如何基于R中最接近的匹配时间来压缩数据帧

Question

我有一个数据框,目前包含两个HH:MM:SS格式的'时间'列.我想压缩这个数据框,这样我每个唯一的'id'值只有一行.我想保留每个唯一'id'值的行,该值的'time1'值与'time2'值最接近.但是,'time1'需要大于'time2'.

这是一个简单的例子:

> dput(df)
structure(list(id = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 
3L, 3L, 4L, 4L, 4L, 4L), count = c(23L, 23L, 23L, 23L, 45L, 45L, 
45L, 45L, 67L, 67L, 67L, 67L, 88L, 88L, 88L, 88L), time1 = structure(c(1L, 
1L, 1L, 1L, 2L, 2L, 2L, 2L, 4L, 4L, 4L, 4L, 3L, 3L, 3L, 3L), .Label = c("00:13:00", 
"01:13:00", "07:18:00", "18:14:00"), class = "factor"), time2 = structure(c(4L, 
1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 2L, 3L), .Label = c("00:00:00", 
"06:00:00", "12:00:00", "18:00:00"), class = "factor"), afn = c(3.36, 
0.63, 1.77, 3.89, 3.36, 0.63, 1.77, 3.89, 3.36, 0.63, 1.77, 3.89, 
3.36, 0.63, 1.77, 3.89), dfn = c(201.67, 157.27, 103.55, 191.41, 
201.67, 157.27, 103.55, 191.41, 201.67, 157.27, 103.55, 191.41, 
201.67, 157.27, 103.55, 191.41)), .Names = c("id", "count", "time1", 
"time2", "afn", "dfn"), class = "data.frame", row.names = c(NA, 
-16L))

> df
   id count    time1    time2  afn    dfn
1   1    23 00:13:00 18:00:00 3.36 201.67
2   1    23 00:13:00 00:00:00 0.63 157.27
3   1    23 00:13:00 06:00:00 1.77 103.55
4   1    23 00:13:00 12:00:00 3.89 191.41
5   2    45 01:13:00 18:00:00 3.36 201.67
6   2    45 01:13:00 00:00:00 0.63 157.27
7   2    45 01:13:00 06:00:00 1.77 103.55
8   2    45 01:13:00 12:00:00 3.89 191.41
9   3    67 18:14:00 18:00:00 3.36 201.67
10  3    67 18:14:00 00:00:00 0.63 157.27
11  3    67 18:14:00 06:00:00 1.77 103.55
12  3    67 18:14:00 12:00:00 3.89 191.41
13  4    88 07:18:00 18:00:00 3.36 201.67
14  4    88 07:18:00 00:00:00 0.63 157.27
15  4    88 07:18:00 06:00:00 1.77 103.55
16  4    88 07:18:00 12:00:00 3.89 191.41

在上述情况下,我想最终得到这个矩阵:

id  count   time1       time2       afn     dfn
1   23      00:13:00    00:00:00    0.63    157.27
2   45      01:13:00    00:00:00    0.63    157.27
3   67      18:14:00    18:00:00    3.36    201.67
4   88      07:18:00    06:00:00    1.77    103.55

我曾使用ddply()函数来压缩过去的数据帧,但没有使用合并的匹配规则.我必须应用这是一个包含大量列的数据框(比这里给出的简单示例更多)所以任何关于如何做到这一点的建议都会非常棒.任何帮助将不胜感激.非常感谢!

Answer 1

以下是一些解决方案.

1)AVE这使用克隆氏病times以及subset和ave从R的基:

library(chron)

delta <- as.vector(times(df$time1) - times(df$time2))
df2 <- subset(df, delta > 0)
df2[ave(delta, df2$id, FUN = function(delta) delta == min(delta)) == 1, ]

2)dplyr这使用chron times和dplyr包:

library(chron)
library(dplyr) 

df %.% 
   mutate(delta = as.vector(times(time1) - times(time2))) %.% 
   filter(delta > 0) %.% 
   group_by(id) %.% 
   filter(delta == min(delta)) %.% 
   select(- delta)

3)sqldf

library(sqldf)

sqldf("select *, min(strftime('%s', time1) - strftime('%s', time2)) delta
  from (select * from df where strftime('%s', time1) > strftime('%s', time2))
  group by id")[seq_along(df)]

或者这个变化我们delta在R中计算然后使用sqldf:

library(sqldf)
library(chron)

df2 = transform(df, delta = as.vector(times(time1) - times(time2)))

sqldf("select *, min(delta) delta
  from (select * from df2 where delta > 0)
  group by id")[-ncol(df2)]

4)data.table

library(data.table)
library(chron)

DT <- data.table(df)
DT[, delta := times(time1) - times(time2)
 ][delta > 0
 ][, .SD[delta == min(delta)], by = id
 ][, seq_along(df), with = FALSE]

增加了额外的解决方案.更正library和subset声明.小改进.