集

3 个答案: