通过减少data.table进行排序 [英] sequence by reducing data.table

问题描述

require(data.table)    
set.seed(333)
t <- data.table(old=1002:2001, dif=sample(1:10,1000, replace=TRUE))
t$new <- t$old + t$dif; t$foo <- rnorm(1000); t$dif <- NULL
i <- data.table(id=1:3, start=sample(1000:1990,3))


> i
   id start
1:  1  1002
2:  2  1744
3:  3  1656

> head(t)
    old  new        foo
1: 1002 1007 -0.7889534
2: 1003 1004  0.3901869
3: 1004 1014  0.7907947
4: 1005 1011  2.0964612
5: 1006 1007  1.1834171
6: 1007 1015  1.1397910

我想从points中删除时间点，以便仅将那些行保留在new[i] = old[i-1]的位置，从而给出一些固定数量的时间点的连续序列.理想情况下，这将同时对i中的所有id完成，其中start给出起点.例如，如果选择n=5，则应获取

I would like to delete time points from points such that only those rows remain where new[i] = old[i-1], giving a continuous sequence of some fixed number of time points. Ideally, this would be done for all id in i simultaneously, where start gives the starting points. For example, if we choose n=5, we should obtain

> head(ans)
    id old  new        foo
1:   1 1002 1007 -0.7889534
2:   1 1007 1015  1.1397910
3:   1 1015 1022 -1.2193670
4:   1 1022 1024  1.2039050
5:   1 1024 1026  0.4388586
6:   2 1744 1750 -0.1368320

无法在上面推断出第3至6行，而foo代表需要保留的其他变量.

where lines 3 to 6 cannot be inferred above and foo is a stand in for other variables that need to be kept.

例如通过巧妙地结合使用联接，是否可以在data.table中有效地完成此任务?

Can this be done efficiently in data.table, for example, using a clever combination of joins?

PS.这个问题有点类似于我的早期版本，但我已修改情况以使其更加清晰.

PS. This question is somewhat similar to an an earlier one of mine but I have modified the situation to make it clearer.

推荐答案

在我看来，您需要图形算法的帮助.如果要以1002开头，可以尝试:

It seems to me that you need help from graph algorithms. If you want to start with 1002, you can try:

require(igraph)
g <- graph_from_edgelist(as.matrix(t[,1:2]))
t[old %in% subcomponent(g,"1002","out")]
#  1: 1002 1007 -0.78895338
#  2: 1007 1015  1.13979100
#  3: 1015 1022 -1.21936662
#  4: 1022 1024  1.20390482
#  5: 1024 1026  0.43885860
# ---                      
#191: 1981 1988 -0.22054875
#192: 1988 1989 -0.22812175
#193: 1989 1995 -0.04687776
#194: 1995 2000  2.41349730
#195: 2000 2002 -1.23425666

当然，您可以对所需的每个start执行上述操作，并限制前n行的结果.例如，我们可以lapply在i$start位置上，然后将rbindlist所有值放在一起，用i$id值声明一个id列.像这样:

Of course you can do the above for each start you want and limiting the results for the first n rows. For instance, we can lapply over the i$start positions and then rbindlist all the values together, declaring an id column with the i$id values. Something like:

n <- 5
rbindlist(
    setNames(lapply(i$start, function(x) t[old %in% subcomponent(g,x,"out")[1:n]]), i$id),
    idcol="id")
#    id  old  new        foo
# 1:  1 1002 1007 -0.7889534
# 2:  1 1007 1015  1.1397910
# 3:  1 1015 1022 -1.2193666
# 4:  1 1022 1024  1.2039048
# 5:  1 1024 1026  0.4388586
# 6:  2 1744 1750 -0.1368320
# 7:  2 1750 1758  0.3331686
# 8:  2 1758 1763  1.3040357
# 9:  2 1763 1767 -1.1715528
#10:  2 1767 1775  0.2841251
#11:  3 1656 1659 -0.1556208
#12:  3 1659 1663  0.1663042
#13:  3 1663 1669  0.3781835
#14:  3 1669 1670  0.2760948
#15:  3 1670 1675  0.3745026

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