Question

我正在使用包含data.table的R，我想通过运行（时间）间隔或重叠的bin来对data.table进行分组。对于这些运行间隔中的每一个，我想找到相等数据对的出现。更多这些＆＃34;相等的数据对＆＃34;应该不完全相等，但在某些间隔范围内也是如此。

问题的简单版本如下：

#Time   X   Y Counts
# ... ... ...      1
#I would like to do:
DT[, sum(counts), by = list(Time, X, Y)]
#with Time, X and Y being in overlapping intervals.

findintervals()会给我带有＆＃34;硬边框＆＃34;的垃圾箱，而不是重叠的。

更详细的问题：我们假设我有一个data.table：

Time    <- c(1,1,2,4,5,5,6,7,8,8,8,8,12,13)
#more equal time values are allowed.
X       <- c(6,6,7,10,5,7,6,3,9,10,6,3,3,6)
Y       <- c(2,6,10,3,4,6,6,9,4,9,6,6,9,9)
DT      <- data.table(Time, X, Y)

    Time  X  Y
 1:    1  6  2
 2:    1  6  6
 3:    2  7 10
 4:    4 10  3
 5:    5  5  4
 6:    5  7  6
 7:    6  6  6
 8:    7  3  9
 9:    8  9  4
10:    8 10  9
11:    8  6  6
12:    8  3  6
13:   12  3  9
14:   13  6  9

一些预定义的间隔大小：

Timeinterval      <- 5
#for a time value of 10 this means to look from 10-5 to 10+5
RangeX.percentage <- 0.5 
RangeY.percentage <- 0.5

结果应该给我一个额外的专栏，让我们称之为＆＃34;计数＆＃34;考虑到X和Y的范围，同时出现数据对X和Y.

我想过按时间间隔进行某种分组，比如

c(1, 1, 2, 4, 5, 5, 6) #for the first item: (1-5):(1+5)
c(1, 1, 2, 4, 5, 5, 6, 7) # for the second item: (1-5):(1+5)
c(1, 1, 2, 4, 5, 5, 6, 7, 8, 8, 8, 8) #for the third item (2-5):(2+5)
#...
c(8, 8, 8, 8, 12, 13) # for the last item (13-5):(13+5)

以及数据的以下条件（但也许该部分有一个更简单的版本）：

编辑：要清除结果应该是什么样的：

Ranges <- DT[ , list(
             X* (1 + RangeX.percentage), X* (1 - RangeX.percentage),
             Y* (1 + RangeY.percentage), Y* (1 - RangeY.percentage))]
DT2 <- cbind(DT, Ranges, count = rep(1, nrow(DT)))
setnames(DT2, c("Time","X","Y","XR1","XR2","YR1","YR2","count"))
for (i in 1:nrow(DT2)){
  #main part of the question how to get this done within data.table:
  DT2.subset <- DT2[which(abs(Time - DT2[i]$Time) < Timeinterval)]
  #subsequent comparison of X and Y:
  DT[i]$Count<- length(which(DT2.subset$X < DT2[i]$XR1 & 
                             DT2.subset$X > DT2[i]$XR2 &
                             DT2.subset$Y < DT2[i]$YR1 & 
                             DT2.subset$Y > DT2[i]$YR2))
}
 DT2
    Time  X  Y  XR1 XR2  YR1 YR2 count
 1:    1  6  2  9.0 3.0  3.0 1.0     1
 2:    1  6  6  9.0 3.0  9.0 3.0     3
 3:    2  7 10 10.5 3.5 15.0 5.0     4
 4:    4 10  3 15.0 5.0  4.5 1.5     3
 5:    5  5  4  7.5 2.5  6.0 2.0     1
 6:    5  7  6 10.5 3.5  9.0 3.0     6
 7:    6  6  6  9.0 3.0  9.0 3.0     4
 8:    7  3  9  4.5 1.5 13.5 4.5     2
 9:    8  9  4 13.5 4.5  6.0 2.0     3
10:    8 10  9 15.0 5.0 13.5 4.5     4
11:    8  6  6  9.0 3.0  9.0 3.0     4
12:    8  3  6  4.5 1.5  9.0 3.0     1
13:   12  3  9  4.5 1.5 13.5 4.5     2
14:   13  6  9  9.0 3.0 13.5 4.5     1

由于我的完整data.table包含超过一百万行，因此检查每行的所有DT $时间在计算时间方面是一团糟。

Answer 1

您可以尝试data.table::foverlaps。我们将像您一样创建Ranges，只需添加Time范围和行索引（以便稍后聚合）。这里的主要问题是你不希望＆lt; =或＆gt; =而是＆lt;和＆gt;，所以我们必须在Time间隔加上+ -1。然后，我们也会向Time添加DT间隔，键，然后运行foverlaps。最后阶段是计算每行的观察结果。

DT[, Time2 := Time] ## Add higher interval to DT
setkey(DT, Time, Time2) ## key (for foverlaps)

Ranges <- 
  DT[ , .(Time = Time - Timeinterval + 1, ## Add lower Time interval
          Time2 = Time + Timeinterval - 1, ## Add higher Time interval
          XR1 = X* (1 - RangeX.percentage), 
          XR2 = X* (1 + RangeX.percentage),
          YR1 = Y* (1 - RangeY.percentage), 
          YR2 = Y* (1 + RangeY.percentage),
          indx = .I)] ## Add row index

# Run foverlaps and count incidences by condition while updating DT by reference
DT[, 
   count := foverlaps(Ranges, DT)[X > XR1 & X < XR2 & Y > YR1 & Y < YR2,
                                   .N, 
                                   keyby = indx]$N]  
DT
#     Time  X  Y Time2  count
#  1:    1  6  2     1      1
#  2:    1  6  6     1      3
#  3:    2  7 10     2      4
#  4:    4 10  3     4      3
#  5:    5  5  4     5      1
#  6:    5  7  6     5      6
#  7:    6  6  6     6      4
#  8:    7  3  9     7      2
#  9:    8  9  4     8      3
# 10:    8 10  9     8      4
# 11:    8  6  6     8      4
# 12:    8  3  6     8      1
# 13:   12  3  9    12      2
# 14:   13  6  9    13      1

通过运行间隔对data.table进行分组

1 个答案: