我正在使用一个数据框架，该数据框架在住院期间保存患者的位置。它的格式化方式是每一行代表该人（= ID）在特定时间段（从BeginTim到EndTime）的新位置（=部门，房间，床）。
以下是初始数据框的示例：
print(data_perlocation[1:10,])
ID department BeginTime EndTime room Bed
<dbl> <chr> <chr> <chr> <dbl> <dbl>
1 2156864 L14B 2016-03-02 09:40:00.0000000 2016-03-02 15:20:00.0000000 102 3
2 2161034 B51A 2016-06-07 00:00:00.0000000 2016-06-07 11:02:00.0000000 109 2
3 2161034 B51A 2016-06-06 09:00:00.0000000 2016-06-06 10:27:00.0000000 109 2
4 2161034 B51A 2016-06-06 12:47:00.0000000 2016-06-07 00:00:00.0000000 109 2
5 2161034 B51A 2016-06-06 10:27:00.0000000 2016-06-06 12:47:00.0000000 103 3
6 2176442 L14B 2016-02-04 07:15:00.0000000 2016-02-04 13:47:00.0000000 101 4
7 2176754 B61A 2016-03-15 07:16:00.0000000 2016-03-15 14:56:00.0000000 109 3
8 2176754 B61A 2016-03-16 08:10:00.0000000 2016-03-17 00:00:00.0000000 109 3
9 2176754 B61A 2016-03-15 14:56:00.0000000 2016-03-16 08:10:00.0000000 109 2
10 2176754 B61A 2016-03-17 00:00:00.0000000 2016-03-17 11:18:00.0000000 109 3
11 2184060 B61A 2016-03-10 20:25:00.0000000 2016-03-11 00:00:00.0000000 105 2
12 2184060 B61A 2016-03-10 20:01:00.0000000 2016-03-10 20:25:00.0000000 105 1
13 2184060 B61A 2016-03-11 00:00:00.0000000 2016-03-12 00:00:00.0000000 105 2
14 2184060 B61A 2016-03-12 00:00:00.0000000 2016-03-12 14:00:00.0000000 105 2
我想转换这个数据框，以便每天都有一行。因此，我创建了一个新的数据框，其中包含ID和入场日。像这样：
ID Date
1 2156864 2016-03-02
2 2161034 2016-06-06
3 2161034 2016-06-07
4 2176442 2016-02-04
5 2176754 2016-03-15
6 2176754 2016-03-16
7 2176754 2016-03-17
8 2184060 2016-03-10
9 2184060 2016-03-11
10 2184060 2016-03-12
现在，我想将data_bylocation数据框中每天存在的（多个）位置添加到匹配data_byday的{{1}}行，并ID 1}}匹配日期。
我最终结合了一个for和两个if语句。到目前为止，我的尝试并没有给出任何接近预期结果的东西，我认为它们必须是一种更容易实现的方法。我的最后一次尝试结束了这样的事情：
beginDate期望的结果如下所示：
data_perday[,3] <- NA
for (index in 1:nrow(data_perlocation)){
if (data_perlocation$ID[index]==data_perday$ID & as.Date(as.character(data_perlocation$BeginTime[index]), format="%Y-%m-%d")==as.Date(data_perday$Date, format="%Y-%m-%d")) {
if (is.na(data_perday[index,3])){
##code to assign location and time of for that day
} else {
##code to assign second location and time of for that day and place
}}}
我对R很新，还在学习。我一直坚持这个问题。因此，非常感谢任何正确方向的提示！
编辑：
可重复的例子：
ID Date BeginTime1 EndTime1 department1 room1 bed1 BeginTime2 EndTime2 department2 room2 bed2 [3rd location, etc]
1 2156864 2016-03-02 [first location of this day] [second location of this day]
2 2161034 2016-06-06
3 2161034 2016-06-07
4 2176442 2016-02-04
5 2176754 2016-03-15
6 2176754 2016-03-16
7 2176754 2016-03-17
8 2184060 2016-03-10
9 2184060 2016-03-11
10 2184060 2016-03-12
第二个例子：
data_byday <- structure(list(ID = c(2156864, 2161034, 2161034, 2176442, 2176754, 2176754, 2176754, 2184060, 2184060, 2184060), Date = c("2016-03-02", "2016-06-06", "2016-06-07", "2016-02-04", "2016-03-15", "2016-03-16", "2016-03-17", "2016-03-10", "2016-03-11", "2016-03-12")), .Names = c("ID", "Date"), row.names = c(NA, 10L), class = "data.frame")
data_bylocation <- structure(list(ID = c(2156864, 2161034, 2161034, 2161034, 2161034, 2176442, 2176754, 2176754, 2176754, 2176754, 2184060, 2184060, 2184060, 2184060), department = c("L14B", "B51A", "B51A", "B51A", "B51A", "L14B", "B61A", "B61A", "B61A", "B61A", "B61A", "B61A", "B61A", "B61A"), BeginTime = c("2016-03-02 09:40:00.0000000", "2016-06-07 00:00:00.0000000", "2016-06-06 09:00:00.0000000", "2016-06-06 12:47:00.0000000", "2016-06-06 10:27:00.0000000", "2016-02-04 07:15:00.0000000", "2016-03-15 07:16:00.0000000", "2016-03-16 08:10:00.0000000", "2016-03-15 14:56:00.0000000", "2016-03-17 00:00:00.0000000", "2016-03-10 20:25:00.0000000", "2016-03-10 20:01:00.0000000", "2016-03-11 00:00:00.0000000", "2016-03-12 00:00:00.0000000"), EndTime = c("2016-03-02 15:20:00.0000000", "2016-06-07 11:02:00.0000000", "2016-06-06 10:27:00.0000000", "2016-06-07 00:00:00.0000000", "2016-06-06 12:47:00.0000000", "2016-02-04 13:47:00.0000000", "2016-03-15 14:56:00.0000000", "2016-03-17 00:00:00.0000000", "2016-03-16 08:10:00.0000000", "2016-03-17 11:18:00.0000000", "2016-03-11 00:00:00.0000000", "2016-03-10 20:25:00.0000000", "2016-03-12 00:00:00.0000000", "2016-03-12 14:00:00.0000000"), room = c(102, 109, 109, 109, 103, 101, 109, 109, 109, 109, 105, 105, 105, 105), Bed = c(3, 2, 2, 2, 3, 4, 3, 3, 2, 3, 2, 1, 2, 2)), .Names = c("ID", "department", "BeginTime", "EndTime", "room", "Bed"), row.names = c(NA, -14L), class = c("tbl_df", "tbl", "data.frame"))
答案 0 :(得分：1)
以下是Collections::max和tidyverse的解决方案：
lubridate使用library(dplyr)
library(tidyr)
library(lubridate)
data_bylocation %>%
mutate(Date = date(BeginTime)) %>%
gather(variable, value, -ID, -Date) %>%
group_by(ID, Date, temp = variable) %>%
mutate(index = row_number(),
variable = paste0(variable, index)) %>%
ungroup() %>%
select(-(temp:index)) %>%
spread(variable, value) %>%
select(ID, Date, c(6, 12, 9, 15, 3), c(6, 12, 9, 15, 3)+1, c(6, 12, 9, 15, 3)+2)
：
data.table<强>结果：
library(data.table)
data = setDT(data_bylocation)
newDT = dcast(melt(data[, date := lubridate::date(BeginTime)],
id.vars = c("ID", "date"))[
,.(variable2 = paste0(variable, 1:.N), value),
by = .(ID, date, variable)],
ID + date ~ variable2, value.var = "value")
setcolorder(newDT, c(1:2, c(6, 12, 9, 15, 3), c(6, 12, 9, 15, 3)+1, c(6, 12, 9, 15, 3)+2))
注意：
我仍在尝试考虑重新排列列的更好方法。这是我现在能做的最好的事情。
答案 1 :(得分：0)
OP已要求将患者的位置按天制表，即使他们在特定地点停留的时间超过一天。
这需要将最长停留时间分成一天。这可以通过foverlaps()函数来完成。要从宽格式转换为长格式，可以使用data.table::dcast() 同时重塑多个列的功能。
library(data.table)
# coerce time columns to POSIXct
cols <- c("BeginTime", "EndTime")
setDT(data_bylocation2)[, (cols) := lapply(.SD, as.POSIXct), .SDcols = cols]
# create sequence of days which cover the whole period
time_seq <- data_bylocation2[, seq(lubridate::floor_date(min(BeginTime), "day"),
lubridate::ceiling_date(max(EndTime), "day"),
by = "days")]
# chop longer stays into one day chunks
mDT <- foverlaps(data.table(Date = head(time_seq, -1L), end = tail(time_seq, -1L)),
setkey(data_bylocation2, BeginTime, EndTime), by.x = c("Date", "end"), nomatch = 0L)[
, (cols) := .(pmax(BeginTime, Date), pmin(EndTime, end))][
EndTime > BeginTime][
order(ID, Date, BeginTime)]
# reshape from long to wide form
val_cols <- c("BeginTime", "EndTime", "department", "room", "bed")
dcast(mDT, ID + Date ~ rowid(ID, Date), value.var = val_cols)[
# reorder columns
, setcolorder(.SD, c("ID", "Date",
# create create column names in expected order using a cross join
mDT[, CJ(seq_len(max(rowid(ID, Date))), val_cols)[
, paste(V2, V1, sep = "_")]]))]
ID Date BeginTime_1 EndTime_1 bed_1 department_1 room_1 BeginTime_2 EndTime_2 bed_2 department_2 room_2 1: 2224003 2016-02-12 2016-02-12 08:00:00 2016-02-12 13:40:00 3 B12A 205 2016-02-12 13:40:00 2016-02-13 1 B12A 209 2: 2224003 2016-02-13 2016-02-13 00:00:00 2016-02-14 00:00:00 1 B12A 209 <NA> <NA> NA NA NA 3: 2224003 2016-02-14 2016-02-14 00:00:00 2016-02-15 00:00:00 1 B12A 209 <NA> <NA> NA NA NA 4: 2224003 2016-02-15 2016-02-15 00:00:00 2016-02-15 16:17:00 1 B12A 209 <NA> <NA> NA NA NA 5: 2248787 2016-04-20 2016-04-20 10:00:00 2016-04-21 00:00:00 2 B53A 306 <NA> <NA> NA NA NA 6: 2248787 2016-04-21 2016-04-21 00:00:00 2016-04-22 00:00:00 2 B53A 306 <NA> <NA> NA NA NA 7: 2248787 2016-04-22 2016-04-22 00:00:00 2016-04-23 00:00:00 2 B53A 306 <NA> <NA> NA NA NA 8: 2248787 2016-04-23 2016-04-23 00:00:00 2016-04-24 00:00:00 2 B53A 306 <NA> <NA> NA NA NA 9: 2248787 2016-04-24 2016-04-24 00:00:00 2016-04-24 11:47:00 2 B53A 306 2016-04-24 11:47:00 2016-04-25 4 B53A 311 10: 2248787 2016-04-25 2016-04-25 00:00:00 2016-04-26 00:00:00 4 B53A 311 <NA> <NA> NA NA NA 11: 2248787 2016-04-26 2016-04-26 00:00:00 2016-04-26 16:00:00 4 B53A 311 <NA> <NA> NA NA NA