Question

我陷入了这个问题，想不出任何简单的dplyr来解决这个问题：

我有两个data.frames df1和df2。我想将time2列的值从df2“绑定”到df1，但前提是user_id和place_id匹配：< / p>

> head(df1)
                time1 user_id   place_id
1 2018-06-09 12:56:12  sdkID1  place_ID1
2 2018-06-24 05:15:07  sdkID1  place_ID1
3 2018-06-12 04:15:21  sdkID1 place_ID10
4 2018-06-12 14:56:42  sdkID1 place_ID17
5 2018-05-16 18:21:51  sdkID1 place_ID20
6 2018-07-11 12:19:27  sdkID1 place_ID21
> head(df2)
                time2 user_id   place_id
1 2018-06-09 13:12:39  sdkID1  place_ID1
2 2018-06-24 06:52:51  sdkID1  place_ID1
3 2018-06-12 05:50:19  sdkID1 place_ID10
4 2018-05-16 19:42:59  sdkID1 place_ID20
5 2018-07-11 12:23:44  sdkID1 place_ID21
6 2018-06-13 11:56:05  sdkID1 place_ID34

但是，我没有任何id变量来检查time2中的df2是否属于df1。为了使事情变得有趣，对于某些事件，我没有任何时间戳可以匹配到df1。

我想要类似的东西：

> head(result)
                time1 user_id   place_id               time2 
1 2018-06-09 12:56:12  sdkID1  place_ID1 2018-06-09 13:12:39 
2 2018-06-24 05:15:07  sdkID1  place_ID1 2018-06-24 06:52:51 
3 2018-06-12 04:15:21  sdkID1 place_ID10 2018-06-12 05:50:19 
4 2018-06-12 14:56:42  sdkID1 place_ID17                  NA
5 2018-05-16 18:21:51  sdkID1 place_ID20 2018-05-16 19:42:59 
6 2018-07-11 12:19:27  sdkID1 place_ID21 2018-07-11 12:23:44

有没有办法采取time2-time1只保留行与   正时差？我知道有但是后来我有机会   像前两行一样，它们具有相同的user_id和place_id   这样我得到2018-06-24 06:52:51-2018-06-24 05:15:07的结果   和 2018-06-24 06:52:51-2018-06-09 12:56:12。我只需要第一个区别。

想象time1是到达，time2是离开。基本上，我的问题归结为找出正在运行的火车或飞机。我需要某种方式来了解2018-06-24 06:52:51-2018-06-24 05:15:07是相同的，并且   2018-06-24 06:52:51-2018-06-09 12:56:12不是同一火车/飞机。

由于我想将代码转换为SQL，因此解决方案必须基于dplyr。我尝试过类似df1 %>% group_by(user_id,place_id)的操作，但现在肯定会卡住。这是一些示例数据

set.seed(42)
u <- runif(1000, 0, 60) # "noise" to add or subtract from some timepoint
df1<-data.frame(time1=as.POSIXlt(sort(u)*100000, origin = "2018-05-03 08:00:00"),
                user_id=sample(rep(paste0('sdkID',1:60)),1000,replace=TRUE),
                place_id=sample(rep(paste0('place_ID',1:60)),1000,replace=TRUE))

df1=df1[order(df1$user_id,df1$place_id,df1$time1),]

df2=df1[-sample(1:1000,200),]  
df2$time1<-df2$time1+u[-sample(1:1000,200)]*100

## cleaning up
colnames(df2)[1]='time2'
rownames(df1)=1:1000
rownames(df2)=1:800

Answer 1

建议使用lubridate：在R中使用日期和时间来计算最小时差。

library(dplyr)
library(lubridate)

# Codes Given
set.seed(42)
u <- runif(1000, 0, 60) # "noise" to add or subtract from some timepoint
df1<-data.frame(time1=as.POSIXlt(sort(u)*100000, origin = "2018-05-03 08:00:00"),
                user_id=sample(rep(paste0('sdkID',1:60)),1000,replace=TRUE),
                place_id=sample(rep(paste0('place_ID',1:60)),1000,replace=TRUE))

df1=df1[order(df1$user_id,df1$place_id,df1$time1),]

df2=df1[-sample(1:1000,200),]  
df2$time1<-df2$time1+u[-sample(1:1000,200)]*100

# dplyr operations
df_3 = df1 %>% left_join(df2, by = c('user_id', 'place_id'))
df_3$time_diff = abs(ymd_hms(df_3$time1.x) - ymd_hms(df_3$time1.y))
df_3 %>% 
    arrange(-desc(user_id), -desc(place_id), -desc(time_diff)) %>% 
    group_by(user_id, place_id) %>%
    slice(which.min(time_diff))

其他资源：

Answer 2

我相信以下内容可以解决您的问题。

library(dplyr)

result <- df1 %>%
  left_join(df2, by = c("user_id", "place_id")) %>%
  mutate(Diff = difftime(time1.y, time1.x, units = "secs"),
         Diff = as.numeric(Diff)) %>%
  filter(Diff > 0) %>%
  arrange(user_id, place_id, time1.x) %>%
  group_by(time1.x) %>%
  mutate(time1 = first(time1.x), time2 = time1.y) %>%
  ungroup() %>%
  select(-Diff, -time1.x, -time1.y)

head(result)
## A tibble: 6 x 4
#  user_id place_id   time1               time2              
#  <fct>   <fct>      <dttm>              <dttm>             
#1 sdkID1  place_ID1  2018-05-14 06:53:01 2018-05-14 08:24:30
#2 sdkID1  place_ID18 2018-06-05 04:38:53 2018-06-05 06:12:35
#3 sdkID1  place_ID19 2018-05-22 19:20:40 2018-05-22 19:49:17
#4 sdkID1  place_ID25 2018-06-15 08:55:55 2018-06-15 10:18:58
#5 sdkID1  place_ID27 2018-05-06 17:34:40 2018-05-15 17:17:48
#6 sdkID1  place_ID27 2018-05-06 17:34:40 2018-06-11 15:14:07

Answer 3

基于@RuiBarradas和@kon_u的答案，我设法解决了我的问题。由于两者都仅部分理解了该问题（部分原因是由于我对问题描述的阐述不够清楚），因此我在这里分享了完整的解决方案：

result<-df1 %>%
       left_join(df2, by = c("user_id", "place_id")) %>%
       mutate(Diff = difftime(time2, time1, units = "secs"),
                           Diff = as.numeric(Diff)) %>%
       filter(Diff > 0) %>%
       arrange(user_id, place_id, time1,time2) %>%
       group_by(user_id, place_id,time2) %>% 
       filter(Diff==min(Diff)) %>%
      right_join(df1,by=c("user_id", "place_id","time1"))

dplyr：基于时间戳绑定列

3 个答案: