Question

我一直试图弄清楚这一点并没有成功。我已经尝试过使用data.table，merge和apply函数，但是我是初学者，可能错过了一些东西。基本上我想使用来自一个数据集的datetime（“dt”）信息和唯一事件标识符（“uid”）来创建一个列，告诉我条件是否得到满足以及发生了哪个“sid”。复杂的是，对于给定的“dt”和“uid”，可能会出现不止一个“sid”并且我想知道这一点。我的数据集如下所示......

     > head(D1)
      X                  dt       tid     rid     sid   uid  nutm     eutm
    1 1 2013-09-06 08:55:00 A69-1601-27 VR2W-11 BUIS_09  1 1932024 389346.7
    2 2 2013-09-06 08:55:00 A69-1601-27 VR2W-12 BUIS_08  1 1932024 389346.7
    3 3 2013-09-06 08:56:00 A69-1601-27 VR2W-11 BUIS_09  1 1932024 389346.7
    4 4 2013-09-06 08:57:00 A69-1601-27 VR2W-11 BUIS_09  1 1932024 389346.7
    5 5 2013-09-06 08:57:00 A69-1601-27 VR2W-12 BUIS_08  1 1932024 389346.7
    6 6 2013-09-06 08:57:00 A69-1601-27 VR2W-11 BUIS_09  1 1932024 389346.7

> head(D2)
  X                  dt uid
1 1 09/06/2013 08:57:00   1
2 2 09/06/2013 08:58:00   1
3 3 09/06/2013 08:59:00   1
4 4 09/06/2013 09:00:00   1
5 5 09/06/2013 09:01:00   1
6 6 09/06/2013 09:02:00   1

我想运行一个函数，从D2的第一行开始查看D1的每一行的“uid”和“dt”，并创建一个列，说明是否满足这些条件以及相关的“sid”条件得到满足。所以我的输出看起来像这样...

head(stimes)
  X                  dt uid  freq sid     tid            rid     nutm    eutm
1 1 09/06/2013 08:57:00   1    T  BUIS_09 A69-1601-27 VR2W-11 1932024 389346.7
1 2 09/06/2013 08:57:00   1    T  BUIS_08 A69-1601-27 VR2W-12 1967024 389346.7

更新：dput（head（D1））如下所示。这有点困难，因为有大量独特的日期时间，但总的来说......

        "2014-05-03 09:12:00", "2014-05-03 09:13:00", "2014-05-03 09:14:00", 
        "2014-05-03 09:15:00", "2014-05-03 09:16:00", "2014-05-03 09:17:00", 
        "2014-05-03 09:18:00", "2014-05-03 09:19:00", "2014-05-03 09:20:00", 
        "2014-05-03 09:21:00", "2014-05-03 09:22:00"), class = "factor"),     
     tid = structure(c(2L, 2L, 3L, 2L, 2L, 2L), .Label = c("A69-1303-54", 
        "A69-1601-27", "A69-1601-275", "A69-1601-31"), class = "factor"), 
        rid = structure(c(30L, 30L, 30L, 30L, 30L, 30L), .Label = c("VR2W-102913", 
        "VR2W-914", "VR2W-916", "VR2W-102917", "VR2W-210", 
        "VR2W-710", "VR2W-711", "VR2W-103821", "VR2W-088", 
        "VR2W-620", "VR2W-621", "VR2W-104622", "VR2W-938", 
        "VR2W-320", "VR2W-145", "VR2W-110204", "VR2W-205", 
        "VR2W-207", "VR2W-208", "VR2W-110210", "VR2W-634", 
        "VR2W-118", "VR2W-119", "VR2W-114120", "VR2W-121", 
        "VR2W-122", "VR2W-123", "VR2W-114124", "VR2W-125", 
        "VR2W-498", "VR2W-499", "VR2W-115500", "VR2W-501", 
        "VR2W-502", "VR2W-503", "VR2W-115504", "VR2W-925", 
        "VR2W-272", "VR2W-273", "VR2W-120967", "VR2W-585"
        ), class = "factor"), sid = structure(c(9L, 9L, 9L, 9L, 9L, 
        9L), .Label = c("BUIS_01", "BUIS_02", "BUIS_03", "BUIS_04", 
        "BUIS_05", "BUIS_06", "BUIS_07", "BUIS_08", "BUIS_09", "BUIS_10", 
        "BUIS_11", "BUIS_12", "BUIS_13", "BUIS_14", "BUIS_15", "BUIS_16", 
        "BUIS_17", "BUIS_18", "BUIS_19", "BUIS_20", "BUIS_22", "BUIS_26", 
        "BUIS_27", "BUIS_31", "BUIS_33", "BUIS_34", "BUIS_35", "BUIS_36", 
        "BUIS_37", "BUIS_38", "BUIS_41", "BUIS_42", "BUIS_43", "BUIS_46"
        ), class = "factor"), uid = c(NA, NA, NA, 1L, 1L, 1L), nutm = c(NA, 
        NA, NA, 1967024.456, 1967024.456, 1967024.456), eutm = c(NA, 
        NA, NA, 328346.6957, 328346.6957, 328346.6957)), .Names = c("X", 
    "dt", "tid", "rid", "sid", "uid", "nutm", "eutm"), row.names = c(NA, 
    6L), class = "data.frame")

Answer 1

如果这些是data.tables（它们不是），

setkey(D1,dt,uid)
setkey(D2,dt,uid)

D1[D2,nomatch=0]

可能会这样做。

根据条件合并R中的数据表信息

1 个答案: