Question

我的数据如下：

    ID1     ID2     Time            diff
1:  1958616 P209576 4/15/2016 7:46  NA mins
2:  1958493 P209580 3/23/2016 9:41  -33005.16793 mins
3:  1958493 P209580 3/25/2016 15:41 3240.09742 mins
4:  1958493 P209580 3/30/2016 10:22 6880.65360 mins
5:  1958492 P209580 3/30/2016 11:31 69.00078 mins
6:  1958493 P209580 4/11/2016 10:07 17196.62313 mins

我想要做的是将所有在彼此的8小时时间窗口内发生的ID分组，并在分组后返回不同ID1的数量。在上面的例子中，第4行和第4行5将从abs(diff) < 60*8开始分组。

我使用data[, diff := TIME - shift(TIME)]生成diff列。

我的理想输出看起来像这样

    num_of_unique_id1   ID2     Initial_time
1:  1                   P209576 4/15/2016 7:46
2:  1                   P209580 3/23/2016 9:41
3:  1                   P209580 3/25/2016 15:41
4:  2                   P209580 3/30/2016 10:22
5:  1                   P209580 4/11/2016 10:07

我认为可以使用num_of_unique_id1和.SD创建length(unique(ID1))，但不知道如何为by =参数创建列。

我知道也会出现边界问题（A在B的8小时内，B在C的8小时内，但A和C相隔超过8小时）在这些情况下，我想我想将它们分成一行。

dput(data)

的输出

structure(list(ID1 = c("1958616", "1958493", "1958493", "1958493",
"1958492", "1958493"), ID2 = c("P209576", "P209580", "P209580",
"P209580", "P209580", "P209580"), Time = structure(c(1460706387.438,
1458726077.362, 1458920483.207, 1459333322.423, 1459337462.47,
1460369259.858), class = c("POSIXct", "POSIXt"), tzone = "GMT"),
    diff = structure(c(NA, -33005.1679333329, 3240.09741666714,
    6880.65360000133, 69.0007833321889, 17196.6231333335), units = "mins", class = "difftime")), .Names = c("ID1",
"ID2", "Time", "diff"), class = c("data.table", "data.frame"), row.names = c(NA,
-6L), .internal.selfref = <pointer: 0x1ce9a28>)

Answer 1

如果数据按时排序，我们可以使用cumsum计算差异并分配唯一的组。

data <- data[order(Time)]
data[ , diff := NULL]  # we will re-compute diff in hours
data[ , diff_hours := as.numeric(c(0, diff(Time)))]
##        ID1     ID2                Time diff_hours
## 1: 1958493 P209580 2016-03-23 09:41:17   0.000000
## 2: 1958493 P209580 2016-03-25 15:41:23  54.001624
## 3: 1958493 P209580 2016-03-30 10:22:02 114.677560
## 4: 1958492 P209580 2016-03-30 11:31:02   1.150013
## 5: 1958493 P209580 2016-04-11 10:07:39 286.610386
## 6: 1958616 P209576 2016-04-15 07:46:27  93.646550

window <- 8  # the time window in hours
data[ , group := cumsum(diff_hours > window) + 1]
data[ , num_of_unique_id1 := uniqueN(ID1), by = group]
##        ID1     ID2                Time diff_hours group num_of_unique_id1
## 1: 1958493 P209580 2016-03-23 09:41:17   0.000000     1                 1
## 2: 1958493 P209580 2016-03-25 15:41:23  54.001624     2                 1
## 3: 1958493 P209580 2016-03-30 10:22:02 114.677560     3                 2
## 4: 1958492 P209580 2016-03-30 11:31:02   1.150013     3                 2
## 5: 1958493 P209580 2016-04-11 10:07:39 286.610386     4                 1
## 6: 1958616 P209576 2016-04-15 07:46:27  93.646550     5                 1

请注意，2016-03-30内彼此约一小时内的两个数据点分配相同的group，num_of_unique_id1（每组）为2，而所有其他数据积分在他们自己的小组中。

如何使用data.tables按附近值进行分组

1 个答案: