Question

我有两张桌子。 table1看起来像这样

  date       hour     data
2010-05-01     3        5
2010-05-02     7        7
2010-05-02     10       8
2010-07-03     18       3
2011-12-09     22       1
2012-05-01     3        0

它存储为data.table，其密钥设置在date和hour上。我有另一张桌子，看起来像这样。这是我的outages表。

 resource        date_out                date_back
   joey       2010-04-30 4:00:00      2010-05-02 8:30:00
   billy      2009-04-20 7:00:00      2009-02-02 5:30:00
   bob        2011-11-15 12:20:00     2010-12-09 23:00:00
   joey       2012-04-28 1:00:00      2012-05-02 17:00:00

我想将列添加到table1，其中这些列是outages表中的资源。我希望这些列中的值在没有中断时为0，在有时为1时为。

此示例的结果应为。

  date       hour     data     joey      billy      bob
2010-05-01     3        5       1          0         0        
2010-05-02     7        7       1          0         0 
2010-05-02     10       8       0          0         0 
2010-07-03     18       3       0          0         0 
2011-12-09     22       1       0          0         1
2012-05-01     3        0       1          0         0

实际上我的table1有大约2500行，而我的outages表有19000.我能想到的唯一方法是循环遍历outages表的每一行，然后在正确的位置将{1}插入table1。我的代码依赖于table1的顺序，所以至少它不必为outages的每一行扫描该表的100％。但是，我的数据下面需要4个小时。

for (out in 1:length(outages$resource)) {
  a<-as.character(outages[out]$resource)
  #if column doesn't exist then create it
  if (a %in% colnames(table1)==FALSE) {
    table1$new<-0
    setnames(table1, "new", a)
    }
  midpoint<-round(length(table1$date)/2,0)
  if (table1$date[midpoint]+table1$hour[midpoint]*60*60>=outages[out]$due_out && table1$date[midpoint]+table1$hour[midpoint]*60*60<=outages    [out]$due_back)
  {
    while(table1$date[midpoint]+table1$hour[midpoint]*60*60>=outages[out]$due_out && midpoint>=1 && midpoint<=length(table1$date)) {
      table1[midpoint,a:=1,with=FALSE]
      midpoint<-midpoint-1
    }
    midpoint<-round(length(table1$date)/2,0)
    while(table1$date[midpoint]+table1$hour[midpoint]*60*60<=outages[out]$due_back && midpoint>=1 && midpoint<=length(table1$date)) {
      table1[midpoint,a:=1,with=FALSE]
      midpoint<-midpoint+1
    }
  } else {
    if (table1$date[midpoint]+table1$hour[midpoint]*60*60>outages[out]$due_back) {
      while(table1$date[midpoint]+table1$hour[midpoint]*60*60>outages[out]$due_back && midpoint>=1 && midpoint<=length(table1$date)) {
        midpoint<-midpoint-1
      }
      while(table1$date[midpoint]+table1$hour[midpoint]*60*60>=outages[out]$due_out && midpoint>=1 && midpoint<=length(table1$date)) {
        table1[midpoint,a:=1,with=FALSE]
        midpoint<-midpoint-1
      }
    } 
    midpoint<-round(length(table1$date)/2,0)
    if (table1$date[midpoint]+table1$hour[midpoint]*60*60<outages[out]$due_out) {
      while(table1$date[midpoint]+table1$hour[midpoint]*60*60<outages[out]$due_out && midpoint>=1 && midpoint<=length(table1$date)) {
        midpoint<-midpoint+1
      }
      while(table1$date[midpoint]+table1$hour[midpoint]*60*60<=outages[out]$due_back && midpoint>=1 && midpoint<=length(table1$date)) {
        table1[midpoint,a:=1,with=FALSE]
        midpoint<-midpoint+1
 }
 }
 }
if (sum(table1[,a,with=FALSE])==0) {
  table1[,a:=NULL,with=FALSE]
}
}

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Answer 1

这是实现你想要的方式。假设您的table1时间精度为1小时。虽然它可以被修改为任意精度，但它会在更大的时间间隔内表现更好，因为它构造了date_out - date_back范围内可能时间的完整序列。注意，我使用与OP略有不同的表来说明重叠间隔并纠正OP中的一些错误。

table1 = data.table(date = c("2010-05-01", "2010-05-02", "2010-05-02", "2010-07-03", "2011-12-09", "2012-05-01"), hour = c(3,7,10,18,22,3), data = c(5,7,8,3,1,0))
outages = data.table(resource = c("joey", "bob", "billy", "bob", "joey"), date_out = c("2010-04-30 4:00:00", "2010-04-30 4:00:00", "2009-04-20 7:00:00", "2011-11-15 12:20:00", "2012-04-28 1:00:00"), date_back=c("2010-05-02 8:30:00", "2010-05-02 8:30:00", "2009-06-02 5:30:00", "2011-12-09 23:00:00", "2012-05-02 17:00:00"))

# round up date_out and round down date_back
# and create a sequence in-between spaced by 1 hour
outages[, list(datetime = seq(as.POSIXct(round(as.POSIXct(date_out) + 30*60-1, "hours")),
                              as.POSIXct(round(as.POSIXct(date_back) - 30*60, "hours")),
                              60*60)),
          by = list(resource, date_out)] -> outages.expanded
setkey(outages.expanded, datetime)

# merge with the original table, then run "table" to get the frequencies/occurences
# and cbind back with the original table
cbind(table1, unclass(table(
                outages.expanded[table1[, list(datetime=as.POSIXct(paste0(date, " ", hour, ":00:00")))],
                                 resource])))

#         date hour data bob joey
#1: 2010-05-01    3    5   1    1
#2: 2010-05-02    7    7   1    1
#3: 2010-05-02   10    8   0    0
#4: 2010-07-03   18    3   0    0
#5: 2011-12-09   22    1   1    0
#6: 2012-05-01    3    0   0    1

根据另一个表中的日期范围在一个表中创建虚拟变量

1 个答案: