Bas*_*ien 11 r data.table
我正在尝试在两个表之间进行复杂的非equi连接.我在最后一次使用R2016的演示中受到启发(https://channel9.msdn.com/events/useR-international-R-User-conference/useR2016/Efficient-in-memory-non-equi-joins-using-datatable这使我相信它将是data.table的合适任务.我的表1看起来像:
library(data.table)
sp <- c("SAB","SAB","SAB","SAB","EPN","EPN","BOP","BOP","BOP","BOP","BOP","PET","PET","PET")
dbh <- c(10,12,16,22,12,16,10,12,14,20,26,12,16,18)
dt1 <- data.table(sp,dbh)
dt1
sp dbh
1: SAB 10
2: SAB 12
3: SAB 16
4: SAB 22
5: EPN 12
6: EPN 16
7: BOP 10
8: BOP 12
9: BOP 14
10: BOP 20
11: BOP 26
12: PET 12
13: PET 16
14: PET 18
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这是dbh的树木列表.我的第二个表(下面)给出了一个通用表,它为每个树种提供了一系列dbh来对大小类或树进行分类:
gr_sp <- c("RES","RES","RES","RES","RES","RES", "DEC", "DEC", "DEC", "DEC", "DEC", "DEC")
sp <- c("SAB","SAB", "SAB", "EPN", "EPN", "EPN", "BOP", "BOP", "BOP", "PET", "PET", "PET")
dbh_min <- c(10, 16, 22, 10, 14, 20, 10, 18, 24, 10, 20, 26)
dbh_max <- c(14, 20, 30, 12, 18, 30, 16, 22, 30, 18, 24, 30)
dhb_clas <- c("s", "m", "l", "s", "m", "l", "s", "m", "l", "s", "m", "l")
dt2 <- data.table(gr_sp, sp, dbh_min, dbh_max, dhb_clas)
dt2
gr_sp sp dbh_min dbh_max dhb_clas
1: RES SAB 10 14 s
2: RES SAB 16 20 m
3: RES SAB 22 30 l
4: RES EPN 10 12 s
5: RES EPN 14 18 m
6: RES EPN 20 30 l
7: DEC BOP 10 16 s
8: DEC BOP 18 22 m
9: DEC BOP 24 30 l
10: DEC PET 10 18 s
11: DEC PET 20 24 m
12: DEC PET 26 30 l
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我希望我的最终表格是两个表格的种类("sp"字段)的连接,并且在"DBH_MIN"和"DBH_MAX"所述的dhb范围内.这会使我的表看起来像:
data.table(dt1, gr_sp = c("RES","RES","RES","RES","RES","RES","DEC","DEC","DEC","DEC","DEC","DEC","DEC","DEC"), dhb_clas = c("s","s","m","l","s","m","s","s","s","m","l","s","s","s"))
sp dbh gr_sp dhb_clas
1: SAB 10 RES s
2: SAB 12 RES s
3: SAB 16 RES m
4: SAB 22 RES l
5: EPN 12 RES s
6: EPN 16 RES m
7: BOP 10 DEC s
8: BOP 12 DEC s
9: BOP 14 DEC s
10: BOP 20 DEC m
11: BOP 26 DEC l
12: PET 12 DEC s
13: PET 16 DEC s
14: PET 18 DEC s
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我尝试过类似的东西:
dt1[dt2, on=.(sp=sp, dbh>=dbh_min, dbh<=dbh_max)]
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这给了太多行......
谢谢你的帮助
Bas*_*ien 14
所以我非常接近.我遇到了2个问题,首先是data.table包的安装不好(数据表错误找不到函数".")导致了一个模糊的错误.
解决了这个问题后,我发现了一个发现:
dt1[dt2, on=.(sp=sp, dbh>=dbh_min, dbh<=dbh_max), nomatch=0]
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用一个糟糕的dbh列给了我想要的东西.使用以下命令反转命令:
dt2[dt1, on=.(sp=sp, dbh_min<=dbh, dbh_max>=dbh)]
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用一个无用的额外列修复了问题.