找到兩個數據幀之間的最短時間差

Question

假設我有兩個數據幀，

df1

id        time1
1         2016-04-07 21:39:10
1         2016-04-05 11:19:17
2         2016-04-03 10:58:25
2         2016-04-02 21:39:10

df2

id        time2
1         2016-04-07 21:39:11
1         2016-04-05 11:19:18
1         2016-04-06 21:39:11
1         2016-04-04 11:19:18
2         2016-04-03 10:58:26
2         2016-04-02 21:39:11
2         2016-04-04 10:58:26
2         2016-04-05 21:39:11

我想為df1中的每個條目找到df2中最短的時間差。 假設我們取第一個條目，它有id 1，所以我想遍歷df2，過濾id 1，然后檢查df1的一個條目和df2的剩余條目之間的時間差，找到最短的差異並獲取相應的條目。 我的樣本輸出應該是

id        time                   time2                    diff(in secs)
1         2016-04-07 21:39:10    2016-04-07 21:39:10        1
1         2016-04-05 11:19:17    2016-04-05 11:19:17        1
2         2016-04-03 10:58:25    2016-04-03 10:58:25        1
2         2016-04-02 21:39:10    2016-04-02 21:39:10        1

以下是我的嘗試，

for(i in unique(df1$id)){
  temp1 = df1[df1$id == i,]
  temp2 = df2[df2$id == i,]
  for(j in unique(df1$time1){
     for(k in unique(df2$time2){
        diff = abs(df1$time1[j] - df2$time2[k]
        print(diff)}}}

在此之后我無法進步，遇到很多錯誤。 任何人都可以幫我糾正這個嗎？ 可能會建議一個更有效的方法來做到這一點？ 任何幫助，將不勝感激。

更新：

可再現數據：

    df1 <- data.frame(
        id = c(1,1,2,2),
        time1 = c('2016-04-07 21:39:10', '2016-04-05 11:19:17', '2016-04-03 10:58:25', '2016-04-02 21:39:10')
    )

    df2 <- data.frame(
        id = c(1,1,1,1,2,2,2,2),
        time2 = c('2016-04-07 21:39:11', '2016-04-05 11:19:18','2016-04-07 21:39:11', '2016-04-05 11:19:18', '2016-04-03 10:58:26', '2016-04-02 21:39:11','2016-04-03 10:58:26', '2016-04-02 21:39:11')
    )

df1$time1 =  as.POSIXct(df1$time1)
df2$time2 = as.POSIXct(df2$time2)

Answer 1

您可以使用dplyr實現此dplyr 。 基本上這個想法是因為我們想要生成一個條目，我們將為df1的每個元素分配一個新的id（在本例中我稱之為rowname）。

在此之后，我們感興趣的是加入id上的兩個數據幀並根據最小絕對差值對它們進行過濾。

library(dplyr)

df1$time1 <- as.POSIXct(as.character(df1$time1))
df2$time2 <- as.POSIXct(as.character(df2$time2))

df1 %>% 
  add_rownames("rowname") %>%
  left_join(df2, "id") %>% 
  mutate(diff=time2-time1) %>%
  group_by(rowname) %>%
  filter(min(abs(diff)) == abs(diff)) %>% 
  distinct

這是我的輸出：

Source: local data frame [4 x 5]
Groups: rowname [4]

  rowname    id               time1               time2   diff
    (chr) (dbl)              (time)              (time) (dfft)
1       1     1 2016-04-07 21:39:10 2016-04-07 21:39:11 1 secs
2       2     1 2016-04-05 11:19:17 2016-04-05 11:19:18 1 secs
3       3     2 2016-04-03 10:58:25 2016-04-03 10:58:26 1 secs
4       4     2 2016-04-02 21:39:10 2016-04-02 21:39:11 1 secs      

Answer 2

您也可以在基礎R中執行此操作。要生成隨機日期（有用），我從StackOverflow上的其他地方借用並編輯了一個很好的函數：

latemail <- function(N, st="2011/01/01", et="2016/12/31") {
  st <- as.POSIXct(as.Date(st))
  et <- as.POSIXct(as.Date(et))
  dt <- as.numeric(difftime(et,st,unit="sec"))
  ev <- sort(runif(N, 0, dt))
  return(st + ev)
}
df1 <- data.frame(id=c(1,1,2,2), time1=latemail(4))
df2 <- data.frame(id=c(rep(1,4), rep(2,4)), time2=latemail(8))

然后您的答案可以分為兩行：

shortest <- sapply(df1$time1, function(x) which(abs(df2$time2 - x) == min(abs(df2$time2 - x))))
cbind(df1, df2[shortest,])

輸出：

id               time1 id               time2
 1 2011-10-08 02:00:21  1 2011-08-17 18:07:47
 1 2012-05-06 17:49:03  1 2012-09-04 19:52:40
 2 2013-10-29 13:14:51  1 2012-10-29 20:09:31
 2 2016-06-17 19:23:43  2 2015-11-24 02:07:15

Answer 3

如果您使用data.table ：

library(data.table)
df1 <- data.table(
  id = c(1,1,2,2),
  time1 = c('2016-04-07 21:39:10', '2016-04-05 11:19:17', '2016-04-03 10:58:25', '2016-04-02 21:39:10')
)

df2 <- data.table(
  id = c(1,1,1,1,2,2,2,2),
  time2 = c('2016-04-07 21:39:11', '2016-04-05 11:19:18','2016-04-07 21:39:11', '2016-04-05 11:19:18', '2016-04-03 10:58:26', '2016-04-02 21:39:11','2016-04-03 10:58:26', '2016-04-02 21:39:11')
)

df1$time1 =  as.POSIXct(df1$time1)
df2$time2 = as.POSIXct(df2$time2)

res <- df1[df2, .(time1, time2), by  = .EACHI, on = "id"][, diff:= abs(time2 -time1)]
setkey(res, id, time1, diff)
res <- res[, row := seq_along(.I), by = .(id, time1)][row == 1]