Question

我想识别并标记我的GPS项圈使用日期时间列切换修复率时间表。

我的项圈在5分钟的时间表上交替持续2天，然后每小时交替持续6天，持续几个月然后关闭。理想情况下，我只会使用原始时间表，但是项圈并不总是立即响应时间表的变化。

在过去，我计算了一个时差列，使用datetime [{datetime2-datetime1）* 24 * 60]的数字版本隔离5分钟的连续位置，但这会导致失去＆＃34;错过修复＆＃34 ;数据。现在我想要包含这些数据，将所有数据标记为5分钟数据，以及在每小时数据时将标记标记。

示例（包括我需要的列）：

> dput(example)
structure(list(OBJECTID = c(1L, 2L, 3L, 4L, 5L, 8L, 9L, 10L, 
11L, 12L, 13L, 14L, 15L, 35L, 36L, 37L, 38L, 39L, 40L, 41L, 42L, 
43L, 44L, 45L, 46L), DT_LMT = structure(c(1L, 2L, 3L, 4L, 5L, 
6L, 7L, 8L, 9L, 10L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L, 
22L, 11L, 12L, 13L, 23L, 24L, 25L), .Label = c("28/05/2013 1:00", 
"28/05/2013 1:05", "28/05/2013 1:10", "28/05/2013 1:15", "28/05/2013 1:20", 
"28/05/2013 1:35", "28/05/2013 1:40", "28/05/2013 1:45", "28/05/2013 1:50", 
"28/05/2013 1:55", "28/05/2013 10:00", "28/05/2013 11:00", "28/05/2013 12:00", 
"28/05/2013 2:00", "28/05/2013 2:05", "28/05/2013 2:10", "28/05/2013 3:50", 
"28/05/2013 3:55", "28/05/2013 4:00", "28/05/2013 5:00", "28/05/2013 6:00", 
"28/05/2013 7:00", "29/05/2013 12:00", "30/05/2013 12:00", "31/05/2013 12:00"
), class = "factor"), DT_LMT.1 = c(41422.0416666667, 41422.0451388889, 
41422.0486111111, 41422.0520833333, 41422.0555555556, 41422.0659722222, 
41422.0694444444, 41422.0729166667, 41422.0763888889, 41422.0798611111, 
41422.0833333333, 41422.0868055556, 41422.0902777778, 41422.1597222222, 
41422.1631944444, 41422.1666666667, 41422.2083333333, 41422.25, 
41422.2916666667, 41422.4166666667, 41422.4583333333, 41422.5, 
41423.5, 41424.5, 41425.5), TIMEDIFF = c(5L, 5L, 5L, 5L, 15L, 
5L, 5L, 5L, 5L, 5L, 5L, 5L, 100L, 5L, 5L, 60L, 60L, 60L, 180L, 
60L, 60L, 1440L, 1440L, 1440L, NA), FIVE = c(1L, 1L, 1L, 1L, 
0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, NA), HOUR = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 0L, 1L, 1L, 0L, 0L, 0L, 
NA), NEED = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 1L), .Label = c("", 
"FIVE", "HOUR", "OFF"), class = "factor")), .Names = c("OBJECTID", 
"DT_LMT", "DT_LMT.1", "TIMEDIFF", "FIVE", "HOUR", "NEED"), class = "data.frame", row.names = c(NA, 
-25L))

在一些帮助下，这是我到目前为止所做的，除非它最终被绊倒：

data<-data[-25,]
c=1
DTD<-1
for (i in 1:24){
if(data$TIMEDIFF[i]<=25){
    data[i,9]<-5}
else{
    j=1
    DTD<-1
    while (j < 100){
        if (data$DT_LMT.1[i+j]<=data$DT_LMT.1[5]+4){
            if (data$TIMEDIFF[i+j]<=25){
                DTD<-0
                j=101
                print (j)}
            else{j=j+1}
        }   
        else{j=101}
    }
    if (DTD==0){data[i,9]<-5}
    else{data[i,9]<-0}
}
}
data

Answer 1

对于想要做这类事情的人来说，这似乎有用：

min.days<-3
summerexample<-summerexample[,-60]
summerexample[,60]<-5
invalid.time.diffs<-which(summerexample$TIMEDIFF>25)#the indices of potential scheduale changes
i<-1
while (i<(length(invalid.time.diffs))){  
  #check for a single min.days gap
  if ((summerexample$DT_LMT[invalid.time.diffs[i]]-summerexample$DT_LMT[invalid.time.diffs[i]-1])>3){
summerexample[invalid.time.diffs[i],60]<-0
i<-i+1
}  
else{    
#j is the number of consecutive invalid time differences
j<-1
while (((i+j)<length(invalid.time.diffs))&
         ((invalid.time.diffs[i+j]-invalid.time.diffs[i])==j)){
  j<-j+1
}    
#if the last consecutive invalid time difference is more than min.days after the first one 
if ((summerexample$DT_LMT[invalid.time.diffs[i+(j-1)]]-summerexample$DT_LMT[invalid.time.diffs[i]])>3){
  summerexample[invalid.time.diffs[i:(i+(j-1))],60]<-0
  i<-i+j
}
else {
  i<-i+1
}

}  
}

使用R识别GPS数据中的不同固定间隔

1 个答案: