在data.table下面,我有关于参与项目的团队组成的信息.该变量id告诉团队ID,而变量event给出项目编号.该变量freqrel描述了团队的组成(您可以看到freqrel在每个团队中最多加1).
structure(list(id = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L,
4L, 4L, 5L, 5L, 5L), event = c("127b", "127b", "127b", "127b",
"127b", "127b", "127b", "127b", "127b", "125t", "125t", "125t",
"125t", "125t", "125t"), membr = c("engineer", "mathematician",
"physicist", "mathematician", "physicist", "surgeon", "dentist",
"mathematician", "programmer", "physicist", "sociologist", "surgeon",
"musician", "sociologist", "surgeon"), freqrel = c(0.4, 0.4,
0.2, 0.166666666666667, 0.5, 0.333333333333333, 0.333333333333333,
0.5, 0.166666666666667, 0.75, 0.125, 0.125, 0.444444444444444,
0.444444444444444, 0.111111111111111)), .Names = c("id", "event",
"membr", "freqrel"), row.names = c(NA, -15L), class = c("data.table",
"data.frame"), sorted = c("id", "event"), .internal.selfref = <pointer: 0x039a24a0>)
我看到数据的方式被拆分为嵌套组.第一个分区发生在项目级别(直线),第二个分支发生在团队级别(虚线).
id event membr freqrel
1: 1 127b engineer 0.4000000
2: 1 127b mathematician 0.4000000
3: 1 127b physicist 0.2000000
--------------------------------------
4: 2 127b mathematician 0.1666667
5: 2 127b physicist 0.5000000
6: 2 127b surgeon 0.3333333
--------------------------------------
7: 3 127b dentist 0.3333333
8: 3 127b mathematician 0.5000000
9: 3 127b programmer 0.1666667
_____________________________________
10: 4 125t physicist 0.7500000
11: 4 125t sociologist 0.1250000
12: 4 125t surgeon 0.1250000
--------------------------------------
13: 5 125t musician 0.4444444
14: 5 125t sociologist 0.4444444
15: 5 125t surgeon 0.1111111
从这个起始条件开始,我想让同一个项目中的团队完全可比,通过向他们每个人添加membr团队没有特征的类型,将他们分配为freqrel = 0.结果应该是这样的:
id event membr freqrel
1: 1 127b dentist 0.0000000
2: 1 127b engineer 0.4000000
3: 1 127b mathematician 0.4000000
4: 1 127b physicist 0.2000000
5: 1 127b programmer 0.0000000
6: 1 127b surgeon 0.0000000
--------------------------------------
7: 2 127b dentist 0.0000000
8: 2 127b engineer 0.0000000
9: 2 127b mathematician 0.1666667
10: 2 127b physicist 0.5000000
11: 2 127b programmer 0.0000000
12: 2 127b surgeon 0.3333333
--------------------------------------
13: 3 127b dentist 0.3333333
14: 3 127b engineer 0.0000000
15: 3 127b mathematician 0.5000000
16: 3 127b physicist 0.0000000
17: 3 127b programmer 0.1666667
18: 3 127b surgeon 0.0000000
_____________________________________
19: 4 125t musician 0.0000000
20: 4 125t physicist 0.7500000
21: 4 125t sociologist 0.1250000
22: 4 125t surgeon 0.1250000
--------------------------------------
23: 5 125t musician 0.4444444
24: 5 125t physicist 0.0000000
25: 5 125t sociologist 0.4444444
26: 5 125t surgeon 0.1111111
换句话说,将所述数据与后by使用event作为密钥,我需要划分的第二时间,并比较与所述第二分束而获得的数据的块.但问题在于我不知道如何引用获得的第一个块by,然后再如何分割并在数据库的各个部分之间进行比较.你知道我怎么能解决这个问题吗？
如果你能帮助我,我会非常感激.真.
这是一个简单的方法:
setkey(dt, id, membr)
ans <- dt[, .SD[CJ(unique(id), unique(membr))], by=list(event)]
然后,您可以将NA0 替换为0,如下所示:
ans[is.na(freqrel), freqrel := 0.0]
一些解释:您的问题归结为此 - 对于每一个event,您需要所有可能的组合,id, membr以便您可以在该分组中使用此组合执行此组合的连接.SD.
所以,首先我们分组event,并在其中,我们首先得到所有组合id, membr的帮助CJ(默认情况下将键设置为所有列).但是,要执行连接,我们需要为其设置密钥.SD.因此,我们设置key了dt对id, membr前期.因此,我们在每个组中执行连接,并为您提供预期的结果.希望这个对你有帮助.