我有一些数据:
我正在尝试根据入口,出口,建筑物ID和唯一标识符(车号)来计算花费在建筑物中的时间。
现在,我按唯一ID对数据框进行排序,然后对日期进行排序,然后应用此算法:
For each row {
if row.type = 'entry' and nextRow.type = 'exit' and row.uid = nextRow.uid {
Calculate time difference and add this data to another df.
}
}
尽管我只有6000行,但仍需要一些时间来运行... 我对R不太熟悉,我认为有很多方法可以加快速度...
代码如下:
# Sort rows:
BldActivity <- BldActivity[order(BldActivity$UniqueId, BldActivity$DateOfEvent),]
df = data.frame(NULL)
DurationOfStay <- data.frame(NULL)
for(i in 1:nrow(BldActivity)) {
row <- BldActivity[i,]
# do stuff with row
if(row$Type == 'entry') {
rowNext <- BldActivity[i+1,]
if(!is.na(rowNext$Type)) {
if(rowNext$Type == 'exit' && row$UniqueId == rowNext$UniqueId)
{
newRow <- data.frame( Entry_DateOfEvent = row$DateOfEvent,
Exit_DateOfEvent = rowNext$DateOfEvent,
BuildingID = row$BuildingID,
BuildingName = row$`Building Name`,
UniqueId = row$UniqueId,
DurationOfStay = difftime(rowNext$DateOfEvent, row$DateOfEvent, units="mins")
)
DurationOfStay <- rbind(DurationOfStay,newRow)
}
}
}
}
能否请您指出可能的改进之处?
这是一个输入示例:
DateOfEvent Type UniqueId BuildingID Building Name
2019/03/22 09:15:43 entry 04352e5b6051c311048a5803f8716700 1e98f5c0e699 Building 2
2019/03/22 09:51:45 exit 04352e5b6051c311048a5803f8716700 1e98f5c0e699 Building 2
2019/03/22 10:31:28 entry 066b9a3995acd495318ad70e0d876f00 062e933d6b9f Building 1
2019/03/22 11:15:02 exit 066b9a3995acd495318ad70e0d876f00 062e933d6b9f Building 1
2019/03/22 11:11:42 entry 0e027aba359aaecbe8fe3eaf5a1bbb00 062e933d6b9f Building 1
2019/03/22 14:44:27 exit 0e027aba359aaecbe8fe3eaf5a1bbb00 062e933d6b9f Building 1
2019/03/22 09:55:03 entry 1747dbaef11176b9ab90f2cfbf056210 1e98f5c0e699 Building 2
2019/03/22 18:13:08 exit 1747dbaef11176b9ab90f2cfbf056210 1e98f5c0e699 Building 2
2019/03/21 14:23:53 entry 3e0d2c4b1b159a24f4dc5fa084b59f00 1e98f5c0e699 Building 2
2019/03/21 15:36:31 exit 3e0d2c4b1b159a24f4dc5fa084b59f00 1e98f5c0e699 Building 2
输出只是IN / OUT和计算的持续时间的列值。
谢谢
菲利普
答案 0 :(得分:0)
library(data.table)
setDT(BldActivity)
DurationOfStay <- dcast(BldActivity, UniqueId + BuildingID + `Building Name` ~ Type, value.var = "DateOfEvent")
DurationOfStay[, DurationOfStay := difftime(exit, entry, units="mins"), ]
答案 1 :(得分:0)
感谢您的代码,它看起来很有希望。 我确实有2个问题:
首先,聚合函数出现问题,引发错误:
聚合函数应采用矢量输入并返回单个值(长度= 1)。
我通过添加聚合函数解决了这个问题
fun.aggregate = function(x) {
lubridate::as_datetime(ifelse(Type == 'entry', min(x), max(x)))
}
我还添加了一个标识符,用于根据唯一ID,建筑物ID和类型(进入/退出)对进入/退出进行分组
这是新代码:
setDT(BldActivity)
BldActivity[, ID_Stay := seq_len(.N), by=list(UniqueId, BuildingID, Type)]
DwellTime <- dcast(BldActivity, UniqueId + BuildingID + `Building Name` ~ Type, value.var = "DateOfEvent",
fun.aggregate = function(x) {
lubridate::as_datetime(ifelse(Type == 'entry', min(x), max(x)))},
fill = 0)
DurationOfStay[, DurationOfStay := difftime(exit, entry, units="mins"), ]
但是我有非常奇怪的值...原因是,如果我有2个条目并且中间没有出口,那么整个序列就搞砸了。
谢谢
菲利普